Boost converter, inductor size?

[Bill Bowden]

On Jul 16, 5:41 pm, Tim Wescott <t...@seemywebsite.com> wrote:

On 07/16/2010 04:18 PM, Bill Bowden wrote:



On Jul 15, 8:09 pm, Tim Wescott<t...@seemywebsite.com>  wrote:
On 07/15/2010 07:33 PM, Bill Bowden wrote:

I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7,  transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH.  But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average).  And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator.....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time.  (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak.  The more
inductance the more the inductor peak (and trough) will approach the
average 1A.  So 1.75A is possible.

This is where I get confused. If the inductor current just kisses
zero, then it has released all the energy into the load and must
recharge during the next 30% time frame. If the power in is equal to
the power out, then the inductor must ramp from 0 to 6 amps during the
30% time to average 3 amps for 30% of the time, or a 1 amp average
continuous input all the time.

I think that where you get confused is when you forget that the inductor
is permanently attached to the source.

The inductor current goes from 0 to 2A during the transistor on time,
and from 2A to zero during the transistor off time.  The source is
_always_ delivering current to the inductor, but the inductor is only
delivering current to the load 70% of the time.

Now, if a very large inductance brings the current peaks and valleys
closer together so they are much the same, then it seems the inductor
current would be 3 amps, so the transistor can switch on for 30% of
the time and supply 3 amps, or 1 amp average. So, I don't see how
increasing the inductance can ever reduce the current below 3 amps.

See above.  With an infinite inductance flowing 1A the input current
will _always_ be 1A exactly, and the output will be 0 30% of the time
and 1A 70% of the time, for 700mA.

HTH.

Yes, that makes it clearer. So, if the transistor were removed, the
load would still get current through the diode all the time, and the
output voltage would be around 9 volts. The extra 5 volts is obtained
by the transistor pulling the inductor to ground for 30% of the time
and adding current to the inductor. But in the case of the infinite
inductance, the current doesn't change.

I was thinking all the power was switched by the transistor, but it's
actually only
30%. Good explanation.

-Bill

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details athttp://www.wescottdesign.com/actfes/actfes.html

 

[Jon Kirwan]

On Fri, 16 Jul 2010 13:53:13 -0700, Tim Wescott
<tim@seemywebsite.com> wrote:

On 07/16/2010 11:46 AM, Jon Kirwan wrote:
On Fri, 16 Jul 2010 09:44:57 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 04:59 AM, Jon Kirwan wrote:
On Thu, 15 Jul 2010 19:33:31 -0700 (PDT), Bill Bowden
wrongaddress@att.net> wrote:

I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator.phtml

Are you even sure that is the right approach? Does this
solar panel work well driving an inductor along a sloped ramp
from I=0 to I=some_peak_value and then nothing at all while
the switcher goes into its t_off period?

As mentioned elsewhere, you use smoothing caps for this.

That does NOT seem optimal to me. I do think it works great
on tiny photovore robots because it is cheap and easy to do
-- but that isn't a case where we are talking watts of power
and getting the most out of it. And at 10W, I wonder at the
sizes required anyway.

And how do things vary with solar motion across the sky?

In the obvious way: power drops with falling light. The less light, or
the more obliquely light falls on the panel, the less power.

Is the 9.65V@1A simply a peak? Or?

It's probably the peak.

I kind of figured. But it would help to be sure.

You can model a solar cell pretty well as a
current source (from photon impingement) in parallel with a silicon
diode.

As a current source and diode, you'd probably NOT want the
voltage to reach enough to cause significant leakage. In
short, you might want something like a transimpedance amp.
Except for the fact that a transimpendance amp would seek
close to zero voltage, and that times lots of current isn't
"much power." So you really want to allow some middling
point where voltage and current (product) is at a peak. I
would guess some kind of odd shape here with a peak power
position that is difficult to track.

A capacitor isn't my first idea of a good tracking circuit.

As the voltage across the cell rises the internal current across
the junction increases, ultimately until the cell has no external
current at all. As the voltage across the cell falls the power lost to
ohmic heating in the cell rises and the amount of power you can extract
goes down.

I think I gather.

And what kind of load should it "see?"

Just the right one. Tracking the maximum power point of the array is
not trivial, particularly when you take the variation of the diode
characteristics with temperature into account.

Yes. Hence why I don't think slapping a cap across the panel
is optimal.

You're not using the cap for tracking for heaven's sake! You're using
it to smooth out the current draw from the panel. It's all or part of
the input filter that any wise designer always puts on the front end of
a switcher.

Thanks, Tim.

But would you really want to paste capacitors across a solar
panel? Slapped across the input side of a closed loop
control?

Jon

 

[Jon Kirwan]

On Fri, 16 Jul 2010 13:51:39 -0700, Tim Wescott
<tim@seemywebsite.com> wrote:

On 07/16/2010 11:40 AM, Jon Kirwan wrote:
On Fri, 16 Jul 2010 08:23:19 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 04:53 AM, Jon Kirwan wrote:
On Thu, 15 Jul 2010 20:09:35 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

snip
If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak.
snip

Hi, Tim. That would ONLY seem correct to me in the extreme
case where f=t_on. (An impossible case.) Not in the case
where, let's say, t_on equals t_off so that D=50%, for
example, where it would seem to be pushed to 4X, not 2X.

In any case, very fundamental considerations would suggest
(and these completely ignore some other relationships that
are necessary, such as nearly fixed limitations on t_off due
to the allowable voltage across the L when transferring
energy to the cap):

I_peak = SQRT[ (2*P) / (f*L) ]

This is simple to observe, since it is nothing more than
figuring out the (1/2)*I^2*L energy at I_peak, times the
number of such pulses allowed in a second, which must match
the input power available, I'd think. (Power is energy per
unit time, after all.)

Well...

Note my constraint: if things are arranged (i.e. if frequency and
inductance is selected such that the inductor current goes to zero just
as the transistor switches on).

I made two points before your comment here, so I'm not sure
which of the two you are addressing.

The first was about 2X vs 4X and evolves from V = L*dI/dt and
depends exactly the point you make where the inductor current
goes to zero. Without intuition, here's the algebra:

V = L*dI/dt
Assuming inductor current goes to zero and dt=t_on:
V_in = L * I_peak / t_on
Now assuming that f=1/t_on (which we know isn't true):
V_in = L * I_peak * f
And therefore,
f = V_in/(L*I_peak)
But, with P=V_in*I_in (power):
I_peak = SQRT[ (2*P) / (f*L) ]
So, substituting for f from above:
I_peak = SQRT[ (2*P) / ([V_in/(L*I_peak)]*L) ]
I_peak = SQRT[ (2*P) / (V_in/I_peak) ]
I_peak^2 = (2*P) / (V_in/I_peak)
I_peak = (2*P) / V_in
But P=V_in*I_in, with 100% efficiency, so:
I_peak = (2*V_in*I_in) / V_in
Which works out conveniently to:
I_peak = 2*I_in

Of course, the problem in the above flow is that we assumed
f=1/t_on. Clearly, it isn't, as t_off is non-zero. If you
use the above and use f=1(2*t_on), assuming D=50%, then you
will get:
I_peak = 4*I_in

The second point I'd made above was about the basic idea of
multiplying power by time to get energy and then observing
what that _must_ mean regarding inductor current, once again
also assuming that it goes to zero each cycle. So this also
does not violate the assumption you called out, again. In
fact, I took it as a given.

That's all well and good, but it simply cannot be right. If the diode
conducts for the entire time that the transistor is off, then the output
side of the inductor is always either connected to the load (13.8V) or
ground (0V). The input side is constant, so the inductor current must
be a sawtooth. The average current is going to be the mean of the peak
and trough of this sawtooth. So if the average is 1A and the trough is
0A, then there's absolutely nothing left for the peak to be but 2A.

If your fancy math comes up with an answer that doesn't match, it means
that either your math or your assumptions are incorrect.

Perhaps you should simulate this on LTSpice and see what it says...

I need to do that, thanks.

When an inductor has a constant voltage across it, the current ramps at
a constant rate (assuming constant inductance with current, which isn't
always valid in power electronics -- close your eyes to that).

Agreed.

If the
current ramps up from zero to 2A, then 2A to zero, then _immediately_
goes from zero to 2A again, the average current is 1A.

I'm not sure I read this correctly. There is the highlighted
word _immediately_ that suggests 0-2A, but I know that _time_
is required to do that. So _immediately_ cannot mean zero
time. Which leaves me confused.

In any case, what I think needs to be focused upon is energy
and time, not average current.


However, without that intuition, the algebra above arrives at


9.65W (the 9.65V time 1A) available on input (and all this
assumes such a switcher even makes sense at all for a solar
panel, which I'm not convinced of because it wants to supply
a constant current and not a wildly swinging one)

That's what input caps are for -- to smooth out the current seen by the
panel, by letting the capacitor supply the AC portion.

I guess I failed to note the OP writing that part. I thought
about it, though. Just didn't see it. In any case, I wonder
about __proper__ design for solar panels here. Not the
simple stuff used in "photovore" robots, for example. But
real, meaningful design with a solar panel and something more
on the order of 10 watts and more varying downwards as the
sun moves.

divided by
the OP's 25kHz yields 386uJ as the energy storage required in
the inductor. From that, I get 3.93A for 50uH and 3.78A for
54uH. Neither of which are 2A and both of which are much
closer to the 4X factor, using D=50%.

Even then, it's probably higher still because there are other
considerations I can imagine. For one example, D will be
more like 30%, as the OP mentioned, and the t_off time will
likely be longer than the t_on, at steady state when the V
across the inductor during t_off will be 4V or so as opposed
to the t_on V of about 9.5V.

But very basic energy-in/energy-out considerations without
any duty cycle or other limitation considerations will
suggest that at 25kHz and 50uH (assuming those are even
tenable with each other, which is yet another question not
answered) the current needs to be higher than 2A here just to
meet the energy transfer per unit time requirement.

Or maybe I'm missing something. I am a modest hobbyist and
have had zero electronics training, after all.

Not so -- the inductor is not storing all of the energy that gets
transferred. Think about it -- for the 28us that the forward diode is
conducting you are still getting current flow from the input side -- the
inductor is only making up the difference between the 9.65V and the 13.8
(or whatever). Thus, you are overestimating the amount of energy that
must be stored in the inductor.

I'm thinking exclusively about the steady state case --
_after_ the output capacitor has reached it's design voltage.
There is allowable droop. But not in my wildest imagination
did I guess that the diode would be forward conducting every
cycle!! Only in the early startup time. Which I set aside
for thinking purposes.



L D
___
.-------UUU----o---->|-------.
| | |
| | |
Vin /+\ ||-+ /+\ Vout
( ) ||<- ( )
\-/ -||-+ \-/
| | |
| | |
=== === ===
GND GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Please explain how it is possible for the diode to _not_ conduct when
there is forward current in the coil and the transistor is off, as will
happen every single cycle at the moment that the transistor turns off.

It will forward conduct when the transistor is off, Tim.
Because of the inductor. That's never been at issue.

The point here is that the forward conduction is DUE to
energy stored in a magnetic field in the inductor. And
_that_ is the fact at issue in our discussion. This energy
_must_ be factored into any discussion.

Since the input side is at a lower voltage than the output,
at steady state, there is no other method by with power is
transferred, except by way of pulsed magnetic field energy
storage. As such, it's a very simple calculation to know how
much energy must be present in each pulse if there are
exactly 25000 of them per second. (25kHz.)

I'm not sure what is wrong about our dialog, yet. There is
something "cross purposes" about it. And I have to take the
blame here. You are the trained expert, not me.

Once the output, within ripple considerations, is
established, the ONLY way energy gets transferred is when
that diode is forward conducting. And if the output voltage
doesn't droop back down to 9V (which I can't accept it does
do), then this only happens when the inductor dumps. And, I
have to believe, the OP intends this to be just about all 10
watts at this steady state.

And if you're at all smart, you size the inductor so that it takes the
entire off-time of the transistor to "dump". Not just during startup --
all the time.

I'm not all that smart, Tim. I'm just a bystander looking
in.

Yes, I would be over-estimating in the case of "startup."
Obviously. But once the output voltage is established and if
the entire panel output of nearly 10W is supposed to continue
to supply energy to the output side at 100% efficiency (I
think we both take this as a given, for now) then the
inductor is fully involved in achieving that, I think.

Or?

I missed the OP's frequency spec. You know that if conduction is
continuous then the duty cycle has to be 30% just from the voltage
ratio.

Yes.

So you know that the FET is on for 12us, and that it reaches 2A
peak. To do this with that voltage in,
L = (9.65)(12us)/(2A) = 58uH.

If you want to do the energy balance equations go ahead, but take into
account that during forward conduction of the diode some of the energy
is going directly into the output. In fact, it'll be 70% of the input
power.

10W power-in needs to be transferred to 10W power-out, at
100% efficiency. Assuming that the output capacitor is at or
near it's rated output voltage of 12-14V, this seems to
suggest that the 10W must be divided out into (1/2)*L*I^2
pulses, since it cannot be directly flowing via a
reverse-biased freewheeling diode as the V_in is less than
10V and the V_out is decidedly higher.

As I had pointed out, this means almost 400uJ at 25kHz. Could
you help me by a more thorough description explaining by what
mechanism power might be transferred -- once steady state is
achieved? I'm not seeing it and that must be my fault.

So, you've got 9.65V on the input side, 4.15V across the inductor, and
you flow a charge of (1A)(28us) = 28uC, for a total energy of
(13.8V)(28uC) = 386uJ.

Yes, per pulse at 25kHz.

So how much of the energy is delivered by the
inductor, and how much is delivered by the input voltage source?

Okay. I think I'm following, a little.

Keep in mind that the only connection to the input source is the coil,

Okay, yes.

so any
current into the coil comes from the input source, which constrains the
input source current to that of the coil.

Okay. I think I follow the argument you are making here.

In my own words, what you are saying here is that the
inductor provides only "part" of the net voltage and that
both the inductor and the panel together provides the current
so that the effect is a "sharing" of net power. The result
of that is that only a portion of the total transfer of
energy per unit time needs to be present, stored temporarily
as energy in the magnetic field each pulse.

So I'm mistaken to imagine that approximately 10J of energy
needs to divided down by 25000 to get each pulse's energy
because some of the energy arrives directly by way of the
source voltage and current.

Jon

 

[Tim Wescott]

On 07/17/2010 07:40 AM, Jon Kirwan wrote:

On Fri, 16 Jul 2010 13:51:39 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 11:40 AM, Jon Kirwan wrote:
On Fri, 16 Jul 2010 08:23:19 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 04:53 AM, Jon Kirwan wrote:
On Thu, 15 Jul 2010 20:09:35 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

snip
If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak.
snip

Hi, Tim. That would ONLY seem correct to me in the extreme
case where f=t_on. (An impossible case.) Not in the case
where, let's say, t_on equals t_off so that D=50%, for
example, where it would seem to be pushed to 4X, not 2X.

In any case, very fundamental considerations would suggest
(and these completely ignore some other relationships that
are necessary, such as nearly fixed limitations on t_off due
to the allowable voltage across the L when transferring
energy to the cap):

I_peak = SQRT[ (2*P) / (f*L) ]

This is simple to observe, since it is nothing more than
figuring out the (1/2)*I^2*L energy at I_peak, times the
number of such pulses allowed in a second, which must match
the input power available, I'd think. (Power is energy per
unit time, after all.)

Well...

Note my constraint: if things are arranged (i.e. if frequency and
inductance is selected such that the inductor current goes to zero just
as the transistor switches on).

I made two points before your comment here, so I'm not sure
which of the two you are addressing.

The first was about 2X vs 4X and evolves from V = L*dI/dt and
depends exactly the point you make where the inductor current
goes to zero. Without intuition, here's the algebra:

V = L*dI/dt
Assuming inductor current goes to zero and dt=t_on:
V_in = L * I_peak / t_on
Now assuming that f=1/t_on (which we know isn't true):
V_in = L * I_peak * f
And therefore,
f = V_in/(L*I_peak)
But, with P=V_in*I_in (power):
I_peak = SQRT[ (2*P) / (f*L) ]
So, substituting for f from above:
I_peak = SQRT[ (2*P) / ([V_in/(L*I_peak)]*L) ]
I_peak = SQRT[ (2*P) / (V_in/I_peak) ]
I_peak^2 = (2*P) / (V_in/I_peak)
I_peak = (2*P) / V_in
But P=V_in*I_in, with 100% efficiency, so:
I_peak = (2*V_in*I_in) / V_in
Which works out conveniently to:
I_peak = 2*I_in

Of course, the problem in the above flow is that we assumed
f=1/t_on. Clearly, it isn't, as t_off is non-zero. If you
use the above and use f=1(2*t_on), assuming D=50%, then you
will get:
I_peak = 4*I_in

The second point I'd made above was about the basic idea of
multiplying power by time to get energy and then observing
what that _must_ mean regarding inductor current, once again
also assuming that it goes to zero each cycle. So this also
does not violate the assumption you called out, again. In
fact, I took it as a given.

That's all well and good, but it simply cannot be right. If the diode
conducts for the entire time that the transistor is off, then the output
side of the inductor is always either connected to the load (13.8V) or
ground (0V). The input side is constant, so the inductor current must
be a sawtooth. The average current is going to be the mean of the peak
and trough of this sawtooth. So if the average is 1A and the trough is
0A, then there's absolutely nothing left for the peak to be but 2A.

If your fancy math comes up with an answer that doesn't match, it means
that either your math or your assumptions are incorrect.

Perhaps you should simulate this on LTSpice and see what it says...

I need to do that, thanks.

When an inductor has a constant voltage across it, the current ramps at
a constant rate (assuming constant inductance with current, which isn't
always valid in power electronics -- close your eyes to that).

Agreed.

If the
current ramps up from zero to 2A, then 2A to zero, then _immediately_
goes from zero to 2A again, the average current is 1A.

I'm not sure I read this correctly. There is the highlighted
word _immediately_ that suggests 0-2A, but I know that _time_
is required to do that. So _immediately_ cannot mean zero
time. Which leaves me confused.

In any case, what I think needs to be focused upon is energy
and time, not average current.


However, without that intuition, the algebra above arrives at


9.65W (the 9.65V time 1A) available on input (and all this
assumes such a switcher even makes sense at all for a solar
panel, which I'm not convinced of because it wants to supply
a constant current and not a wildly swinging one)

That's what input caps are for -- to smooth out the current seen by the
panel, by letting the capacitor supply the AC portion.

I guess I failed to note the OP writing that part. I thought
about it, though. Just didn't see it. In any case, I wonder
about __proper__ design for solar panels here. Not the
simple stuff used in "photovore" robots, for example. But
real, meaningful design with a solar panel and something more
on the order of 10 watts and more varying downwards as the
sun moves.

divided by
the OP's 25kHz yields 386uJ as the energy storage required in
the inductor. From that, I get 3.93A for 50uH and 3.78A for
54uH. Neither of which are 2A and both of which are much
closer to the 4X factor, using D=50%.

Even then, it's probably higher still because there are other
considerations I can imagine. For one example, D will be
more like 30%, as the OP mentioned, and the t_off time will
likely be longer than the t_on, at steady state when the V
across the inductor during t_off will be 4V or so as opposed
to the t_on V of about 9.5V.

But very basic energy-in/energy-out considerations without
any duty cycle or other limitation considerations will
suggest that at 25kHz and 50uH (assuming those are even
tenable with each other, which is yet another question not
answered) the current needs to be higher than 2A here just to
meet the energy transfer per unit time requirement.

Or maybe I'm missing something. I am a modest hobbyist and
have had zero electronics training, after all.

Not so -- the inductor is not storing all of the energy that gets
transferred. Think about it -- for the 28us that the forward diode is
conducting you are still getting current flow from the input side -- the
inductor is only making up the difference between the 9.65V and the 13.8
(or whatever). Thus, you are overestimating the amount of energy that
must be stored in the inductor.

I'm thinking exclusively about the steady state case --
_after_ the output capacitor has reached it's design voltage.
There is allowable droop. But not in my wildest imagination
did I guess that the diode would be forward conducting every
cycle!! Only in the early startup time. Which I set aside
for thinking purposes.



L D
___
.-------UUU----o---->|-------.
| | |
| | |
Vin /+\ ||-+ /+\ Vout
( ) ||<- ( )
\-/ -||-+ \-/
| | |
| | |
=== === ===
GND GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

Please explain how it is possible for the diode to _not_ conduct when
there is forward current in the coil and the transistor is off, as will
happen every single cycle at the moment that the transistor turns off.

It will forward conduct when the transistor is off, Tim.
Because of the inductor. That's never been at issue.

The point here is that the forward conduction is DUE to
energy stored in a magnetic field in the inductor. And
_that_ is the fact at issue in our discussion. This energy
_must_ be factored into any discussion.

Since the input side is at a lower voltage than the output,
at steady state, there is no other method by with power is
transferred, except by way of pulsed magnetic field energy
storage. As such, it's a very simple calculation to know how
much energy must be present in each pulse if there are
exactly 25000 of them per second. (25kHz.)

I'm not sure what is wrong about our dialog, yet. There is
something "cross purposes" about it. And I have to take the
blame here. You are the trained expert, not me.

Once the output, within ripple considerations, is
established, the ONLY way energy gets transferred is when
that diode is forward conducting. And if the output voltage
doesn't droop back down to 9V (which I can't accept it does
do), then this only happens when the inductor dumps. And, I
have to believe, the OP intends this to be just about all 10
watts at this steady state.

And if you're at all smart, you size the inductor so that it takes the
entire off-time of the transistor to "dump". Not just during startup --
all the time.

I'm not all that smart, Tim. I'm just a bystander looking
in.

Yes, I would be over-estimating in the case of "startup."
Obviously. But once the output voltage is established and if
the entire panel output of nearly 10W is supposed to continue
to supply energy to the output side at 100% efficiency (I
think we both take this as a given, for now) then the
inductor is fully involved in achieving that, I think.

Or?

I missed the OP's frequency spec. You know that if conduction is
continuous then the duty cycle has to be 30% just from the voltage
ratio.

Yes.

So you know that the FET is on for 12us, and that it reaches 2A
peak. To do this with that voltage in,
L = (9.65)(12us)/(2A) = 58uH.

If you want to do the energy balance equations go ahead, but take into
account that during forward conduction of the diode some of the energy
is going directly into the output. In fact, it'll be 70% of the input
power.

10W power-in needs to be transferred to 10W power-out, at
100% efficiency. Assuming that the output capacitor is at or
near it's rated output voltage of 12-14V, this seems to
suggest that the 10W must be divided out into (1/2)*L*I^2
pulses, since it cannot be directly flowing via a
reverse-biased freewheeling diode as the V_in is less than
10V and the V_out is decidedly higher.

As I had pointed out, this means almost 400uJ at 25kHz. Could
you help me by a more thorough description explaining by what
mechanism power might be transferred -- once steady state is
achieved? I'm not seeing it and that must be my fault.

So, you've got 9.65V on the input side, 4.15V across the inductor, and
you flow a charge of (1A)(28us) = 28uC, for a total energy of
(13.8V)(28uC) = 386uJ.

Yes, per pulse at 25kHz.

So how much of the energy is delivered by the
inductor, and how much is delivered by the input voltage source?

Okay. I think I'm following, a little.

Keep in mind that the only connection to the input source is the coil,

Okay, yes.

so any
current into the coil comes from the input source, which constrains the
input source current to that of the coil.

Okay. I think I follow the argument you are making here.

In my own words, what you are saying here is that the
inductor provides only "part" of the net voltage and that
both the inductor and the panel together provides the current
so that the effect is a "sharing" of net power. The result
of that is that only a portion of the total transfer of
energy per unit time needs to be present, stored temporarily
as energy in the magnetic field each pulse.

So I'm mistaken to imagine that approximately 10J of energy
needs to divided down by 25000 to get each pulse's energy
because some of the energy arrives directly by way of the
source voltage and current.

Yup.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Jon Kirwan]

On Sat, 17 Jul 2010 09:47:49 -0700, Tim Wescott
<tim@seemywebsite.com> wrote:

On 07/17/2010 07:40 AM, Jon Kirwan wrote:
snip
Okay. I think I follow the argument you are making here.

In my own words, what you are saying here is that the
inductor provides only "part" of the net voltage and that
both the inductor and the panel together provides the current
so that the effect is a "sharing" of net power. The result
of that is that only a portion of the total transfer of
energy per unit time needs to be present, stored temporarily
as energy in the magnetic field each pulse.

So I'm mistaken to imagine that approximately 10J of energy
needs to divided down by 25000 to get each pulse's energy
because some of the energy arrives directly by way of the
source voltage and current.

Yup.

Okay. So I think I follow, now. I had completely neglected
the 'pressure' that exists on the other side of the inductor
in figuring things. Thanks for the kick in the side.

I just ran a quick behavioral simulation (near perfect
switch, etc) and used a "realistic" schottky diode that
burned a little power and tweaked things so that the output
was a steady 13V at 0.7A. The MBRS1100 I used here presented
about 0.69V when forward-conducting. Input voltage was set
to 9.65V. (Used .IC to init the output node.) The result
is:

Input power: 9584 mW
Output power: 9101 mW
Schottky power: 482 mW
Misc power: 1 mW

Those figures include numerical rounding errors by LTSpice.

Peak inductor current is 2.13A and with 50uH is about 113.5uJ
per pulse. At 25kHz, this works out to about 2836 mJ/s and
it runs nicely with 11us ON time and 29us OFF time. The
inductor current has fallen to zero after a little more than
27us, so there is a tiny period of zero inductor current
before the next pulse starts.

In short, to deliver 9100mJ to the load about 2840mJ of it
appears in pulsed magnetic storage in the same period. About
31% or so.

Working the other way, from theory to practice, I get 13V
plus the schottky's 0.69V for 13.69V, then minus the 9.65V
source voltage, for 4.04V net. This, compared against the
output voltage of 13V gives 31.08%... which is very close to
the required estimate for the inductor energy factor.

I think I get it, now.

Thanks,
Jon

 

[Jasen Betts]

On 2010-07-17, Jon Kirwan <jonk@infinitefactors.org> wrote:

On Fri, 16 Jul 2010 13:53:13 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 11:46 AM, Jon Kirwan wrote:

Yes. Hence why I don't think slapping a cap across the panel
is optimal.

You're not using the cap for tracking for heaven's sake! You're using
it to smooth out the current draw from the panel. It's all or part of
the input filter that any wise designer always puts on the front end of
a switcher.

Thanks, Tim.

But would you really want to paste capacitors across a solar
panel? Slapped across the input side of a closed loop
control?

The control input is from the switchers' output, that's whay it's
called closed-loop



--- news://freenews.netfront.net/ - complaints: news@netfront.net ---

 

[Jon Kirwan]

On 18 Jul 2010 09:44:32 GMT, Jasen Betts <jasen@xnet.co.nz>
wrote:

On 2010-07-17, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Fri, 16 Jul 2010 13:53:13 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 11:46 AM, Jon Kirwan wrote:

Yes. Hence why I don't think slapping a cap across the panel
is optimal.

You're not using the cap for tracking for heaven's sake! You're using
it to smooth out the current draw from the panel. It's all or part of
the input filter that any wise designer always puts on the front end of
a switcher.

Thanks, Tim.

But would you really want to paste capacitors across a solar
panel? Slapped across the input side of a closed loop
control?

The control input is from the switchers' output, that's whay it's
called closed-loop

That much, I know. I'm just wondering about the addition of
more C on that end of the control loop.

Jon

 

[Jon Kirwan]

I suppose it depends, though, on what the closed control loop
observes to perform its function. It just seems to me that a
capacitor is not a panacea for all designs. It may actually
be a problem, for some. This is a question, not a claim.

Jon

 

[Tim Wescott]

On 07/18/2010 02:44 AM, Jasen Betts wrote:

On 2010-07-17, Jon Kirwan<jonk@infinitefactors.org> wrote:
On Fri, 16 Jul 2010 13:53:13 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

On 07/16/2010 11:46 AM, Jon Kirwan wrote:

Yes. Hence why I don't think slapping a cap across the panel
is optimal.

You're not using the cap for tracking for heaven's sake! You're using
it to smooth out the current draw from the panel. It's all or part of
the input filter that any wise designer always puts on the front end of
a switcher.

Thanks, Tim.

But would you really want to paste capacitors across a solar
panel? Slapped across the input side of a closed loop
control?

The control input is from the switchers' output, that's whay it's
called closed-loop

In a photovoltaic array the control system needs to take input from both
sides of the switcher if it's going to suck the maximum possible power
from the array.

And then (to answer Jon's question) you have to take the input
capacitance into account -- but it's not that difficult.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Bill Bowden]

On Jul 16, 9:38 am, Tim Wescott <t...@seemywebsite.com> wrote:

On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator.....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor.  Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot.  There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

Works fairly well at 84% efficiency. I lowered the frequency to 12KHz
to reduce the diode switching losses. The current ramps from about 1
amp minimum to 2.25 peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

-Bill

--

Tim Wescott
Wescott Design Serviceshttp://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details athttp://www.wescottdesign.com/actfes/actfes.html

 

[Jon Kirwan]

On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden
<wrongaddress@att.net> wrote:

On Jul 16, 9:38 am, Tim Wescott <t...@seemywebsite.com> wrote:
On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor.  Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot.  There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1". .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is: r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries. That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.) Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51". From that,
I get about 107uH. Not 250uH. Let's call that 110uH and
talk about this later.

In any case, Tim's point about radiation may be important. An
air inductor has a very widely dispersed magnetic field (it's
not constrained much.) I don't know how to calculate the
loss to far field radiation here. But maybe someone can
discuss that a little bit.

Works fairly well at 84% efficiency.

What are you loading the output with? What is the output
voltage?

I lowered the frequency to 12KHz to reduce the diode
switching losses. The current ramps from about 1 amp
minimum to 2.25 peak.

Given the earlier discussion and what I think I learned from
Tim's comments, let's say your output voltage is 14V and your
diode circa 1A is running about 0.8V. From this, I gather
that the inductor must handle a percentage of the total power
by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%. (The
0.1V is your transistor switch drop and the 0.8V is the diode
drop estimate during conduction.)

With assumed usable input power near 9.5 watts, and 37.5% of
that in the inductor, about 3.56 Joules must be divided out
by 12000 pulses. I get close to 300uJ per pulse, here.
Assuming you actually had built 250uH into your air core,
this would be about 1.56A at the peak. However, using that
110uH I earlier estimated from that multilayer formula and
your other figures, I get about 2.3A at 110uH. Which is much
closer to your own comments about the observed peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe? It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

Jon

 

[Tim Wescott]

On 07/21/2010 02:52 AM, Jon Kirwan wrote:

On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden
wrongaddress@att.net> wrote:

On Jul 16, 9:38 am, Tim Wescott<t...@seemywebsite.com> wrote:
On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor. Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot. There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1". .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is: r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries. That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.) Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51". From that,
I get about 107uH. Not 250uH. Let's call that 110uH and
talk about this later.

In any case, Tim's point about radiation may be important. An
air inductor has a very widely dispersed magnetic field (it's
not constrained much.) I don't know how to calculate the
loss to far field radiation here. But maybe someone can
discuss that a little bit.

Works fairly well at 84% efficiency.

What are you loading the output with? What is the output
voltage?

I lowered the frequency to 12KHz to reduce the diode
switching losses. The current ramps from about 1 amp
minimum to 2.25 peak.

Given the earlier discussion and what I think I learned from
Tim's comments, let's say your output voltage is 14V and your
diode circa 1A is running about 0.8V. From this, I gather
that the inductor must handle a percentage of the total power
by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%. (The
0.1V is your transistor switch drop and the 0.8V is the diode
drop estimate during conduction.)

With assumed usable input power near 9.5 watts, and 37.5% of
that in the inductor, about 3.56 Joules must be divided out
by 12000 pulses. I get close to 300uJ per pulse, here.
Assuming you actually had built 250uH into your air core,
this would be about 1.56A at the peak. However, using that
110uH I earlier estimated from that multilayer formula and
your other figures, I get about 2.3A at 110uH. Which is much
closer to your own comments about the observed peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe? It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

I'd want to know the inter-coil capacitance. What's the self-resonant
frequency of the coil? If you know the self-resonant frequency and the
resonant frequency with some capacitance attached you can model the coil
as an inductance in parallel with a capacitor and assign values.

The closer the self-resonant frequency is to the operating frequency the
more switching losses you're going to have, unless you go to a fancy
resonant-mode controller.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Jon Kirwan]

On Wed, 21 Jul 2010 10:44:22 -0700 (PDT), Bill Bowden
<wrongaddress@att.net> wrote:

On Jul 21, 2:52 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden

snip

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6
layers. It resonates at 20KHz with a 0.22uF cap, so L=288uH.

Okay. I took things the other way. (Thought that the 1" was
the length.) I'll go with your measured values.

snip

What are you loading the output with?  What is the output
voltage?

In normal operation, the load is the battery, which holds the output
voltage constant, regardless of input or output current.

Yes, but what is that voltage it holds?

The duty
cycle is fine tuned for a peak in output current, which seems the best
match.

Okay. Which implies a higher output voltage under load, I
think, as a higher voltage across your battery will mean more
current into it. And all this means, as you already know,
that there is more power delivered, which is the goal.

But as the battery voltage rises, the duty cycle needs a small
adjustment. There is no feedback, so I just leave it set optimum at 13
volts. It's just a 555 timer driving a mosfet.

Okay. Got it.

But using a 16 ohm load in place of the battery gives me 1.14A in at
9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe
83.5% efficiency.

Thanks. I'm enjoying thinking about the problem and reading
about your observations. It's teaching me to apply theory
and think better, generally.

Tim's knock in the head about the inductor's role in the work
involved helped a lot and now I'd like to see if paper and
pencil match experimental result, as it should!

But it was hard to get the same numbers twice since the sunlight kept
changing slightly. I'll run the test again using a power supply instead
of the panel.

Makes sense!!

snip
Bill:
I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe?  It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

I'm not sure of the current ramps, I was using a 0.1 ohm resistor in
series with the coil (for scope monitor) which upsets things a bit.

Given .2 ohms for the inductor itself, it's a 50% increase in
the DC resistance. But you should loose an eighth watt or so
there, so I don't think it's all that terrible. I might have
tried the same thing, since scoping makes observation of the
details clear.

The peak to peak ramp appears to be about 1.25 amps offset by 1 amp,
or 2.25 peak to 1 minimum.

Doesn't that bother you? It just about cannot be.

But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) =
14 volts.

From this, I gather you are setting your 555 timer's pulse
width so that the mosfet is on for 25us, yes?

I think it should be 9.5.

I agree.

Assume you are right that L=288uH. (I'm convinced you've
tested this.) Assume you are right about 25us. (I'm
convinced you can easily see this on the scope, by simply
looking.) You and I both know the voltage should be 9.5V --
it's a given, considering the source you have. So from that
the dI=V*dt/L, or .825A. Since the others are solid givens,
this MUST be true. It cannot be false. So it is clear the
1.25A delta is off in some way.

Probably inaccurate measurements, and
uncalibrated scope. Just ball park figures.

Okay.

It still bothers me about the 1A minimum. I know you can
easily establish where "zero" is at on your scope by simply
connecting the probe tip and ground and adjusting that line
to a graticule line for reference, so I don't think you can
mistake the fact that the pulse voltage bottom is above it.

This bothers me. But I believe you are seeing a baseline
there, too. Not that you have an exact figure for it. But
the fact that it is present.

What is missing in my mind is WHY. What would cause it? Oh!
Obvious. dt=(L/V)*dI, with V selected for the relaxation
time, which will be output+diode-9.5V, or so. At 13V output,
this is maybe 13.7-9.65 or close to 4V. This suggests a dt
of about 60us. Added to 25us, this is 85us and exceeds what
is happening at 12kHz. There just isn't enough time.

Try lowering the frequency a little?

Jon

 

[Bill Bowden]

On Jul 21, 2:52 am, Jon Kirwan <j...@infinitefactors.org> wrote:

On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden



wrongaddr...@att.net> wrote:
On Jul 16, 9:38 am, Tim Wescott <t...@seemywebsite.com> wrote:
On 07/15/2010 08:09 PM, Tim Wescott wrote:

On 07/15/2010 07:33 PM, Bill Bowden wrote:
I have a DIY solar panel that delivers 9.65 volts (24 cells) at about
1 amp and need a boost converter (12-14) to charge a SLA battery. I
want to use a air core inductor (30-60uH) to avoid special ferrite
cores. I think a small spool of copper wire on a plastic core will be
about 50uH.

According to this boost converter calculator, with 9.65 volts in, 13
out, diode drop of 0.7, transistor drop of 0.1 (IRFZ44 mosfet), freq
of 25KHz, output current of 700mA, and inductor ripple current of
300%, the inductor is 54uH. But it also indicates the peak inductor
current is 1.75 amps with a duty cycle of 30%. I'm not sure what the
inductor Ipp (2.1 amps) is?

Anyway, I don't see how the peak current can be only 1.75 amps with a
duty cycle of 30%. If the input current is a constant 1 amp, then it
must be around 3 amps during the 30% time the transistor is on. I can
add a input cap to supply 3 amps during the 30% on time (1 amp
average). And if the current ramps from 0 to a peak and averages 3
amps, the peak would seem to be around 6 amps?

What am I missing?

http://www.daycounter.com/Calculators/Switching-Converter-Calculator.....

A good calculator?

The 30% duty cycle is the on time of the transistor -- the 13.8V side is
getting current about 70% of the time.

The 30% duty cycle sounds about right -- 9.6V is about 70% of 13.8V, so
for 13.8V side to match the 9.6V side it needs to be pulled to zero for
about 30% of the time. (that makes sense, really).

If the inductor current is just kissing zero on each cycle then it needs
to ramp up to twice average, or 2A (ignoring losses) peak. The more
inductance the more the inductor peak (and trough) will approach the
average 1A. So 1.75A is possible.

I forgot to mention -- there's no reason to make an air-core inductor
for this, or to obtain a core for a custom inductor.  Switching supplies
have gotten popular enough that you can almost always buy the inductor
you need from the likes of DigiKey or Mouser.

I suspect that an air core inductor would have problems with parasitics,
as well as being huge and wanting to radiate a lot.  There's a _reason_
people use cores.

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6
layers.
It resonates at 20KHz with a 0.22uF cap, so L=288uH.

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is:  r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries.  That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.)  Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51".  From that,
I get about 107uH.  Not 250uH.  Let's call that 110uH and
talk about this later.

In any case, Tim's point about radiation may be important. An
air inductor has a very widely dispersed magnetic field (it's
not constrained much.)  I don't know how to calculate the
loss to far field radiation here.  But maybe someone can
discuss that a little bit.

Works fairly well at 84% efficiency.

What are you loading the output with?  What is the output
voltage?

In normal operation, the load is the battery, which holds the output

voltage constant, regardless of input or output current. The duty
cycle is fine tuned for a peak in output current, which seems the best
match. But as the battery voltage rises, the duty cycle needs a small
adjustment. There is no feedback, so I just leave it set optimum at 13
volts. It's just a 555 timer driving a mosfet.

But using a 16 ohm load in place of the battery gives me 1.14A in at
9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe
83.5% efficiency.

But it was hard to get the same numbers twice since the sunlight kept
changing
slightly. I'll run the test again using a power supply instead of the
panel.

I lowered the frequency to 12KHz to reduce the diode
switching losses.  The current ramps from about 1 amp
minimum to 2.25 peak.

Given the earlier discussion and what I think I learned from
Tim's comments, let's say your output voltage is 14V and your
diode circa 1A is running about 0.8V.  From this, I gather
that the inductor must handle a percentage of the total power
by the factor of (14V+0.8V-9.65V+0.1V)/14V, or 37.5%.  (The
0.1V is your transistor switch drop and the 0.8V is the diode
drop estimate during conduction.)

With assumed usable input power near 9.5 watts, and 37.5% of
that in the inductor, about 3.56 Joules must be divided out
by 12000 pulses.  I get close to 300uJ per pulse, here.
Assuming you actually had built 250uH into your air core,
this would be about 1.56A at the peak.  However, using that
110uH I earlier estimated from that multilayer formula and
your other figures, I get about 2.3A at 110uH.  Which is much
closer to your own comments about the observed peak.

I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe?  It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

I'm not sure of the current ramps, I was using a 0.1 ohm resistor in
series with the coil (for scope monitor) which upsets things a bit.
The peak to peak ramp appears to be about 1.25 amps offset by 1 amp,
or 2.25 peak to 1 minimum.

But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) 14 volts.
I think it should be 9.5. Probably inaccurate measurements, and
uncalibrated scope. Just ball park figures.

-Bill

> Jon

 

[Jon Kirwan]

On Wed, 21 Jul 2010 12:33:56 -0700, Jon Kirwan
<jonk@infinitefactors.org> wrote:

Try lowering the frequency a little?

Actually, a lot to kill something near 1A baseline. Worse,
though, is that with a lower frequency you need MORE energy
per pulse. And that means MORE ON time. So you need to
adjust that, too.

Let's see:

ON dV = 9.5V
OFF dV = 4V (or so)
V_out = 13V (or so)
V_in = 9.65V
V_sw = 0.1V (roughly)
V_diode= 0.7V (roughly)
P_in = 11W (very roughly)
L = 288uH
V_Lon = V_in - V_sw
V_Loff = V_out + V_diode - (V_in - V_sw)

Percent of power needed in the inductor's magnetic field is
roughly:

PCT = V_Loff / V_out

In this case, (13V+0.7V-(9.65V-0.1V))/13V, or 31.9%. So,

PCT = 0.319

Energy per pulse is then:

E_ind = (P_in * PCT / f) = (1/2)*I_peak^2*L

Assuming a starting point of zero inductor current:

I_peak = V_Lon * t_on / L

Also, assuming enough time to relax back to zero:

t_off = L * I_peak / V_Loff

and now we know,

f = 1 / (t_on + t_off)

So,

t_on+t_off = I_peak^2 * L / (2 * P_in * PCT)

Substituting like crazy and flipping around to remove t_off
and I_peak in the equation and solve for t_on, I get this:

t_on = (1 + V_Lon/V_Loff) * 2 * L * P_in * PCT / V_Lon^2

Dimensional analysis, as a quick check, says:

seconds = Henries * Watts / Volts^2

Which is right. Watt is Joules/second, Volt is Joules/
Coulomb, and Henry is Joule-second^2/Coulomb^2. So it seems
that I didn't make that kind of gross error.

Computing using your L=288uH, PCT=.319, P_in=11W,
V_Lon=9.55V, and V_Loff=4.15V, I get t_on = 73.16us.

This is a lot longer than you've been using. From this, I
also get I_peak=2.426A. This then says t_off=168.36us. So
from all of this, I get t_on+t_off should be about 240us
long.

You might try that. Set t_on+t_off to 240us and t_on to
somewhere in the range of 73-74us and see what happens.

I'm hoping you see a zero-amp baseline, then.

Jon

 

[Bill Bowden]

On Jul 21, 12:33 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

On Wed, 21 Jul 2010 10:44:22 -0700 (PDT), Bill Bowden

wrongaddr...@att.net> wrote:
On Jul 21, 2:52 am, Jon Kirwan <j...@infinitefactors.org> wrote:
On Tue, 20 Jul 2010 21:34:07 -0700 (PDT), Bill Bowden

snip

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

The inner diameter is 1 inch, width is 5/8, 15 turns per layer, 6
layers.  It resonates at 20KHz with a 0.22uF cap, so L=288uH.

Okay.  I took things the other way.  (Thought that the 1" was
the length.)  I'll go with your measured values.

snip

What are you loading the output with?  What is the output
voltage?

In normal operation, the load is the battery, which holds the output
voltage constant, regardless of input or output current.

Yes, but what is that voltage it holds?

The duty
cycle is fine tuned for a peak in output current, which seems the best
match.

Okay.  Which implies a higher output voltage under load, I
think, as a higher voltage across your battery will mean more
current into it.  And all this means, as you already know,
that there is more power delivered, which is the goal.

But as the battery voltage rises, the duty cycle needs a small
adjustment. There is no feedback, so I just leave it set optimum at 13
volts. It's just a 555 timer driving a mosfet.

Okay.  Got it.

But using a 16 ohm load in place of the battery gives me 1.14A in at
9.55 volts and 12.06 volts out, or 9.1 watts out of 10.9, or maybe
83.5% efficiency.

Thanks.  I'm enjoying thinking about the problem and reading
about your observations.  It's teaching me to apply theory
and think better, generally.

Tim's knock in the head about the inductor's role in the work
involved helped a lot and now I'd like to see if paper and
pencil match experimental result, as it should!

But it was hard to get the same numbers twice since the sunlight kept
changing slightly. I'll run the test again using a power supply instead
of the panel.

Makes sense!!



snip
Bill:
I tried a fast recovery diode against a regular rectifier diode and
only got a 1.5 % difference, so I guess diode recovery time doesn't
matter much at 12KHz.

The inductor current appears to ramp from 1 amp minimum to 2.25 amps
peak, or maybe 1.6 average, so inductor wire loss is about 1/2 watt.
Steady state diode loss is about 0.7*0.7 or maybe another 1/2 watt, so
I lose 1 watt out of 10, or maybe 90% efficiency. But I only get 84%,
so I don't where the other 6% went, but it works ok. The battery is
fully charged.

Can you talk more about how you are loading things and what
output voltage you observe?  It's interesting thinking about
this and perhaps Tim can add a few points I'm missing out on,
as well.

I'm not sure of the current ramps, I was using a 0.1 ohm resistor in
series with the coil (for scope monitor) which upsets things a bit.

Given .2 ohms for the inductor itself, it's a 50% increase in
the DC resistance.  But you should loose an eighth watt or so
there, so I don't think it's all that terrible.  I might have
tried the same thing, since scoping makes observation of the
details clear.

The peak to peak ramp appears to be about 1.25 amps offset by 1 amp,
or 2.25 peak to 1 minimum.

Doesn't that bother you?  It just about cannot be.

But that doesn't agree with 280uH if e=L di/dt = .00028* (1.25/25uS) > >14 volts.

From this, I gather you are setting your 555 timer's pulse
width so that the mosfet is on for 25us, yes?

I think it should be 9.5.

I agree.

Assume you are right that L=288uH.  (I'm convinced you've
tested this.)  Assume you are right about 25us.  (I'm
convinced you can easily see this on the scope, by simply
looking.)  You and I both know the voltage should be 9.5V --
it's a given, considering the source you have.  So from that
the dI=V*dt/L, or .825A.  Since the others are solid givens,
this MUST be true.  It cannot be false.  So it is clear the
1.25A delta is off in some way.

Probably inaccurate measurements, and
uncalibrated scope. Just ball park figures.

Okay.

It still bothers me about the 1A minimum.  I know you can
easily establish where "zero" is at on your scope by simply
connecting the probe tip and ground and adjusting that line
to a graticule line for reference, so I don't think you can
mistake the fact that the pulse voltage bottom is above it.

This bothers me.  But I believe you are seeing a baseline
there, too.  Not that you have an exact figure for it.  But
the fact that it is present.

What is missing in my mind is WHY.  What would cause it?  Oh!
Obvious.  dt=(L/V)*dI, with V selected for the relaxation
time, which will be output+diode-9.5V, or so.  At 13V output,
this is maybe 13.7-9.65 or close to 4V.  This suggests a dt
of about 60us.  Added to 25us, this is 85us and exceeds what
is happening at 12kHz.  There just isn't enough time.

Try lowering the frequency a little?

Jon

I got some better figures using a 9.5 volt PS and 16.3 ohm load.

Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts.
efficiency is around 90% which is closer to what I added up
considering 1/2 watt loss in the diode and another 1/2 watt in the
coil. The mosfet resistance is only 28 milliohms, so not much lost
there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33%

Assuming the inductor is 280uH, then di/dt is 9.5 / .00028 = 33929
amps per second, so the current should ramp up 1.36 amps in 40uS. But
the problem is on the down ramp, the voltage should be 12.7 + 0.8 for
the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should
be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But
this doesn't agree with the ramp up value of 1.36, so something is
missing. How would you determine the minimum current point?

I think the idea is to drop the inductance from infinity where the
current is constant to a smaller value where the minimum current gets
close to zero.
So, I might take a few turns off the inductor and save some
resistance, and let the current get closer to zero, but I don't think
the efficiency will go much past 90% that I already have.

-Bill

 

[Jon Kirwan]

On Wed, 21 Jul 2010 17:03:24 -0700 (PDT), Bill Bowden wrote:

snip
I got some better figures using a 9.5 volt PS and 16.3 ohm load.

Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts.
efficiency is around 90% which is closer to what I added up
considering 1/2 watt loss in the diode and another 1/2 watt in the
coil. The mosfet resistance is only 28 milliohms, so not much lost
there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33%

Okay. So about 120us period, 40us ON and 80us OFF. Got it.

Assuming the inductor is 280uH, then di/dt is 9.5 / .00028 = 33929
amps per second, so the current should ramp up 1.36 amps in 40uS. But
the problem is on the down ramp, the voltage should be 12.7 + 0.8 for
the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should
be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But
this doesn't agree with the ramp up value of 1.36, so something is
missing.

This is exactly the point. And I think you've got your
numbers right.

How would you determine the minimum current point?

That's a little trickier. I had to think a moment to get a
clue. I might still be wrong, but here is the argument. If
I'm lucky, I will derive something quantitative below.

You are driving (forcing) the situation using a clock and
duty cycle. The inductor doesn't control any of that. It
is, instead, driven by it. However, it does respond.

Imagine you set your duty cycle to some value that causes a
finite dI to be (t_on*V_on/L). The V_on can't be controlled
-- it just is. The t_on is set by your clock source. It
also just is. The same with L. So dI is fixed. It doesn't
matter what happens on the output. The inductor has no
choice in the matter. The dI will be some value. What isn't
said here is what the baseline I happens to be. Let's call
this baseline current through the inductor as I_0. We hope
and expect I_0 = 0A. But this isn't necessarily so. What we
can compute is dI. So the inductor current will have to rise
from I_0 to I_1 and it _will_ be the case that I_1=I_0+dI.
There is no choice on that matter.

In the steady state case, which we still aren't sure of, I_0
will be stable and so will I_1. We hope I_0 is zero and I_1
is simply dI. But hold that thought for a moment.

Now, your clock source also sets t_off. In the steady state
case, we know what the dI must be for the t_off period, too.
It has to be the same. If not, then it's not steady state.
And it turns out that we should think about that V_off value,
which will be L*dI/t_off. Since dI=(t_on*V_on/L), this works
out to V_off=[V_on*(t_on/t_off)] -- the voltage that _must_
be across L when the switch is off if a steady state exists.

Now, this V_off might not actually be what you imagine. In
other words, you might think it is simply the output voltage
plus a diode drop and less your input voltage because that is
what is needed. But it isn't what happens.

Suppose t_off is shortened a bit. Look at the equation for
V_off. Note that t_off is in the denominator? If t_off gets
smaller, V_off gets larger!! This means the output voltage
also __increases__. It has to (assume for a moment that the
diode voltage drop remains close to the same value as
before.)

But how can that be? Doesn't that mean that the load
(resistive, for now) will use more power? Yes, it does. You
might think that is a good thing. You've upped the output
voltage purely by reducing t_off!! You could reduce it even
more and get a still higher output voltage. But more output
power means more input power. And you still have a fixed
number of clocks per second. So more power must mean more
energy per pulse!

How do you get more energy per pulse?? You've already
determined that there is a fixed dI, completely determined by
L, t_on, and your input source voltage. So that can't
change. What can happen to increase the energy per pulse?

Well, take note that a fixed change in the magnitude of the
inductor current is NOT the same as representing a fixed
change in the magnitude of the energy stored in that magnetic
field.

You have these two circumstances:

E_0 = (1/2) * L * I_0^2
E_1 = (1/2) * L * I_1^2

The difference is then:

dE = E_1 - E_0 = (1/2) * L * (I_1^2 - I_0^2)

But I_1 = I_0+dI. So,

dE = (1/2) * L * ((I_0+dI)^2 - I_0^2)
= (1/2) * L * (I_0^2 + 2*dI*I_0 + dI^2 - I_0^2)
= (1/2) * L * (2*dI*I_0 + dI^2)
= (1/2) * L * dI * (2*I_0 + dI)
= (1/2) * L * dI^2 + [L * dI * I_0]

Note the last term enclosed in brackets? When I_0 is non-
zero and the same sign as dI, there is an additional amount
of energy in each pulse.

Another way of noting this fact is assume dI=1A and then to
compare when I_0=0A and I_0=1A. dE when I_0=0A is some
number we can call X. dE when I_0=1A will be 4X-1X, or 3X.
In other words, more energy is taken up and released in each
pulse when I_0=1A, even though there is the same dI in both
cases.

The point I'm getting to is that the inductor _must_ yield a
larger V_off if you shorten t_off. To do that, the output
voltage increases. That causes more power to be required.
Which, because the frequency is constant, requires more
energy per pulse. But to do that, I_0 must rise upwards so
that each pulse _can_ deliver more energy to meet the power
requirements. So it rises from zero. But this then causes a
fixed DC current in the inductor, which means more loss in
the resistive parts and the diode, as well.

Which is not desired.

The computation of exactly what I_0 must level out as depends
upon the output load's response to a change in the voltage.

for example, let's do some computations using some of your
numbers. R_out=16.3, V_Lon=9.5V, t_on=40us, t_off=80us,
f=8333Hz, L=288uH, etc. We don't know V_out, but we can
compute it from V_Lon*t_on/t_off. This is 4.75V, in short.
Assuming your diode voltage is about 0.7V, this means 4.05V
added to the 9.6V input. Call it 13.6V. However, there is
some resistance you mentioned which at a couple of amps
average (grossly assuming a lot) is maybe .5V? So call it
about 13.1V on the output. The PCT of power in L should be
about 4.75V/13.1V or, say, about 36%. And the power is
13.1V^2/16.3Ohms, times that 36%, or in other words about
3.79W is delivered by the inductor itself. With f=8333, this
means almost 455uJ per pulse.

We know from t_on=40us that the dI=1.32A. Energy from that
part, using (1/2)*L*dI^2, is about 251uJ. Since we need a
total of about 455uJ, this leaves about 204uJ that needs to
be made up in the L*dI*I_0 term. This suggests I_0=.534A.

Which may or may not be about where you are. But that seems
to be where theory takes me, right now.

I think the idea is to drop the inductance from infinity where the
current is constant to a smaller value where the minimum current gets
close to zero.
So, I might take a few turns off the inductor and save some
resistance, and let the current get closer to zero, but I don't think
the efficiency will go much past 90% that I already have.

I think you are losing some efficiency you could get back,
simply because you have a DC current flowing all the time,
increasing the mean power dissipation in resistances. Not to
mention that this would be a problem if you weren't using an
air core inductor.

I hope the above isn't too terrible to follow. It took me
some moments of thought to realize _why_ the minimum inductor
current would rise above zero on its own. But once I
realized that dI is determined by t_on and that the inductor
off-voltage is determined by dI and t_off and that shortening
t_off only means that the off-voltage goes UP, the rest
seemed to fall into place for me and make some sense.

Jon

 

[Don Klipstein]

In article <l88d46h5bl21v35s21t3lvqh02r4n0jv9h@4ax.com>, Jon Kirwan wrote:

On 20 Jul 2010 21:34:07 (PDT), Bill Bowden <wrongaddress@att.net> wrote:

<SNIP what leads to this to edit for space>

Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1". .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is: r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries. That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.) Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51". From that,
I get about 107uH. Not 250uH. Let's call that 110uH and
talk about this later.

If Bill's spool is 1 inch long by 5/8 inch diameter and his
18 AWG wire is .065 inch in overall diameter, such as hookup
wire, then this inductor as described would have inductance
around 110 uH.

I doubt it will have even .01 microfarad of interlayer
capacitance.

Looks like I would want to hear more details about this coil.

For example, if the spool was 1 inch in diameter and 5/8 inch long
and the wire was AWG 18 magnet wire more like .047 inch in diameter,
Bill could have wound 6 layers of 13 turns and a 7th layer of 12 turns.

The radius difference c becomes .33 inch
The average radius a becomes .665 inch
The length b is .625 inch

The inductance in microhenries at this rate would be 222 microhenries.

- Don Klipstein (don@misty.com)

 

[Bill Bowden]

On Jul 22, 1:39 pm, d...@manx.misty.com (Don Klipstein) wrote:

In article <l88d46h5bl21v35s21t3lvqh02r4n0j...@4ax.com>, Jon Kirwan wrote:
On 20 Jul 2010 21:34:07 (PDT), Bill Bowden <wrongaddr...@att.net> wrote:

SNIP what leads to this to edit for space



Well, I wanted to experiment with air core inductors, so I made a
250uH inductor on a 1 inch by 5/8 plastic wire spool, about 90 turns
of #18 wire and 200 milliohms resistance.

90 turns of #18 has to be longer than 1".  .0403" diameter
without insulation times 90 is more than 3.6", so I take it
you stacked things up?

Wheeler's single layer formula, which I gather isn't bad for
low frequency use, is:  r^2*N^2/(9*r+10*l) where r is the
radius, l is the length, and N is the number of turns, with
lengths given in inches and the resulting value given in
micro Henries.  That doesn't seem to cut it for your case, so
I then read that the multilayer formula is: 0.8*a^2*N^2/
(6*a+9*l+10*c) where a is the average radius and c is the
radius difference (span.)  Assuming you did 6 layers by 15
turns, I then estimate .065" for the wire diameter with
insulation and thus c=.39" and l=.98" and a=.51".  From that,
I get about 107uH.  Not 250uH.  Let's call that 110uH and
talk about this later.

  If Bill's spool is 1 inch long by 5/8 inch diameter and his
18 AWG wire is .065 inch in overall diameter, such as hookup
wire, then this inductor as described would have inductance
around 110 uH.

  I doubt it will have even .01 microfarad of interlayer
capacitance.

  Looks like I would want to hear more details about this coil.

  For example, if the spool was 1 inch in diameter and 5/8 inch long
and the wire was AWG 18 magnet wire more like .047 inch in diameter,
Bill could have wound 6 layers of 13 turns and a 7th layer of 12 turns.

  The radius difference c becomes .33  inch
  The average radius    a becomes .665 inch
  The length            b is      .625 inch

  The inductance in microhenries at this rate would be 222 microhenries..

 - Don Klipstein (d...@misty.com)

Yes, the spool is 1 inch inner diameter and 5/8 length. 6 layers of 15
turns each, 90 turns total of #18 enamel copper wire. I used this
calculator:

http://www.pronine.ca/multind.htm

-Bill

 

[Bill Bowden]

On Jul 22, 3:25 am, Jon Kirwan <j...@infinitefactors.org> wrote:

On Wed, 21 Jul 2010 17:03:24 -0700 (PDT), Bill Bowden wrote:
snip
I got some better figures using a 9.5 volt PS and 16.3 ohm load.

Output is 12.7 at 9.9 watts. Input is 9.5 at 1.16 amps, or 11 watts.
efficiency is around 90% which is closer to what I added up
considering 1/2 watt loss in the diode and another 1/2 watt in the
coil. The mosfet resistance is only 28 milliohms, so not much lost
there. Frequency is 8.33KHz, duty cycle is 40uS and 80uS = 33%

Okay.  So about 120us period, 40us ON and 80us OFF.  Got it.

Assuming the inductor is 280uH, then di/dt is 9.5 / .00028  = 33929
amps per second, so the current should ramp up 1.36 amps in 40uS. But
the problem is on the down ramp, the voltage should be 12.7 + 0.8 for
the diode, or 13.5 minus the supply of 9.5 = 4 volts. So di/dt should
be e/L = 4/.00028 = 14286 amps per second, or 1.143 amps in 80uS. But
this doesn't agree with the ramp up value of 1.36, so something is
missing.

This is exactly the point.  And I think you've got your
numbers right.

How would you determine the minimum current point?

That's a little trickier.  I had to think a moment to get a
clue.  I might still be wrong, but here is the argument.  If
I'm lucky, I will derive something quantitative below.

You are driving (forcing) the situation using a clock and
duty cycle.  The inductor doesn't control any of that.  It
is, instead, driven by it.  However, it does respond.

Imagine you set your duty cycle to some value that causes a
finite dI to be (t_on*V_on/L).  The V_on can't be controlled
-- it just is.  The t_on is set by your clock source.  It
also just is.  The same with L.  So dI is fixed.  It doesn't
matter what happens on the output.  The inductor has no
choice in the matter.  The dI will be some value.  What isn't
said here is what the baseline I happens to be.  Let's call
this baseline current through the inductor as I_0.  We hope
and expect I_0 = 0A.  But this isn't necessarily so.  What we
can compute is dI.  So the inductor current will have to rise
from I_0 to I_1 and it _will_ be the case that I_1=I_0+dI.
There is no choice on that matter.

In the steady state case, which we still aren't sure of, I_0
will be stable and so will I_1.  We hope I_0 is zero and I_1
is simply dI.  But hold that thought for a moment.

Now, your clock source also sets t_off.  In the steady state
case, we know what the dI must be for the t_off period, too.
It has to be the same.  If not, then it's not steady state.
And it turns out that we should think about that V_off value,
which will be L*dI/t_off.  Since dI=(t_on*V_on/L), this works
out to V_off=[V_on*(t_on/t_off)] -- the voltage that _must_
be across L when the switch is off if a steady state exists.

Now, this V_off might not actually be what you imagine.  In
other words, you might think it is simply the output voltage
plus a diode drop and less your input voltage because that is
what is needed.  But it isn't what happens.

Suppose t_off is shortened a bit.  Look at the equation for
V_off.  Note that t_off is in the denominator?  If t_off gets
smaller, V_off gets larger!!  This means the output voltage
also __increases__.  It has to (assume for a moment that the
diode voltage drop remains close to the same value as
before.)

But how can that be?  Doesn't that mean that the load
(resistive, for now) will use more power?  Yes, it does.  You
might think that is a good thing.  You've upped the output
voltage purely by reducing t_off!!  You could reduce it even
more and get a still higher output voltage.  But more output
power means more input power.  And you still have a fixed
number of clocks per second.  So more power must mean more
energy per pulse!

How do you get more energy per pulse??  You've already
determined that there is a fixed dI, completely determined by
L, t_on, and your input source voltage.  So that can't
change.  What can happen to increase the energy per pulse?

Well, take note that a fixed change in the magnitude of the
inductor current is NOT the same as representing a fixed
change in the magnitude of the energy stored in that magnetic
field.

You have these two circumstances:

   E_0 = (1/2) * L * I_0^2
   E_1 = (1/2) * L * I_1^2

The difference is then:

   dE = E_1 - E_0 = (1/2) * L * (I_1^2 - I_0^2)

But I_1 = I_0+dI.  So,

   dE = (1/2) * L * ((I_0+dI)^2 - I_0^2)
      = (1/2) * L * (I_0^2 + 2*dI*I_0 + dI^2 - I_0^2)
      = (1/2) * L * (2*dI*I_0 + dI^2)
      = (1/2) * L * dI * (2*I_0 + dI)
      = (1/2) * L * dI^2 + [L * dI * I_0]

Note the last term enclosed in brackets?  When I_0 is non-
zero and the same sign as dI, there is an additional amount
of energy in each pulse.

Another way of noting this fact is assume dI=1A and then to
compare when I_0=0A and I_0=1A.  dE when I_0=0A is some
number we can call X.  dE when I_0=1A will be 4X-1X, or 3X.
In other words, more energy is taken up and released in each
pulse when I_0=1A, even though there is the same dI in both
cases.

The point I'm getting to is that the inductor _must_ yield a
larger V_off if you shorten t_off.  To do that, the output
voltage increases.  That causes more power to be required.
Which, because the frequency is constant, requires more
energy per pulse.  But to do that, I_0 must rise upwards so
that each pulse _can_ deliver more energy to meet the power
requirements.  So it rises from zero.  But this then causes a
fixed DC current in the inductor, which means more loss in
the resistive parts and the diode, as well.

Which is not desired.

The computation of exactly what I_0 must level out as depends
upon the output load's response to a change in the voltage.  

for example, let's do some computations using some of your
numbers.  R_out=16.3, V_Lon=9.5V, t_on=40us, t_off=80us,
f=8333Hz, L=288uH, etc.  We don't know V_out, but we can
compute it from V_Lon*t_on/t_off.  This is 4.75V, in short.
Assuming your diode voltage is about 0.7V, this means 4.05V
added to the 9.6V input.  Call it 13.6V.  However, there is
some resistance you mentioned which at a couple of amps
average (grossly assuming a lot) is maybe .5V?  So call it
about 13.1V on the output.  The PCT of power in L should be
about 4.75V/13.1V or, say, about 36%.  And the power is
13.1V^2/16.3Ohms, times that 36%, or in other words about
3.79W is delivered by the inductor itself.  With f=8333, this
means almost 455uJ per pulse.

We know from t_on=40us that the dI=1.32A.  Energy from that
part, using (1/2)*L*dI^2, is about 251uJ.  Since we need a
total of about 455uJ, this leaves about 204uJ that needs to
be made up in the L*dI*I_0 term.  This suggests I_0=.534A.

Which may or may not be about where you are.  But that seems
to be where theory takes me, right now.

I think the idea is to drop the inductance from infinity where the
current is constant to a smaller value where the minimum current gets
close to zero.
So, I might take a few turns off the inductor and save some
resistance, and let the current get closer to zero, but I don't think
the efficiency will go much past 90% that I already have.

I think you are losing some efficiency you could get back,
simply because you have a DC current flowing all the time,
increasing the mean power dissipation in resistances.  Not to
mention that this would be a problem if you weren't using an
air core inductor.

I hope the above isn't too terrible to follow.  It took me
some moments of thought to realize _why_ the minimum inductor
current would rise above zero on its own.  But once I
realized that dI is determined by t_on and that the inductor
off-voltage is determined by dI and t_off and that shortening
t_off only means that the off-voltage goes UP, the rest
seemed to fall into place for me and make some sense.

Jon

I think I see the error. The scope readings are not exact and the duty
cycle may be 38Us to 82 instead of 40/80. Hard to read a couple
microseconds out of 120.

This makes the current ramps more equal so that 33929 amps per second
on the up side for 38uS is about 1.29 amps and 14286 amps per second
on the down side for 82uS = 1.17 amps. Pretty much the same. So they
look about equal and probably are if I could read the scope
accurately. A small error in duty cycle reading upsets the apple cart.

Now, as to where the peak and valleys should occur, I think we should
consider the case of infinite inductance where the current will be
constant, or 779 ma for 68% of the time, or about 1 amp average. So,
using the smaller inductor, the current waveform should move above the
center by 1.2 amps, or 2.2 amps peak and then back down to 1 amp which
is close to what I saw on the scope.

Just some thoughts.

-Bill