Boost converter, inductor size?

[Bill Bowden]

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

-Bill

 

[Tim Wescott]

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.phtml

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.

--

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

 

[Jasen Betts]

On 2010-07-16, Bill Bowden <wrongaddress@att.net> wrote:

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.

it's an inductor not a resistor. current flows while the transistor is off.
that's the whole idea of a boost converter.










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

 

[Jon Kirwan]

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.)

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) 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.

Jon

 

[Jon Kirwan]

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? And how do things
vary with solar motion across the sky? Is the 9.65V@1A
simply a peak? Or? And what kind of load should it "see?"

Jon

 

[Jon Kirwan]

On Fri, 16 Jul 2010 04:53:01 -0700, Jon Kirwan
<jonk@infinitefactors.org> wrote:

f=t_on

I meant f=1/t_on.

Jon

 

[Tim Wescott]

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).

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). 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.

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.

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 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. 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.

--

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

 

[Tim Wescott]

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.phtml


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.

--

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

 

[Tim Wescott]

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.

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

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.

--

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 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.

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.

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.

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.

Jon

 

[Jon Kirwan]

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.

Jon

 

[Jon Kirwan]

On Fri, 16 Jul 2010 11:40:47 -0700, Jon Kirwan
<jonk@infinitefactors.org> wrote:

However, without that intuition, the algebra above arrives at

Ignore that. It should have simply been removed.

Sorry,
Jon

 

[Jamie]

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.phtml

-Bill
What's wrong with using a charge pumped converter ? With the given

spec's you have supplied here and taken into account for loses that can
occur with such a circuit, it looks like the final output should be
exactly what you need doing a doubler of the 9.65V via a charge pump
which in theory gives you ~ 18V, then you remove ~ 2.5 volts for
lose in areas that can't do rail to rail operating and maybe some
lower output conditions. The net should give you a nice charge pump that
will bring up a 12V SLA in due time.

It's my understanding that pulsing a battery for charging maybe better
than a constant current source. This charge pump circuit would be doing
this of course.

As I think about this, I am visualizing a basic 555 circuit driving a
MOSFET that is part of the boost circuit.. in your case, a CMOS
version of the timer would be better, I would think..

Coming from a solar panel, you really don't have a lot of current to
work with. This is basically how some solar lights that use only one
cell operate the LED, because with a single cell, you aren't going to
see much light from a white LED with out doing something with the
insufficient voltage.


Jamie..

 

[Tim Wescott]

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...

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.

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.

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. So how much of the energy is delivered by the
inductor, and how much is delivered by the input voltage source? Keep
in mind that the only connection to the input source is the coil, so any
current into the coil comes from the input source, which constrains the
input source current to that of the coil.

--

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

 

[Tim Wescott]

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.

--

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

 

[Tim Wescott]

On 07/16/2010 01:42 PM, Jamie wrote:

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.phtml


-Bill
What's wrong with using a charge pumped converter ?

Well, other than the fact that efficiency would absolutely suck, nothing
I guess.

--

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

 

[Tim Wescott]

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.

--

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, 4:59 am, Jon Kirwan <j...@infinitefactors.org> wrote:

On Thu, 15 Jul 2010 19:33:31 -0700 (PDT), Bill Bowden



wrongaddr...@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....

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?  And how do things
vary with solar motion across the sky?  Is the 9.65V@1A
simply a peak?  Or?  And what kind of load should it "see?"

Jon

There is a large capacitor across the panel, so it can deliver high
currents for short times. The output impedance of the panel is close
to zero for 50uS or more. The 9.65 at 1 amp is the peak in bright
sunlight. I got a bargain on 25 chipped cells on ebay. The seller was
smart enough not to give me 36 cells that would be ideal for a 12 volt
panel. He wants me to buy 2 sets of 25 and have a few left over.

-Bill

 

[Bill Bowden]

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.

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.

And another problem is if the inductor current was a constant 3 amps,
the output power would be (13X3) or 39 watts for an input of only 9.65
watts?

How do we get the extra power?

-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

 

[Bill Bowden]

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.

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.

And another problem is if the inductor current was a constant 3 amps,
the output power would be (13X3)*.70 or 27 watts for an input of only
9.65
watts?

How do we get the extra power?

-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