Joule Thief - still not working....

[ehsjr]

Jon Kirwan wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:


Jon Kirwan wrote:

On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.


In the sense that Ic_peak is higher.

No. I the sense of what he said: "the value of R
influences both ON time and OFF time." He refers
to *time*, not Ic.

You're losing yourself in the details. Put aside all
the damn math - it's just confusing you - and measure
the TIME.

You need to get on the same page as John to have a
go at the math and make it meaningful.

Ed

The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

In other words, the ratio of OFF time to OFF+ON time is already cast
in concrete by the desired output voltage and the available input
voltage. That's a done deal. Ic_peak is then set by the desired
Iout, taking into account that required ratio. Lprimary is used to
then tweak in the desired frequency of operation (wind or unwind, as
needed.) And then Rbase is designed to compensate for the BJT's
operating beta and the design Ic_peak (tweak up or down, as needed.)

The frequency of operation, within practical bounds set by volt-secs
and reverse transit times to name two, has no apparent impact on power
out. Ic_peak, itself a function of both Vout and Iout, is established
by that. So the frequency (if you only use Lprimary windings to
adjust it) can be moved around independently without changing Ic_peak,
as Rbase has already fixed that in place.

I was so focused on that aspect of it that I didn't realize John was
talking at cross purposes to where my mind was mired.


Go to the bench, put a variable R in the base circuit
and observe for yourself.

By the way, did you ever build the air core version?
It eliminates core saturation from consideration when
trying to understand how the circuit functions.


I will try and get some time to do that, today. Actually, I just
wound up some various ferrite cores, as well (7 of them.) What I
don't have handy

Jon

 

[Jon Kirwan]

On Sat, 25 Jul 2009 11:19:58 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 25, 6:23 pm, default <defa...@defaulter.net> wrote:
On Sat, 25 Jul 2009 07:48:45 -0700 (PDT), fungus

openglMYSO...@artlum.com> wrote:
I think we can definitely assume that the ferrite was
saturating or something

Probably "or something."  



Iron powder usually has way more permeability than ferrite - ferrites
claim to fame is it is more permeable than air and able to handle high
frequencies with lower loss.  Some iron powder alloys can go to 100's
of megahertz, but iron is usually used in the kilohertz range.

I just made another one with my thin wire and an iron powder ring.
I was hoping hoping to get a lot of turns on it to see what happened
but unfortunately it behaved like the ferrite - current started
dropping off after about ten turns and varying R1 makes no difference to the
output of the circuit.

Well, default may have been thinking about the BJT, itself. Elsewhere,
he wrote, "If the transistor heats that's where most of the waste
power is going," so that's why I think he is thinking so. And he
could be right about that. You've mentioned about the BJT heating up
a lot.

However, your new test seems to argue in support of saturation, I
think. More turns means more inductance and a lower frequency of
operation (other things similar.) It also means that Bsat is reached
at lower Ic_peak: B = mu*N*I/l_m. Assuming mu and l_m is the same in
both cores you used, and so is Bsat, then winding more N will push you
closer to Bsat.

In any event, when you select a core you are selecting for the
characteristics you need - and it is usually more accurate to say
"magnetic material, or magnetic material mix" than iron or ferrite.

Trouble is, I don't know the specs of any of these because I pulled
them out of a PSU.

Like most of us hobbyist types.

Ed has suggested pushing towards an air core. We all breathe the same
air, so you pretty much know what you have there. Have you tried
that? I'm going to.

Jon

 

[Jon Kirwan]

On Sat, 25 Jul 2009 19:08:00 GMT, ehsjr <ehsjr@NOSPAMverizon.net>
wrote:

Jon Kirwan wrote:
On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:

On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher.

No. I the sense of what he said: "the value of R
influences both ON time and OFF time." He refers
to *time*, not Ic.

I know and you pointed this out. So I agree that when we are talking
about time and talking about changing Rbase then it does affect both.
Didn't I just write this? But I also explained where I was coming
from, earlier, too.

You're losing yourself in the details. Put aside all
the damn math - it's just confusing you - and measure
the TIME.

You need to get on the same page as John to have a
go at the math and make it meaningful.
snip

I think I now understand his point, though. And I think I said so,
just a moment ago.

Didn't I?

Jon

 

[Jon Kirwan]

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

Maybe I'm not getting the point, though. Never been trained in this
stuff... just read and do, sometimes.

Jon

 

[fungus]

On Jul 25, 6:23 pm, default <defa...@defaulter.net> wrote:

On Sat, 25 Jul 2009 07:48:45 -0700 (PDT), fungus

openglMYSO...@artlum.com> wrote:
I think we can definitely assume that the ferrite was
saturating or something

Probably "or something."  

Iron powder usually has way more permeability than ferrite - ferrites
claim to fame is it is more permeable than air and able to handle high
frequencies with lower loss.  Some iron powder alloys can go to 100's
of megahertz, but iron is usually used in the kilohertz range.

I just made another one with my thin wire and an iron powder ring.
I was hoping hoping to get a lot of turns on it to see what happened
but unfortunately it behaved like the ferrite - current started
dropping
off.after about ten turns and varying R1 makes no difference to the
output of the circuit.

In any event, when you select a core you are selecting for the
characteristics you need - and it is usually more accurate to say
"magnetic material, or magnetic material mix" than iron or ferrite.

Trouble is, I don't know the specs of any of these because I pulled
them out of a PSU.

 

[John Larkin]

On Sat, 25 Jul 2009 19:41:33 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

To turn a transistor off fast, you have to suck current out of the
base. This circuit doesn't do that. At best is applies a wimpy
negative current into the base at turnoff time.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

The "current source bahavior" is the enemy of speed. One more resistor
can help a lot.

John

 

[Jon Kirwan]

On Thu, 23 Jul 2009 09:24:55 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

That's a horrible circuit. Too many conflicting parameters depend on
the value of R1. A proper blocking oscillator uses an RC time constant
to set the rep rate, and a separate resistor to limit the base
current.

ftp://jjlarkin.lmi.net/BlockOsc.JPG

I still can't make sense of that one. It bugs me a lot because I just
can't see why it would work well.

This one does make more sense to me, though:

,-------------------+----------,
| | |
| | |
| | |
| \ )|
| / R1 )| L2
| \ )|
| / )|o
| | |
| ,----+------+ | D8 ,-------, ,--,
--- | | | +--|>|---+ | | |
- V1 | | | | | --- | ---
--- | | )|o | | \ / D6 | \ / D3
- | | )| | | --- | ---
| | | )| L1 | | | | |
| | | )| | | | | |
| | | | | --- C1 --- | ---
| | | | |/c Q1 --- \ / D5 | \ / D2
| | | +--------| | --- | ---
| \ | | |>e | | | |
| R2 / --- C2 | | | | | |
| \ --- _|_ D7 | | --- | ---
| / | /_\ | | \ / D4 | \ / D1
| | | | | | --- | ---
| | | | | | | | |
| | | | | | '-----' |
gnd gnd gnd gnd gnd gnd gnd

Is that what you were thinking of, instead?

Jon

 

[Jon Kirwan]

On Sat, 25 Jul 2009 13:43:14 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 19:41:33 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

To turn a transistor off fast, you have to suck current out of the
base. This circuit doesn't do that. At best is applies a wimpy
negative current into the base at turnoff time.

Let me try and think about this more closely. Assume for a moment
that we are driving six 3.3V LEDs. Output voltage is 20V. Battery
voltage is three 1.5V's for a total of 4.5V. When the transformer
flies back, there is about (20-4.5+.35) across the collector winding,
or 15.85V... call it 16V. This is reflected back by flux changes to
the base winding as ... 16V to oppose the 4.5V battery. 11V or so,
net. That's a fair push to sweep out charge, isn't it, even with Rb?

Okay. So maybe the point arrives to me, now. What you are saying is
that the base drive current during the ON phase is of a similar size
to the current available during the reverse transit time, which really
needs to be a LOT higher to be much faster. But then I'd argue that
it isn't all that important because _with_ only that current, the OFF
transition is still in the area of the reverse transit time of the BJT
and that with a 2N2222, that's only about 100ns or so. Making that
faster is fine, but if the frequency is only 100kHz or so, it's not
going to make a huge difference.

But then... okay. Maybe I'm not getting the point, at all.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

The "current source bahavior" is the enemy of speed. One more resistor
can help a lot.

To ground from the base?

Jon

 

[John Larkin]

On Sat, 25 Jul 2009 21:10:42 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 13:43:14 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 19:41:33 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

To turn a transistor off fast, you have to suck current out of the
base. This circuit doesn't do that. At best is applies a wimpy
negative current into the base at turnoff time.

Let me try and think about this more closely. Assume for a moment
that we are driving six 3.3V LEDs. Output voltage is 20V. Battery
voltage is three 1.5V's for a total of 4.5V. When the transformer
flies back, there is about (20-4.5+.35) across the collector winding,
or 15.85V... call it 16V. This is reflected back by flux changes to
the base winding as ... 16V to oppose the 4.5V battery. 11V or so,
net. That's a fair push to sweep out charge, isn't it, even with Rb?

Are you assuming a 1:1 transformer? Back-biasing the base by 11 volts
will zener the b-e junction. That degrades the transistor beta over
time.

But it's charge we're trying to pull out of the base, and that takes
current, not just voltage.

Okay. So maybe the point arrives to me, now. What you are saying is
that the base drive current during the ON phase is of a similar size
to the current available during the reverse transit time, which really
needs to be a LOT higher to be much faster. But then I'd argue that
it isn't all that important because _with_ only that current, the OFF
transition is still in the area of the reverse transit time of the BJT
and that with a 2N2222, that's only about 100ns or so. Making that
faster is fine, but if the frequency is only 100kHz or so, it's not
going to make a huge difference.


But then... okay. Maybe I'm not getting the point, at all.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

The "current source bahavior" is the enemy of speed. One more resistor
can help a lot.

To ground from the base?

No. Make a voltage divider from Vbat to ground. Its output voltage
should be +0.8 or so, enough to ensure powerup turn-on. Connect the
feedback winding between that and the base. Scale the turns ratio to
give maybe 2 volts p-p across the feedback winding. Scale the divider
resistors so that the thevenin impedance of the divider results in the
desired turn-on base current. Now we have a low-voltage, low-impedance
base drive circuit. When the transistor is on, most of the drive
current is coming from ground, not Vbatt. And there's plenty of stiff
drive to turn the transistor off.

More complex circuits can do even better.

John

 

[Jon Kirwan]

On Sat, 25 Jul 2009 14:33:16 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 21:10:42 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 13:43:14 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 19:41:33 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

To turn a transistor off fast, you have to suck current out of the
base. This circuit doesn't do that. At best is applies a wimpy
negative current into the base at turnoff time.

Let me try and think about this more closely. Assume for a moment
that we are driving six 3.3V LEDs. Output voltage is 20V. Battery
voltage is three 1.5V's for a total of 4.5V. When the transformer
flies back, there is about (20-4.5+.35) across the collector winding,
or 15.85V... call it 16V. This is reflected back by flux changes to
the base winding as ... 16V to oppose the 4.5V battery. 11V or so,
net. That's a fair push to sweep out charge, isn't it, even with Rb?

Are you assuming a 1:1 transformer? Back-biasing the base by 11 volts
will zener the b-e junction. That degrades the transistor beta over
time.

Oh yes, I know. I've mentioned this issue a number of times in the
thread. It's why I suggested a diode there.

But it's charge we're trying to pull out of the base, and that takes
current, not just voltage.

Agreed. I still need to scope out some cases in practice and see.
That may show me something that will force me to change my current
state of mental theory about it. LTSpice shows me exactly what I
expect from theory, so far, so it's not helping me see something you
are pointing towards.

Okay. So maybe the point arrives to me, now. What you are saying is
that the base drive current during the ON phase is of a similar size
to the current available during the reverse transit time, which really
needs to be a LOT higher to be much faster. But then I'd argue that
it isn't all that important because _with_ only that current, the OFF
transition is still in the area of the reverse transit time of the BJT
and that with a 2N2222, that's only about 100ns or so. Making that
faster is fine, but if the frequency is only 100kHz or so, it's not
going to make a huge difference.

But then... okay. Maybe I'm not getting the point, at all.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

The "current source bahavior" is the enemy of speed. One more resistor
can help a lot.

To ground from the base?

No. Make a voltage divider from Vbat to ground. Its output voltage
should be +0.8 or so, enough to ensure powerup turn-on. Connect the
feedback winding between that and the base. Scale the turns ratio to
give maybe 2 volts p-p across the feedback winding. Scale the divider
resistors so that the thevenin impedance of the divider results in the
desired turn-on base current. Now we have a low-voltage, low-impedance
base drive circuit. When the transistor is on, most of the drive
current is coming from ground, not Vbatt. And there's plenty of stiff
drive to turn the transistor off.

I follow this, I think. I posted something to you, elsewhere, that
may be similar to what you are suggesting here. (I had thought of it
as a correction to your posted GIF, which I still cannot make sense of
and which [to me] seems different from your above paragraph that does
seem to make sense to me.)

More complex circuits can do even better.

Understood.

Thanks for the thoughts and time,
Jon

 

[John Larkin]

On Sat, 25 Jul 2009 21:40:35 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 14:33:16 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 21:10:42 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 13:43:14 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 19:41:33 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 12:01:01 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 25 Jul 2009 18:38:34 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 15:40:36 GMT, ehsjr <ehsjr@NOSPAMverizon.net
wrote:

Jon Kirwan wrote:
On Fri, 24 Jul 2009 08:51:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


snip

At any
rate, the value of R influences both the ON time and the OFF time.

snip

The OFF time is NOT determined by R1, so far as I can tell.

John Larkin is right.

In the sense that Ic_peak is higher. The off time equation is:

t_off = Ic_peak * Lprimary / (Vout + Vschottky - Vbattery)
t_on = Ic_peak * Lprimary / (Vbattery - Vcesat)

I suppose I failed to follow John's point because I was looking at
these as separate processes and the fact that while frequency is
indeed influenced by Rbase, that it isn't important because it is also
influenced by Lprimary and that can be adjusted, as needed.

So I like to look at the design process here as choosing the necessary
Ic_peak from what greg and I were talking about earlier:

N = (Vout+Vd-Vcesat)/(Vin-Vcesat)
Ic_peak = 2 * Iout * (N + 1)

And then using Lprimary by selecting a desired 'f':

Lprimary = (Vin-Vcesat)*(Vout+Vd-Vin)/(Ic_peak*(Vout+Vd-Vcesat)*f)

Rbase isn't selected due to off time, on time, etc., but instead based
upon the BJT and Ic_peak:

Rbase = 2*beta*(Vin-Vbeon)/Ic_peak

That's the thing I don't like about this circuit, the large value of
Rb, which wastes power and seriously slows down transistor switching
time, wasting more power. All this gets worse as Vin increases.

I'll look at that aspect. But on first glance it doesn't seem that
much. For example, at 2mA during the ON time and figuring 4.5k,
that's 18mW even assuming that the ON time is 100% of the total, which
it isn't. Meanwhile, the circuit is delivering more than 600mW to the
load, according to a spice simulation here. Frequency is 100kHz.

Transistor turnoff is especially ugly.

The reverse transit time seems a significant problem as the frequency
goes up.

To turn a transistor off fast, you have to suck current out of the
base. This circuit doesn't do that. At best is applies a wimpy
negative current into the base at turnoff time.

Let me try and think about this more closely. Assume for a moment
that we are driving six 3.3V LEDs. Output voltage is 20V. Battery
voltage is three 1.5V's for a total of 4.5V. When the transformer
flies back, there is about (20-4.5+.35) across the collector winding,
or 15.85V... call it 16V. This is reflected back by flux changes to
the base winding as ... 16V to oppose the 4.5V battery. 11V or so,
net. That's a fair push to sweep out charge, isn't it, even with Rb?

Are you assuming a 1:1 transformer? Back-biasing the base by 11 volts
will zener the b-e junction. That degrades the transistor beta over
time.

Oh yes, I know. I've mentioned this issue a number of times in the
thread. It's why I suggested a diode there.

A diode in series with the base drive circuit? Then there's zero base
turnoff current. Transistor storage and Miller effect will really slow
down turnoff and increase losses.

But it's charge we're trying to pull out of the base, and that takes
current, not just voltage.

Agreed. I still need to scope out some cases in practice and see.
That may show me something that will force me to change my current
state of mental theory about it. LTSpice shows me exactly what I
expect from theory, so far, so it's not helping me see something you
are pointing towards.

Okay. So maybe the point arrives to me, now. What you are saying is
that the base drive current during the ON phase is of a similar size
to the current available during the reverse transit time, which really
needs to be a LOT higher to be much faster. But then I'd argue that
it isn't all that important because _with_ only that current, the OFF
transition is still in the area of the reverse transit time of the BJT
and that with a 2N2222, that's only about 100ns or so. Making that
faster is fine, but if the frequency is only 100kHz or so, it's not
going to make a huge difference.

But then... okay. Maybe I'm not getting the point, at all.

The base only needs 0.7 volts or so; it's silly to get that from a 5
volt supply, or more like 10 when the transformer feedback winding
signal is added.

Actually, that helps, doesn't it? First, doubling the voltage means
the battery can be usefully used down to lower values. Second, a
higher voltage implies more of a current source behavior via Rb, as
Vbe droop variations are a lower percent of the total.

The "current source bahavior" is the enemy of speed. One more resistor
can help a lot.

To ground from the base?

No. Make a voltage divider from Vbat to ground. Its output voltage
should be +0.8 or so, enough to ensure powerup turn-on. Connect the
feedback winding between that and the base. Scale the turns ratio to
give maybe 2 volts p-p across the feedback winding. Scale the divider
resistors so that the thevenin impedance of the divider results in the
desired turn-on base current. Now we have a low-voltage, low-impedance
base drive circuit. When the transistor is on, most of the drive
current is coming from ground, not Vbatt. And there's plenty of stiff
drive to turn the transistor off.

I follow this, I think. I posted something to you, elsewhere, that
may be similar to what you are suggesting here. (I had thought of it
as a correction to your posted GIF, which I still cannot make sense of
and which [to me] seems different from your above paragraph that does
seem to make sense to me.)

The other schematic I posted was a classic discontinuous blocking
oscillator, which allows the duty cycle to be tuned. But it still
benefits from a high transformer ratio and a low-impedance base drive.

Mosfets are simpler to drive... only a tiny amount of Vbatt current is
needed to get them started, and no current limiting is needed.

John

 

[David Eather]

fungus wrote:

On Jul 25, 3:14 am, David Eather <eat...@tpg.com.au> wrote:
Aside: Does wire thickness make any difference?
It won't make much difference at these low power levels. The main
difference in this case is how many turns you can wrap on the ferrite bead.


So the size of the magnetic field is purely down to current?

Not only current, also the number of turns and the type of former the
turns are on etc

But all else being equal more current equals a stronger magnetic field

==================

I've just another experiment and I've got some more fuel for
the raging debate....

Oh, no....

I just did a comparison between a ferrite bead and an iron
powder bead. With the ferrite bead the current through the
LEDs drops off as the number of turns of wire increases.
With seven turns I get 12mA ... with 30 turns I only get 1mA.

For comparison I just tried one of my iron powder rings
and I got completely the *opposite* effect - more turns
gave more output current. At 15 turns I was getting 6mA,
at 30 turns I was getting 12mA.

I was doing this with a thicker piece of wire I pulled from a
transformer in the PSU* so luckily for the transistor I couldn't
physically get more than 30 turns on the ring. The trend was
very clear though - every turn I added produced a measurable
increase in LED current.

nb. The transistor was getting hotter with every extra turn
despite the oscillation frequency going down, also the opposite
of what happens with ferrite. It seems that transistor temperature
is more strongly related to output current than frequency.

Assuming I get the same result with thin wire this seems like
a really easy way to get any desired output current - just keep
adding turns until you get there.

This also assuming we can solve the transistor heating problem.
What would happen if I put two transistors in parallel? Would the
load be halved or would differences in manufacturing tolerance
mean one of them took most of the load?

Exactly correct! manufacturing tolerance would mean one transistor would
take most of the load (and as the transistor gets hotter it will take an
even greater percentage)

You would need to use a couple of small resistors to make sure the
transistors shared the current from L2 and the current into the base
equally.

PS: FWIW I measured the efficiency of this new circuit and it
was 61% - a bit better that the 55% I get with a ferrite circuit
with has similar output


[*] PC PSUs are a real goldmine of parts if you really get in
there....

 

[David Eather]

fungus wrote:

On Jul 25, 3:14 am, David Eather <eat...@tpg.com.au> wrote:
Aside: Does wire thickness make any difference?
It won't make much difference at these low power levels. The main
difference in this case is how many turns you can wrap on the ferrite bead.


So the size of the magnetic field is purely down to current?

==================

I've just another experiment and I've got some more fuel for
the raging debate....

I just did a comparison between a ferrite bead and an iron
powder bead. With the ferrite bead the current through the
LEDs drops off as the number of turns of wire increases.
With seven turns I get 12mA ... with 30 turns I only get 1mA.

For comparison I just tried one of my iron powder rings
and I got completely the *opposite* effect - more turns
gave more output current. At 15 turns I was getting 6mA,
at 30 turns I was getting 12mA.

I was doing this with a thicker piece of wire I pulled from a
transformer in the PSU* so luckily for the transistor I couldn't
physically get more than 30 turns on the ring. The trend was
very clear though - every turn I added produced a measurable
increase in LED current.

nb. The transistor was getting hotter with every extra turn
despite the oscillation frequency going down, also the opposite
of what happens with ferrite. It seems that transistor temperature
is more strongly related to output current than frequency.

Assuming I get the same result with thin wire this seems like
a really easy way to get any desired output current - just keep
adding turns until you get there.

This also assuming we can solve the transistor heating problem.
What would happen if I put two transistors in parallel? Would the
load be halved or would differences in manufacturing tolerance
mean one of them took most of the load?

It shouldn't be necessary the 2n22222 is quite a robust transistor

PS: FWIW I measured the efficiency of this new circuit and it
was 61% - a bit better that the 55% I get with a ferrite circuit
with has similar output


[*] PC PSUs are a real goldmine of parts if you really get in
there....

 

[Jon Kirwan]

On Sat, 25 Jul 2009 14:48:42 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

snip
A diode in series with the base drive circuit?
snip

What??? No!! I may be a hobbyist, but...

Look at some of the schematics I've posted.

Jon

 

[David Eather]

greg wrote:

David Eather wrote:

In your circuit you could also try removing C1 - the LED's will light
just fine with the JT pulses and your eye will make the circuit appear
brighter for the same current

No, it won't. An LED run continuously at 20mA will always
look brighter than one pulsed at 20mA.

There is no current limiting in the LED circuit, it was reasonable to
expect current peaks higher than 20ma. In that case, while the average
current was 20ma, the higher current peaks might mean the output would
be brighter without C1.

It was worth a try.

snip

 

[John Larkin]

On Sun, 26 Jul 2009 00:01:55 GMT, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 14:48:42 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

snip
A diode in series with the base drive circuit?
snip

What??? No!! I may be a hobbyist, but...

Look at some of the schematics I've posted.

Jon

But if you use it to clamp the negative swing at the base, it
increases the power dissipation in R and still doesn't speed up
turnoff.

Transistors are low-impedance/small-swing input,
high-impedance/large-swing output devices. The transformer windings
should correspond. Why not?

John

 

[Jon Kirwan]

On Sat, 25 Jul 2009 20:36:25 +1200, greg <greg@cosc.canterbury.ac.nz>
wrote:

David Eather wrote:

In your circuit you could also try removing C1 - the LED's will light
just fine with the JT pulses and your eye will make the circuit appear
brighter for the same current

No, it won't. An LED run continuously at 20mA will always
look brighter than one pulsed at 20mA.

Pulsing LEDs only helps if the average current available
from your supply is much less than the rated current of
the LEDs. E.g. if you can only manage 2mA continuously,
then 20mA at 10% duty cycle will give slightly more
light overall, due to nonlinearity of the LED's light
vs. current curve.
snip

HP put out an extensive treatise on LEDs and includes a page or two on
this point, in particular. It's called "Optoelectronics: Fiber-Optics
Applications Manual," 2nd edition, 1981. It was hard-bound and cost
$27.50 when new or else was given away by HP, if they liked you.

They provide a curve on page 5.20 (for high efficiency red LEDs) that
shows a step climb from 1 (100%) at 5mA peak current (which I assume
is their benchmark continuous current, since the curve doesn't go to
lower peak currents) up towards about 1.6 where it flattens out and
doesn't change much, at about 40mA peak current. They hold the time
averaged current fixed for the curve, so at 40mA peak current this
implies that the duty cycle is 12.5%. (They also take a moment to say
that all this only works before the junction reaches saturation. After
that, additional current will obviously reduce this efficiency ratio.)

This suggests to me that there is a limit to the advantage in
increasing pulsed currents. At some point, gains aren't to be had.
But also, HP's curve of efficiency is focused on current. It doesn't
take respect of the fact that the voltage also increases, as well. So
although efficiency vs pulsed current _in terms of current_ does rise
up to a point, _in terms of power_ in the device the curve would
likely be different as the drive voltage increases with higher pulsed
currents. I'm betting that this creates a kind of parabolic shape
with a peak, once voltage is added back into the mix.

Another aspect though is that (using this particular curve) the value
for 20mA is 1.43 while the value for 50mA is 1.61. The ratio of
these, 1.61/1.43 = 1.13, suggests the relative value gained by pulsing
2.5 times the current with a 40% duty cycle. Not a lot for the work
involved. (All this is in candelas, so I'm assuming this is relative
to human perception -- actual radiant intensity ratios may be somewhat
different.) And if the voltage drive were also taken into account
(let's say, 1/3rd of a volt more so that it is 3 1/3 V and 3 2/3 V in
the two cases), then this works out to about 1.1 times more _power_,
which is might darned close to the 1.13 relative efficiency calc. Or,
in other words, from a battery's perspective... almost no gain at all.

Does anyone else have this book, by the way? It's pretty thorough, if
not entirely modern. And if there is another to replace it, with
newer technologies and covering as much in one volume, I'd like to
know about it.

Jon

 

[Jon Kirwan]

On Sat, 25 Jul 2009 17:20:40 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sun, 26 Jul 2009 00:01:55 GMT, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Sat, 25 Jul 2009 14:48:42 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

snip
A diode in series with the base drive circuit?
snip

What??? No!! I may be a hobbyist, but...

Look at some of the schematics I've posted.

Jon

But if you use it to clamp the negative swing at the base, it
increases the power dissipation in R and still doesn't speed up
turnoff.

Agreed. But it does protect the poor BJT, which in the big priority
scheme of important things to worry about comes just a little bit
ahead of worrying over power dissipation. When jacking things up with
a higher Vin (more batteries), the whole thing goes bad real quick.
The diode makes it just a bit more idiot proof.

Transistors are low-impedance/small-swing input,
high-impedance/large-swing output devices. The transformer windings
should correspond. Why not?

I agree. It's just that it is very nice for a hobbyist to use just a
few parts they can grasp without too much difficulty and get something
neat working out of it. It's actually kind of nice to just tweak up
Ic_peak by adjusting Rb, then wrap or unwrap wires on the transformer
until the frequency is about right. Anyone can understand that much
and just go with it. And without a well-equipped lab, soldering more
wires and parts and placing all of them into something is... well...
more work. Less is more.

On the other hand, I'm very glad you brought up some of this because
I'm being forced to think more and learn something in the process. So
I'm happy you mention such things. I'm studying the new circuit I
posted at you, earlier today.

Jon

 

[fungus]

On Jul 25, 11:33 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

Are you assuming a 1:1 transformer? Back-biasing the base by 11 volts
will zener the b-e junction. That degrades the transistor beta over
time.

Yep, that's why D7 was added to the circuit.

 

[fungus]

On Jul 26, 2:35 am, fungus <openglMYSO...@artlum.com> wrote:

On Jul 25, 6:10 pm, default <defa...@defaulter.net> wrote:



That looks way too "clean." What is the bandwidth of your scope?

Probably not very high - it's 1970's tech, no digital storage or
dual beams or any of that stuff.

I think a lot of fine detail is also being lost in the photo because
the camera's a bit out of focus.