No. I completely understand that behavior and it's NOT
Looking back on notes I wrote a long time ago, I find
On Sun, 14 Jul 2013 20:16:42 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 14 Jul 2013 15:01:25 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Sun, 14 Jul 2013 12:05:32 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 14 Jul 2013 11:10:54 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Sun, 14 Jul 2013 08:56:02 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sat, 13 Jul 2013 23:15:32 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Sat, 13 Jul 2013 15:11:40 -0500, John Fields
jfields@austininstruments.com> wrote:
snip
http://www.edn.com/design/power-management/4324682/Current-mirror-drives-multiple-LEDs-from-a-low-supply-voltage
snip
Nice link. Using a diff-amp in a controlled feedback loop.
Thanks.
It wastes 25% of the supply power just to sense the LED current.
A current mirror wastes current. That's what they do.
And it depends on transistor matching.
All current mirrors depend on matching Is, less with
degeneration. From my limited experience, this isn't a big
problem for LED driving if using parts from the same reel.
And it's pretty easy to test for, anyway.
I don't see how it avoids thermal runaway
by pulsing.
Yeah. If one of them heats up more than others, it's Vbe
declines by 2.3mV/C or something like that and this pulls the
others down with it, lowering their Ic. And if the feedback
BJT lowers its Ic the feedback circuit supplies more drive
current... and so on. I've epoxied parts together with
modestly okay effect here. But I've also bought arrays on
ICs, too (which at the prices I pay aren't exactly matched
well but again aren't terrible, either.)
Degeneration would seem to be a partial answer here, too.
A point in this circuit is low overhead drivers. What would
you do if designing a discrete, bipolar solution here? I'd
learn something from a good answer.
Usual EDN standard of quality. They need filler between the ads.
I do NOT like the capacitor there. When inactive (OFF), C1
charges up to Vcc across it, if C1 is small enough (if C1 is
big, then it never reaches Vcc across it.) When active (ON),
the common node between C1 and R4 jumps up towards 1.2V below
Vcc and in doing so causes the collector of Q2 to be driven
below ground, Vcc-1.2V-Vc1=-1.2V since Vc1=Vcc. As C1
discharges it actually borrows current from Q3's collector
(rather than from R3) causing a lower current in R3, yielding
a lower voltage potential for Q2's base... leading to
diversion of R2's current towards Q1 and thus MORE drive
current into Q4 through Q7. The upshot is that upon turn-on,
the peak current in the LEDs is actually HIGHER than designed
and it then gently settles down to the desired level as C1
discharges to a new (lower) potential across its leads. This
is spikey behavior instead of a "soft start." Make C1 bigger
and it gets worse, too. The sharp leading edge will reach
even higher initial currents in the LEDs that way.
So it's value should be kept small. I'm not yet convinced
about the danger of oscillations or the source of them in
this circuit. There are very low impedances on every node of
Q2. It just doesn't seem necessary to me. I don't see how the
small Cbc value and the value of Rb' can conspire to make a
problem here, even with the collector making a small jump at
turn on. And it actually seems to have a downside where I'd
rather simply remove it.
I might also invert the whole structure and delete R1 and D1,
if my LED supply was a volt or more above the micro rail
voltage. The pin output won't be loaded much and will
probably be a very close reflection of the micro rail
voltage. Of course, I'd adjust values, accordingly.
For example, if my micro were running on 3.6V and my LED
supply were 6V (hypothetically) and I wanted 100mA and 5
chains of them (not 4), then I might invert the polarities
and use R3=36 ohms, R2=470 ohms, and R4=270 ohms, kill C1,
and just drive the base of Q1 directly.
I'd start the design by estimating the total base currents
required for Q3 through Q7 (600mA/beta=200, or about 3mA.)
Then I'd double that for the diff-amp pair to 6mA. Given 3.6V
drive and an estimated 0.7V drop on Q1, I get 2.9V at Q1's
emitter. So 2.9V/6mA is 483 ohms. 470 is a standard value and
gives a slight increase in estimated current, so I'd use that
value instead. I also want 3.6V across R3 at 100mA, so 36
ohms. Changing that to 33 or 39 ohms would change the LED
currents by 10% or more. Luckily, 5% resistors do come in 36
ohms. So I'd stick with that. Then, since R4 should be
(assuming 3mA in each branch, hopefully) at about one 100mA
sized Vbe (about 110mV added to 700mV as a guess), so R4
should be about 0.81/3mA, or 270 ohms. Nice. Standard value.
So that's where I'd get my values.
I'm still stuck trying to understand C1's benefit.
It probably keeps the loop from oscillating.
That's what they say. But I still don't see it, probably
because I'm blind and not necessarily because it doesn't do
that. It's why I said I can't see it -- hoping someone would
explain it in detail to me. If you can see the exact
mechanism here, I'd like to see a detailed explanation.
Jon
https://dl.dropboxusercontent.com/u/53724080/Circuits/Current_Sources/CC_Driver.JPG
Not an explanation of the oscillation issue, which is what I
seek.
Jon
It's all about control theory. Some closed loops oscillate, some don't.
A high-gain loop with two active elements, each with similar frequency rolloffs,
is a prime candidate for oscillation. The added cap radically slows down one of
those two elements.
I'm looking for the mathematical explanation -- quantitative
-- as well as the theoretical underpinning. Can you elucidate
that?
Jon
I have all of the mathematics required (which just happens to
I'm fairly well informed about the models. I cannot say I
I understand poles and zeros -- very simple thing
Just develop the simple poles-zeros equation for me and show
---
JF
---On Sun, 14 Jul 2013 08:56:02 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sat, 13 Jul 2013 23:15:32 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Sat, 13 Jul 2013 15:11:40 -0500, John Fields
jfields@austininstruments.com> wrote:
snip
http://www.edn.com/design/power-management/4324682/Current-mirror-drives-multiple-LEDs-from-a-low-supply-voltage
snip
Nice link. Using a diff-amp in a controlled feedback loop.
Thanks.
It wastes 25% of the supply power just to sense the LED current.
A current mirror wastes current. That's what they do.
And it depends on transistor matching.
All current mirrors depend on matching Is, less with
degeneration. From my limited experience, this isn't a big
problem for LED driving if using parts from the same reel.
And it's pretty easy to test for, anyway.
I don't see how it avoids thermal runaway
by pulsing.
Yeah. If one of them heats up more than others, it's Vbe
declines by 2.3mV/C or something like that and this pulls the
others down with it, lowering their Ic. And if the feedback
BJT lowers its Ic the feedback circuit supplies more drive
current... and so on. I've epoxied parts together with
modestly okay effect here. But I've also bought arrays on
ICs, too (which at the prices I pay aren't exactly matched
well but again aren't terrible, either.)
Degeneration would seem to be a partial answer here, too.
A point in this circuit is low overhead drivers. What would
you do if designing a discrete, bipolar solution here? I'd
learn something from a good answer.
Usual EDN standard of quality. They need filler between the ads.
I do NOT like the capacitor there. When inactive (OFF), C1
charges up to Vcc across it, if C1 is small enough (if C1 is
big, then it never reaches Vcc across it.) When active (ON),
the common node between C1 and R4 jumps up towards 1.2V below
Vcc and in doing so causes the collector of Q2 to be driven
below ground, Vcc-1.2V-Vc1=-1.2V since Vc1=Vcc. As C1
discharges it actually borrows current from Q3's collector
(rather than from R3) causing a lower current in R3, yielding
a lower voltage potential for Q2's base... leading to
diversion of R2's current towards Q1 and thus MORE drive
current into Q4 through Q7. The upshot is that upon turn-on,
the peak current in the LEDs is actually HIGHER than designed
and it then gently settles down to the desired level as C1
discharges to a new (lower) potential across its leads. This
is spikey behavior instead of a "soft start." Make C1 bigger
and it gets worse, too. The sharp leading edge will reach
even higher initial currents in the LEDs that way.
So it's value should be kept small. I'm not yet convinced
about the danger of oscillations or the source of them in
this circuit. There are very low impedances on every node of
Q2. It just doesn't seem necessary to me. I don't see how the
small Cbc value and the value of Rb' can conspire to make a
problem here, even with the collector making a small jump at
turn on. And it actually seems to have a downside where I'd
rather simply remove it.
I might also invert the whole structure and delete R1 and D1,
if my LED supply was a volt or more above the micro rail
voltage. The pin output won't be loaded much and will
probably be a very close reflection of the micro rail
voltage. Of course, I'd adjust values, accordingly.
For example, if my micro were running on 3.6V and my LED
supply were 6V (hypothetically) and I wanted 100mA and 5
chains of them (not 4), then I might invert the polarities
and use R3=36 ohms, R2=470 ohms, and R4=270 ohms, kill C1,
and just drive the base of Q1 directly.
I'd start the design by estimating the total base currents
required for Q3 through Q7 (600mA/beta=200, or about 3mA.)
Then I'd double that for the diff-amp pair to 6mA. Given 3.6V
drive and an estimated 0.7V drop on Q1, I get 2.9V at Q1's
emitter. So 2.9V/6mA is 483 ohms. 470 is a standard value and
gives a slight increase in estimated current, so I'd use that
value instead. I also want 3.6V across R3 at 100mA, so 36
ohms. Changing that to 33 or 39 ohms would change the LED
currents by 10% or more. Luckily, 5% resistors do come in 36
ohms. So I'd stick with that. Then, since R4 should be
(assuming 3mA in each branch, hopefully) at about one 100mA
sized Vbe (about 110mV added to 700mV as a guess), so R4
should be about 0.81/3mA, or 270 ohms. Nice. Standard value.
So that's where I'd get my values.
I'm still stuck trying to understand C1's benefit.
It probably keeps the loop from oscillating.
That's what they say. But I still don't see it, probably
because I'm blind and not necessarily because it doesn't do
that. It's why I said I can't see it -- hoping someone would
explain it in detail to me. If you can see the exact
mechanism here, I'd like to see a detailed explanation.
Jon
Sure, that is an assumption one can make. But it doesn't helpOn Sun, 14 Jul 2013 11:10:54 -0700, Jon Kirwan
jonk@infinitefactors.org> wrote:
On Sun, 14 Jul 2013 08:56:02 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sat, 13 Jul 2013 23:15:32 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Sat, 13 Jul 2013 15:11:40 -0500, John Fields
jfields@austininstruments.com> wrote:
snip
http://www.edn.com/design/power-management/4324682/Current-mirror-drives-multiple-LEDs-from-a-low-supply-voltage
snip
Nice link. Using a diff-amp in a controlled feedback loop.
Thanks.
It wastes 25% of the supply power just to sense the LED current.
A current mirror wastes current. That's what they do.
And it depends on transistor matching.
All current mirrors depend on matching Is, less with
degeneration. From my limited experience, this isn't a big
problem for LED driving if using parts from the same reel.
And it's pretty easy to test for, anyway.
I don't see how it avoids thermal runaway
by pulsing.
Yeah. If one of them heats up more than others, it's Vbe
declines by 2.3mV/C or something like that and this pulls the
others down with it, lowering their Ic. And if the feedback
BJT lowers its Ic the feedback circuit supplies more drive
current... and so on. I've epoxied parts together with
modestly okay effect here. But I've also bought arrays on
ICs, too (which at the prices I pay aren't exactly matched
well but again aren't terrible, either.)
Degeneration would seem to be a partial answer here, too.
A point in this circuit is low overhead drivers. What would
you do if designing a discrete, bipolar solution here? I'd
learn something from a good answer.
Usual EDN standard of quality. They need filler between the ads.
I do NOT like the capacitor there. When inactive (OFF), C1
charges up to Vcc across it, if C1 is small enough (if C1 is
big, then it never reaches Vcc across it.) When active (ON),
the common node between C1 and R4 jumps up towards 1.2V below
Vcc and in doing so causes the collector of Q2 to be driven
below ground, Vcc-1.2V-Vc1=-1.2V since Vc1=Vcc. As C1
discharges it actually borrows current from Q3's collector
(rather than from R3) causing a lower current in R3, yielding
a lower voltage potential for Q2's base... leading to
diversion of R2's current towards Q1 and thus MORE drive
current into Q4 through Q7. The upshot is that upon turn-on,
the peak current in the LEDs is actually HIGHER than designed
and it then gently settles down to the desired level as C1
discharges to a new (lower) potential across its leads. This
is spikey behavior instead of a "soft start." Make C1 bigger
and it gets worse, too. The sharp leading edge will reach
even higher initial currents in the LEDs that way.
So it's value should be kept small. I'm not yet convinced
about the danger of oscillations or the source of them in
this circuit. There are very low impedances on every node of
Q2. It just doesn't seem necessary to me. I don't see how the
small Cbc value and the value of Rb' can conspire to make a
problem here, even with the collector making a small jump at
turn on. And it actually seems to have a downside where I'd
rather simply remove it.
I might also invert the whole structure and delete R1 and D1,
if my LED supply was a volt or more above the micro rail
voltage. The pin output won't be loaded much and will
probably be a very close reflection of the micro rail
voltage. Of course, I'd adjust values, accordingly.
For example, if my micro were running on 3.6V and my LED
supply were 6V (hypothetically) and I wanted 100mA and 5
chains of them (not 4), then I might invert the polarities
and use R3=36 ohms, R2=470 ohms, and R4=270 ohms, kill C1,
and just drive the base of Q1 directly.
I'd start the design by estimating the total base currents
required for Q3 through Q7 (600mA/beta=200, or about 3mA.)
Then I'd double that for the diff-amp pair to 6mA. Given 3.6V
drive and an estimated 0.7V drop on Q1, I get 2.9V at Q1's
emitter. So 2.9V/6mA is 483 ohms. 470 is a standard value and
gives a slight increase in estimated current, so I'd use that
value instead. I also want 3.6V across R3 at 100mA, so 36
ohms. Changing that to 33 or 39 ohms would change the LED
currents by 10% or more. Luckily, 5% resistors do come in 36
ohms. So I'd stick with that. Then, since R4 should be
(assuming 3mA in each branch, hopefully) at about one 100mA
sized Vbe (about 110mV added to 700mV as a guess), so R4
should be about 0.81/3mA, or 270 ohms. Nice. Standard value.
So that's where I'd get my values.
I'm still stuck trying to understand C1's benefit.
It probably keeps the loop from oscillating.
That's what they say. But I still don't see it, probably
because I'm blind and not necessarily because it doesn't do
that. It's why I said I can't see it -- hoping someone would
explain it in detail to me. If you can see the exact
mechanism here, I'd like to see a detailed explanation.
Jon
---
My guess would be that when he built it he found ringing at turn-on
and turn-off which he cured by connecting a cap across various points
in the circuit until the ringing went away.
What I told you is that loop analysis of a thing like this is far tooOn Mon, 15 Jul 2013 19:18:14 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 14 Jul 2013 11:10:54 -0700, Jon Kirwan
jonk@infinitefactors.org> wrote:
On Sun, 14 Jul 2013 08:56:02 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sat, 13 Jul 2013 23:15:32 -0700, Jon Kirwan <jonk@infinitefactors.org> wrote:
On Sat, 13 Jul 2013 15:11:40 -0500, John Fields
jfields@austininstruments.com> wrote:
snip
http://www.edn.com/design/power-management/4324682/Current-mirror-drives-multiple-LEDs-from-a-low-supply-voltage
snip
Nice link. Using a diff-amp in a controlled feedback loop.
Thanks.
It wastes 25% of the supply power just to sense the LED current.
A current mirror wastes current. That's what they do.
And it depends on transistor matching.
All current mirrors depend on matching Is, less with
degeneration. From my limited experience, this isn't a big
problem for LED driving if using parts from the same reel.
And it's pretty easy to test for, anyway.
I don't see how it avoids thermal runaway
by pulsing.
Yeah. If one of them heats up more than others, it's Vbe
declines by 2.3mV/C or something like that and this pulls the
others down with it, lowering their Ic. And if the feedback
BJT lowers its Ic the feedback circuit supplies more drive
current... and so on. I've epoxied parts together with
modestly okay effect here. But I've also bought arrays on
ICs, too (which at the prices I pay aren't exactly matched
well but again aren't terrible, either.)
Degeneration would seem to be a partial answer here, too.
A point in this circuit is low overhead drivers. What would
you do if designing a discrete, bipolar solution here? I'd
learn something from a good answer.
Usual EDN standard of quality. They need filler between the ads.
I do NOT like the capacitor there. When inactive (OFF), C1
charges up to Vcc across it, if C1 is small enough (if C1 is
big, then it never reaches Vcc across it.) When active (ON),
the common node between C1 and R4 jumps up towards 1.2V below
Vcc and in doing so causes the collector of Q2 to be driven
below ground, Vcc-1.2V-Vc1=-1.2V since Vc1=Vcc. As C1
discharges it actually borrows current from Q3's collector
(rather than from R3) causing a lower current in R3, yielding
a lower voltage potential for Q2's base... leading to
diversion of R2's current towards Q1 and thus MORE drive
current into Q4 through Q7. The upshot is that upon turn-on,
the peak current in the LEDs is actually HIGHER than designed
and it then gently settles down to the desired level as C1
discharges to a new (lower) potential across its leads. This
is spikey behavior instead of a "soft start." Make C1 bigger
and it gets worse, too. The sharp leading edge will reach
even higher initial currents in the LEDs that way.
So it's value should be kept small. I'm not yet convinced
about the danger of oscillations or the source of them in
this circuit. There are very low impedances on every node of
Q2. It just doesn't seem necessary to me. I don't see how the
small Cbc value and the value of Rb' can conspire to make a
problem here, even with the collector making a small jump at
turn on. And it actually seems to have a downside where I'd
rather simply remove it.
I might also invert the whole structure and delete R1 and D1,
if my LED supply was a volt or more above the micro rail
voltage. The pin output won't be loaded much and will
probably be a very close reflection of the micro rail
voltage. Of course, I'd adjust values, accordingly.
For example, if my micro were running on 3.6V and my LED
supply were 6V (hypothetically) and I wanted 100mA and 5
chains of them (not 4), then I might invert the polarities
and use R3=36 ohms, R2=470 ohms, and R4=270 ohms, kill C1,
and just drive the base of Q1 directly.
I'd start the design by estimating the total base currents
required for Q3 through Q7 (600mA/beta=200, or about 3mA.)
Then I'd double that for the diff-amp pair to 6mA. Given 3.6V
drive and an estimated 0.7V drop on Q1, I get 2.9V at Q1's
emitter. So 2.9V/6mA is 483 ohms. 470 is a standard value and
gives a slight increase in estimated current, so I'd use that
value instead. I also want 3.6V across R3 at 100mA, so 36
ohms. Changing that to 33 or 39 ohms would change the LED
currents by 10% or more. Luckily, 5% resistors do come in 36
ohms. So I'd stick with that. Then, since R4 should be
(assuming 3mA in each branch, hopefully) at about one 100mA
sized Vbe (about 110mV added to 700mV as a guess), so R4
should be about 0.81/3mA, or 270 ohms. Nice. Standard value.
So that's where I'd get my values.
I'm still stuck trying to understand C1's benefit.
It probably keeps the loop from oscillating.
That's what they say. But I still don't see it, probably
because I'm blind and not necessarily because it doesn't do
that. It's why I said I can't see it -- hoping someone would
explain it in detail to me. If you can see the exact
mechanism here, I'd like to see a detailed explanation.
Jon
---
My guess would be that when he built it he found ringing at turn-on
and turn-off which he cured by connecting a cap across various points
in the circuit until the ringing went away.
Sure, that is an assumption one can make. But it doesn't help
me learn anything at all.
I already explained, in detail, what I can SEE that C1
actually does do. It makes the circuit WORSE, from one
perspective. I will CAUSE a higher than normal leading edge
current peak. And I completely understand exactly WHY it does
that. I can actually sit down and easily compute the time
constant and the peak excess that will occur for various
component values, even, and get both results right (to the
degree that Spice agrees, anyway.) So I do understand one
thing that C1 does do and I frankly don't like it.
The problem for me remains, in that I still don't see the
"other problem" nor why C1 fixes it. (It's hard to see how
something gets fixed if you can't even see the problem in the
first place, I suppose.)
Anyway, I want to be able to (1) see the problem, (2) COMPUTE
the problem, quantitatively, (3) recognize cases where the
problem is significant enough to be worth fixing, and (4)
understand the effect C1 has on resolving that problem in the
circumstances where it is a significant issue. I can follow a
good argument, if someone would just make one.
So far, I'm clueless. And so far, John Larkin hasn't done
anything to educate me. I'm not complaining. He owes me
nothing. But he has wasted his valuable time on me, in
effect, because what he has written is worse than just vague.
....
I'm certain that if I take this to a professor at a
I have simulated it. But one doesn't (and shouldn't) learn
It was completely useless ... to me.
Yes. But this really isn't appropriate for the design group.
That point is right, as far as the linked circuit goes and
Speaking of which, any thoughts on this issue:
I would plan to design the power supply to provide about 1.5V
I know guys who are working on PhDs in control theory. It's usually a junior or
I have done serious LaPlace type math for control loops, but not lately. Simple
No argument there.
It makes sense to me, and I use the concept often, in real designs. You can
How about an analog switch and an opamp per transistor? Opamp sections cost,
Why not opamps?
The fundamental answer is because I want to learn how to do
Here's the best I can do with my meager imagination, John.
It's basically a diff-amp driven through a level shifter.
Sorry, a 2N7002 costs about 2 cents, about the same as a BCX70.
I have some, of course. Cheapest mosfet I could find.
Perhaps. But I buy PN2222A for less than 0.4 cents each. And
Okay. So...
The +0.2 could be a shared rail. Current ewquirement is low.
No. It should work.
With your attitude, you don't deserve much.
Explain why.
Saturated BJT in the diff-amp?
(The base drive will have to be significant, relative to the
