how to get ground referenced amplified V difference from high side(250VDC) shunt?

[John Larkin]

You don't what you're talking about. Deconvolution no more manufactures
lost bits than pre-emphasis 'cheats' line loss. Deconvolving a linear
circuit is the simplest possible use.

What do you propose to deconvolve in this situation? If you lowpass
the data, you just kill bandwidth, which is hard to come by here. If
you try to extend the bandwidth, you magnify the noise... a linear
transfer function can't create precision where it doesn't exist.

Deconvolution is one of the family of "ill-posed problems", in the
sense that, given an imperfect signal A, and an ideal signal B, what
is the transfer function T that satisfies

B = A ** T

where ** is convolution.

Forests have died making papers about this problem.

The obvious Fourier solution (f.transform A and B, divide fB/fA,
reverse transform) tends to have singularities (ie, zero lines in A)
or bad noise problems (near-zero ditto). This ain't turf for the faint
of heart, or wikipedia bs artists, either.

I've developed an iterative time-domain method, and it's fun to play
with. A little iterating is pretty good, but if you get greedy and try
to equalize too well, namely insist on making B perfect, it blows up.

http://s2.supload.com/free/demo.jpg/view/


What sort of work have you done in deconvolution?

John

 

[Michael]

 Actually, my circuit above is a short-changing cheat!
 It has no standing bias for the critical transistor,
 Q1, and hence cannot perform to the desired bandwidth
 when Iout is near 0mA.  Adding a bias like the original
 non-opamp BJT current-mirror circuit, boosts the parts
 count to 15, exactly the same as the all-BJT circuit.

 So it's a tie.  Go to the trouble of finding a suitable
 low-current high-side opamp and get better zero-current
 accuracy without trimming.  Or something like that.

   high-voltage fast high-side current monitor

                   10.0       +250V    Iout
  ------+------+---/\/\----+----o ---> 0 to 10mA load
        |      |           |
        /      '--+--------|---,
        \ 200     |        |   |
        /  1%     |    200 /   |
        |         |     1% \   |
        +---------|----,   /   +---,
        |        _|_   |   |   |   | 7.5V zener
    Q1  e       /  -|--'   |  _|_/ |
 mpsa92   b ---<    |      |  /_\ _|_
   PNP  c       \__+|------+   |  --- 0.1uF
        | LT1783  |        |   |   |
        |         '--------|---+---+---,
        /                  |          |
        \ 1k          1.0M |           /  330k
        /                  /          \ 0.5W
        | Ib = 0.25        \ Ia        /
        | to 0.75mA        /           |
        |                  |          gnd
        |      10.0k       |
        +--/\/\---,        | 10.0k
        |   __    |        +--/\/\---,  0 to +5V
        '--|- \   | 10.0k  |   __    |  out, for
           |   >--+--/\/\--+--|- \   |  0 to 10mA
    gnd ---|+_/               |   >--+------
                       gnd ---|+_/-

Hi Winfield - my response is so slow that you may miss it altogeteher
- but I finally had time to work through your circuits. (getting stuck
in the airport for a day and a half really frees you up!). To me, it
looks like the 200 ohm resistor on the emitter of the transistor
should be a 210 ohm resistor. Otherwise, by my maths, you get a
constant (assuming the HV supply is constant) offset as well as a vout
not quite equal to 500*ILoad. It looks to me like you are assuming
that the current through the shunt is almost entirely the load
current, which at the low currents we're talking about is not entirely
accurate, as the resistive divider on the non-inverting terminal of
the LT1783 draws about a quarter of a ma through the shunt.

By the way, if I wanted much higher bandwidth (say 100MHz) - do you
think this circuit might still be able to work? Certainly the the
LT1783 would need to be swapped out. Can you tell me why exactly you
chose that part? I don't see why an over-the-top op-amp is needed.

Thanks!

-Michael

 

[Winfield]

Hi Winfield - my response is so slow that you may miss it altogether
- but I finally had time to work through your circuits. (getting stuck
in the airport for a day and a half really frees you up!). To me, it
looks like the 200 ohm resistor on the emitter of the transistor
should be a 210 ohm resistor. Otherwise, by my maths, you get a
constant (assuming the HV supply is constant) offset as well as a vout
not quite equal to 500*ILoad. It looks to me like you are assuming
that the current through the shunt is almost entirely the load
current, which at the low currents we're talking about is not entirely
accurate, as the resistive divider on the non-inverting terminal of
the LT1783 draws about a quarter of a ma through the shunt.

Yes, you are a little slow at reading the postings! You are
correct, or at least partially so. In earlier postings, I
caught my error and changed the other resistor to 190 ohms.
That's the way to preserve a 1/20 mirror gain while keeping
a zero dc-offset from the bias current.

By the way, if I wanted much higher bandwidth (say 100MHz) - do you
think this circuit might still be able to work? Certainly the the
LT1783 would need to be swapped out. Can you tell me why exactly you
chose that part? I don't see why an over-the-top op-amp is needed.

It was a convenient sot-23 low-current over-1MHz
rail-rail part. Hopefully there are better ones.

As for higher bandwidths, yes, certainly you can, with
lower resistor values, and at higher mirror currents.
But at frequencies well below 100MHz the sense resistor's
inductance, etc., will be a big problem. For moderately-
high RF-frequencies I suggest trying a current transformer.

 

[Michael]

 Actually, my circuit above is a short-changing cheat!
 It has no standing bias for the critical transistor,
 Q1, and hence cannot perform to the desired bandwidth
 when Iout is near 0mA.  Adding a bias like the original
 non-opamp BJT current-mirror circuit, boosts the parts
 count to 15, exactly the same as the all-BJT circuit.

 So it's a tie.  Go to the trouble of finding a suitable
 low-current high-side opamp and get better zero-current
 accuracy without trimming.  Or something like that.

   high-voltage fast high-side current monitor

                   10.0       +250V    Iout
  ------+------+---/\/\----+----o ---> 0 to 10mA load
        |      |           |
        /      '--+--------|---,
        \ 200     |        |   |
        /  1%     |    200 /   |
        |         |     1% \   |
        +---------|----,   /   +---,
        |        _|_   |   |   |   | 7.5V zener
    Q1  e       /  -|--'   |  _|_/ |
 mpsa92   b ---<    |      |  /_\ _|_
   PNP  c       \__+|------+   |  --- 0.1uF
        | LT1783  |        |   |   |
        |         '--------|---+---+---,
        /                  |          |
        \ 1k          1.0M |           /  330k
        /                  /          \ 0.5W
        | Ib = 0.25        \ Ia        /
        | to 0.75mA        /           |
        |                  |          gnd
        |      10.0k       |
        +--/\/\---,        | 10.0k
        |   __    |        +--/\/\---,  0 to +5V
        '--|- \   | 10.0k  |   __    |  out, for
           |   >--+--/\/\--+--|- \   |  0 to 10mA
    gnd ---|+_/               |   >--+------
                       gnd ---|+_/-

Hello again Winfield - This circuit was a little tricky for me to
understand at first, so then I just started making assumptions to
simplify things and I think I figured out what you were going for, but
I was hoping you might be able to address a couple questions/worries I
have about it:

1. It looks like you are assuming that the base to emitter voltages of
the two matched transistors are the same. This would require the base
currents to be the same, and the collector to emitter voltages to be
the same. This clearly isn't the case, however. Am I wrong in assuming
that you are treating the VBEs of the two matched transistors to be
the same? Or is the error small enough that the approximation will
work?

2. I think the 2N3906 is being used to protect the left matched
transistor from the early effect by providing a fairly constant VCE
across that transistor. It would also provide a path for current to
flow from the matched transistors to ground. Is that right?

3. Another assumption I think you made was that the base currents were
all small enough to ignore. This seems fairly safe to me, but I just
want to be sure that I'm seeing things correctly.

4. The MPSA92 just serves to drop the voltage down to ground, correct?
How is it able to regulate the current through it? It seems to me that
it'll have a relatively constant VBE and VCE. This really confuses me.

5. How did you choose the values of the 270K, 1K, and 4.02K resistors?
I understand that the 1K and 4.02K resistors needed to be chosen with
a specific ratio so as to provide a specific current divider, but why
the magnitudes that you chose? It looks like the left matched
transistor will have .2-2.2ma running through it, while the right
matched transistor will have about 1ma through it, so did you just aim
to place the current of the right matched transistor to be in the
middle of the range of currents going through the left matched
transistor, so as to keep their VBEs as close as possible?

6. I'm getting the final output to look like: (2.49K/4.99) * ILoad -
(1/4.99-1/5.02)*IQ2 (where IQ2 is the current through the right
matched transistor, and should be about 1ma, thus making that second
term approximately 1uV and thus negligible). This is all from the
assumptions mentioned above, thus this is why I think my assumption
about your assumptions is probably right.

Sorry for bugging you with so many questions - this circuit just left
me a bit more confused than the other.

Thanks again!

-Michael

 

[Michael]

Hello again Winfield - This circuit was a little tricky for me to
understand at first, so then I just started making assumptions to
simplify things and I think I figured out what you were going for, but
I was hoping you might be able to address a couple questions/worries I
have about it:

1. It looks like you are assuming that the base to emitter voltages of
the two matched transistors are the same. This would require the base
currents to be the same, and the collector to emitter voltages to be
the same. This clearly isn't the case, however. Am I wrong in assuming
that you are treating the VBEs of the two matched transistors to be
the same? Or is the error small enough that the approximation will
work?

2. I think the 2N3906 is being used to protect the left matched
transistor from the early effect by providing a fairly constant VCE
across that transistor. It would also provide a path for current to
flow from the matched transistors to ground. Is that right?

3. Another assumption I think you made was that the base currents were
all small enough to ignore. This seems fairly safe to me, but I just
want to be sure that I'm seeing things correctly.

4. The MPSA92 just serves to drop the voltage down to ground, correct?
How is it able to regulate the current through it? It seems to me that
it'll have a relatively constant VBE and VCE. This really confuses me.

5. How did you choose the values of the 270K, 1K, and 4.02K resistors?
I understand that the 1K and 4.02K resistors needed to be chosen with
a specific ratio so as to provide a specific current divider, but why
the magnitudes that you chose? It looks like the left matched
transistor will have .2-2.2ma running through it, while the right
matched transistor will have about 1ma through it, so did you just aim
to place the current of the right matched transistor to be in the
middle of the range of currents going through the left matched
transistor, so as to keep their VBEs as close as possible?

6. I'm getting the final output to look like: (2.49K/4.99) * ILoad -
(1/4.99-1/5.02)*IQ2 (where IQ2 is the current through the right
matched transistor, and should be about 1ma, thus making that second
term approximately 1uV and thus negligible). This is all from the
assumptions mentioned above, thus this is why I think my assumption
about your assumptions is probably right.

Sorry for bugging you with so many questions - this circuit just left
me a bit more confused than the other.

Thanks again!

-Michael- Hide quoted text -

- Show quoted text -

Oh shoot - I just realized I quoted the wrong post. My apologies for
what must be a very confusing message. The post I meant to quote is
below.

-Michael

 

[Winfield]

Hello again Winfield - This circuit was a little tricky for me to
understand at first, so then I just started making assumptions to
simplify things and I think I figured out what you were going for, but
I was hoping you might be able to address a couple questions/worries I
have about it:

This is the circuit you were evaluating:

. 10.0 +250V Iout
. ----+---/\/\----+------o ---> 0 to 10mA load
. | |
. / |
. \ 49.9 | all resistors 1%
. / |
. | | Q1-Q2 PNP low-voltage
. e e high beta, matched
. b --+-- b
. Q1 c | c Q2
. | e |
. +-- b |
. | c Q3 2n3906
. | |_____|
. e |
. b --------+ Ia = 1mA
. c |
. | Q4 | 270k 1.00k
. | mpsa92 '--/\/\---+--/\/\--- gnd
. | 0.5W |
. | 5% \ 4.02k
. | Ib /
. | 2.49k \
. | ,--/\/\---, | 2.49k
. | | __ | +--/\/\---, 0 to +5V
. '--+--|- \ | 2.49k | __ | out, for
. | >--+--/\/\--+--|- \ | 0 to 10mA
. gnd ---|+_/ | >--+----
. gnd ---|+_/

1. It looks like you are assuming that the base to emitter voltages of
the two matched transistors are the same. This would require the base
currents to be the same, and the collector to emitter voltages to be
the same. This clearly isn't the case, however. Am I wrong in assuming
that you are treating the VBEs of the two matched transistors to be
the same? Or is the error small enough that the approximation will
work?

Yes, the circuit suffers from errors from the Vbe dependence
on transistor current. Let's see. At 10mA full-scale load
current, we have 11mA through the 10 ohms and 2.2mA though
the 50 ohms, with 0.2mA worth subtracted out by the second
opamp. The Q1 and Q2 transistors are operating at 2.2mA and
1mA, for an offset error of about 20mV or -20%.

At zero current load transistor Q1 would nominally operate at
1mA 10/50 = 0.2mA, ignoring the Vbe issue, which adds an offset
of about 40mV = +40% of full scale. Hmm, doesn't look too good.

2. I think the 2N3906 is being used to protect the left matched
transistor from the early effect by providing a fairly constant VCE
across that transistor. It would also provide a path for current to
flow from the matched transistors to ground. Is that right?

3. Another assumption I think you made was that the base currents were
all small enough to ignore. This seems fairly safe to me, but I just
want to be sure that I'm seeing things correctly.

Yes and yes.

4. The MPSA92 just serves to drop the voltage down to ground, correct?
How is it able to regulate the current through it? It seems to me that
it'll have a relatively constant VBE and VCE. This really confuses me.

That's the common-base connection, where Ic = Ie, minus 1 to 2%
for the base current.

5. How did you choose the values of the 270K, 1K, and 4.02K resistors?
I understand that the 1K and 4.02K resistors needed to be chosen with
a specific ratio so as to provide a specific current divider, but why
the magnitudes that you chose?

That's a 1/5 ratio, matching the 10/50 ohms up top, to subtract
out the effect of the standing 1mA bias current.

It looks like the left matched
transistor will have .2-2.2ma running through it, while the right
matched transistor will have about 1ma through it, so did you just
aim to place the current of the right matched transistor to be in
the middle of the range of currents going through the left matched
transistor, so as to keep their VBEs as close as possible?

Yes, wishful thinking apparently.

6. I'm getting the final output to look like: (2.49K/4.99) * ILoad -
(1/4.99-1/5.02)*IQ2 (where IQ2 is the current through the right
matched transistor, and should be about 1ma, thus making that second
term approximately 1uV and thus negligible). This is all from the
assumptions mentioned above, thus this is why I think my assumption
about your assumptions is probably right.

Sorry for bugging you with so many questions - this circuit just left
me a bit more confused than the other.

Well, I think you were chipping away at the weak foundation
of the matched-voltage emitter-input differential amplifier,
and my only defense is the good performance of the opamp
circuit I posted later, which you replied to in your post.

 

[Michael]

 Yes, you are a little slow at reading the postings!  You are
 correct, or at least partially so.  In earlier postings, I
 caught my error and changed the other resistor to 190 ohms.
 That's the way to preserve a 1/20 mirror gain while keeping
 a zero dc-offset from the bias current.

Ah, I missed that! The 190 ohm change is a much better change.

 It was a convenient sot-23 low-current over-1MHz
 rail-rail part.  Hopefully there are better ones.

 As for higher bandwidths, yes, certainly you can, with
 lower resistor values, and at higher mirror currents.
 But at frequencies well below 100MHz the sense resistor's
 inductance, etc., will be a big problem.  For moderately-
 high RF-frequencies I suggest trying a current transformer.

Current transormers are more for measuring AC, not DC, correct? I
should have mentioned more about my application: I'm looking to be
able to analyze the current through an avalanche photodiode that is
receiving really sharp pulses of light. Specifically, my plan is to
have a very high speed comparator on the output of the current sensor
circuit, as well as an integrator and a peak detector. Thus I think a
current transormer would not work very well, right? Is there a good
way to handle these really high speeds? I'm planning on the pulses
having rise and fall times in the single digit (double digit worst
case) picoseonds, and lengths of probably a nanosecond or two. Thanks,

-Michael

 

[Michael]

 This is the circuit you were evaluating:






 Yes, the circuit suffers from errors from the Vbe dependence
 on transistor current.  Let's see.  At 10mA full-scale load
 current, we have 11mA through the 10 ohms and 2.2mA though
 the 50 ohms, with 0.2mA worth subtracted out by the second
 opamp.  The Q1 and Q2 transistors are operating at 2.2mA and
 1mA, for an offset error of about 20mV or -20%.

 At zero current load transistor Q1 would nominally operate at
 1mA 10/50 = 0.2mA, ignoring the Vbe issue, which adds an offset
 of about 40mV = +40% of full scale.  Hmm, doesn't look too good.

How did you calculate these values? Wouldn't that value depend on the
Is of the transistor chosen?

 Yes and yes.


 That's the common-base connection, where Ic = Ie, minus 1 to 2%
 for the base current.

Interesting - I was not familiar with that type of amplifier. Looks
like I need to do a bit of reading!

 That's a 1/5 ratio, matching the 10/50 ohms up top, to subtract
 out the effect of the standing 1mA bias current.


 Yes, wishful thinking apparently.



 Well, I think you were chipping away at the weak foundation
 of the matched-voltage emitter-input differential amplifier,
 and my only defense is the good performance of the opamp
 circuit I posted later, which you replied to in your post.

To be honest, I prefer the idea behind this circuit. I've always
preferred circuits that you can do without op-amps, and I really liked
how you connected the two bases together. I think it's because there's
this tendency these days just to connect black boxes, and I'm always
trying to avoid that tendency. It's unfortunate that it won't work
perfectly, though.

Thanks again!

-Michael

 

[Michael]

 This is the circuit you were evaluating:






 Yes, the circuit suffers from errors from the Vbe dependence
 on transistor current.  Let's see.  At 10mA full-scale load
 current, we have 11mA through the 10 ohms and 2.2mA though
 the 50 ohms, with 0.2mA worth subtracted out by the second
 opamp.  The Q1 and Q2 transistors are operating at 2.2mA and
 1mA, for an offset error of about 20mV or -20%.

 At zero current load transistor Q1 would nominally operate at
 1mA 10/50 = 0.2mA, ignoring the Vbe issue, which adds an offset
 of about 40mV = +40% of full scale.  Hmm, doesn't look too good.



 Yes and yes.


 That's the common-base connection, where Ic = Ie, minus 1 to 2%
 for the base current.

To follow up on my post from a few minutes ago - is the idea of a
common base amplifier that VBE is high enough that the transistor will
push as much current through itself as possible, but VCE is kept high
enough to keep it active so that the base current stays minimal? Looks
like hFE is around 80 at 25°C with an Ic from 0-10ma, so 1-2% current
loss to the base would make sense. Thanks,

-Michael

 

[Winfield]

How did you calculate these values? Wouldn't that value depend
on the Is of the transistor chosen?

No. It's easy with two identical transistors, you can
take ratios and get a simple equation, dV = kT/q ln Ix/Iy.
See AoE page 91. Figure 2.53 plots dV vs current ratios.

[ snip ]

To be honest, I prefer the idea behind this circuit. I've always
preferred circuits that you can do without op-amps, and I really liked
how you connected the two bases together. I think it's because there's
this tendency these days just to connect black boxes, and I'm always
trying to avoid that tendency. It's unfortunate that it won't work
perfectly, though.

It's not hard to fix the circuit's Vbe-vs-Ic problem, if we
allow ourselves one more transistor, in order to force both
Ic's equal, as I should have done in the first place. You
think about it, and I'll try to find time to post the answer
later today or tomorrow.

 

[Winfield]

How did you calculate these values? Wouldn't that value depend
on the Is of the transistor chosen?

No. It's easy with two identical transistors, you can
take ratios and get a simple equation, dV = kT/q ln Ix/Iy.
See AoE page 91. Figure 2.53 plots dV vs current ratios.

[ snip ]

To be honest, I prefer the idea behind this circuit. I've always
preferred circuits that you can do without op-amps, and I really liked
how you connected the two bases together. I think it's because there's
this tendency these days just to connect black boxes, and I'm always
trying to avoid that tendency. It's unfortunate that it won't work
perfectly, though.

It's not hard to fix the circuit's Vbe-vs-Ic problem, if we
allow ourselves one more transistor, in order to force both
Ic's equal, as I should have done in the first place. You
think about it, and I'll try to find time to post the answer
later today or tomorrow.

The goals for our circuit are to present a 0-5V output
at ground, for a 0-10mA current on a +250V rail, with a
1MHz or greater bandwidth, to an accuracy of say 2%.

Here's my original poor circuit, with very low gain and
a 60mV input error required to get its full-scale output,
With its high errors, this circuit met only two goals.

. 10.0 +250V Iout
. ----+---/\/\----+------o ---> 0 to 10mA load
. | |
. / |
. \ 49.9 | all resistors 1%
. / |
. | | Q1-Q2 PNP low-voltage
. e e high beta, matched
. b --+-- b
. Q1 c | c Q2
. | e |
. +-- b |
. | c Q3 2n3906
. | |_____|
. e |
. b --------+ Ia = 1mA
. c |
. | Q4 | 270k 1.00k
. | mpsa92 '--/\/\---+--/\/\--- gnd
. | 0.5W |
. | 5% \ 4.02k
. | Ib /
. | 2.49k \
. | ,--/\/\---, | 2.49k
. | | __ | +--/\/\---, 0 to +5V
. '--+--|- \ | 2.49k | __ | out, for
. | >--+--/\/\--+--|- \ | 0 to 10mA
. gnd ---|+_/ | >--+----
. gnd ---|+_/

To reduce the input error below say 1%, or 1mV, both of
the input transistors must always operate at the same
current, to within e^dV/25mV = 1.04, or 4% change for
a full-scale output swing. So we need more gain in the
differential stage. As before, it must operate from
signals at the positive rail, and it must be pre-biased,
to have more-than 1MHz bandwidth at zero input signal.

We'll split our bias current into two equal parts, using
new transistors Q3 Q4, to feed input transistors Q1 Q2.
Except for the base current of Q5, the high-voltage PNP
output transistor, Q1 and Q2 operate at equal collector
currents of about 450uA each (assuming similar betas).
The Q5 current ranges from 450uA, to insure high speed,
to 1.45mA at full scale. A 100k collector resistor holds
its power dissipation down to 150mW.

.. +250V Rs 10.0 Iout
.. IN ----+---/\/\----+------o ---> 0 to 10mA load
.. | |
.. R1 R2
.. 100 190 all resistors 1%
.. | | R1 = 10Rs
.. ,-------+ | R2 = 2R1 - Rs
.. | | |
.. | e e
.. | Q1 b --+-- b Q2 Q1 Q2 matched pnp
.. | c | c pnp Q3 Q4 matched npn
.. e | | | low-voltage types
.. b ----+ +-----+
.. c Q5 | | |
.. | mps c Q3 | c Q4
.. | A92 b --+-- b npn
.. | e e
.. | | |
.. | R3 R4 10k 1%
.. | |___________|
.. | |
.. \ Ib/2 | 270k 4.99k
.. / Ib '--/\/\---+--/\/\--- gnd
.. \ 100k 0.5W |
.. | 0.5W 5% \ 4.99k
.. | /
.. | 4.99k \
.. | ,--/\/\---, | 4.99k
.. | | __ | +--/\/\---, 0 to +5V
.. '---+--|- \ | 4.99k | __ | out, for
.. | >--+---/\/\--+--|- \ | 0 to 10mA
.. gnd ---|+_/ | >--+----
.. gnd ---|+_/

If Q5's beta is more than 100 (the datasheet plot says 150),
then Q1's change in Ic is under 10uA, for 0 to full scale,
or half of our 2% budget. So this circuit should fix the
problem.

 

[Winfield]

No. It's easy with two identical transistors, you can
take ratios and get a simple equation, dV = kT/q ln Ix/Iy.
See AoE page 91. Figure 2.53 plots dV vs current ratios.

[ snip ]

To be honest, I prefer the idea behind this circuit. I've always
preferred circuits that you can do without op-amps, and I really liked
how you connected the two bases together. I think it's because there's
this tendency these days just to connect black boxes, and I'm always
trying to avoid that tendency. It's unfortunate that it won't work
perfectly, though.

It's not hard to fix the circuit's Vbe-vs-Ic problem, if we
allow ourselves one more transistor, in order to force both
Ic's equal, as I should have done in the first place. You
think about it, and I'll try to find time to post the answer
later today or tomorrow.

The goals for our circuit are to present a 0-5V output
at ground, for a 0-10mA current on a +250V rail, with a
1MHz or greater bandwidth, to an accuracy of say 2%.

Here's my original poor circuit, with very low gain and
a 60mV input error required to get its full-scale output,
With its high errors, this circuit met only two goals.

. 10.0 +250V Iout
. ----+---/\/\----+------o ---> 0 to 10mA load
. | |
. / |
. \ 49.9 | all resistors 1%
. / |
. | | Q1-Q2 PNP low-voltage
. e e high beta, matched
. b --+-- b
. Q1 c | c Q2
. | e |
. +-- b |
. | c Q3 2n3906
. | |_____|
. e |
. b --------+ Ia = 1mA
. c |
. | Q4 | 270k 1.00k
. | mpsa92 '--/\/\---+--/\/\--- gnd
. | 0.5W |
. | 5% \ 4.02k
. | Ib /
. | 2.49k \
. | ,--/\/\---, | 2.49k
. | | __ | +--/\/\---, 0 to +5V
. '--+--|- \ | 2.49k | __ | out, for
. | >--+--/\/\--+--|- \ | 0 to 10mA
. gnd ---|+_/ | >--+----
. gnd ---|+_/

To reduce the input error below say 1%, or 1mV, both of
the input transistors must always operate at the same
current, to within e^dV/25mV = 1.04, or 4% change for
a full-scale output swing. So we need more gain in the
differential stage. As before, it must operate from
signals at the positive rail, and it must be pre-biased,
to have more-than 1MHz bandwidth at zero input signal.

We'll split our bias current into two equal parts, using
new transistors Q3 Q4, to feed input transistors Q1 Q2.
Except for the base current of Q5, the high-voltage PNP
output transistor, Q1 and Q2 operate at equal collector
currents of about 450uA each (assuming similar betas).
The Q5 current ranges from 450uA, to insure high speed,
to 1.45mA at full scale. A 100k collector resistor holds
its power dissipation down to 150mW. [ snip drawing ]
If Q5's beta is more than 100 (the datasheet plot says 150),
then Q1's change in Ic is under 10uA, for 0 to full scale,
or half of our 2% budget. So this circuit should fix the
problem.

Oops! That was positive, rather than negative
feedback. Sorry about that! Here we go.


.. +250V Rs 10.0 Iout
.. IN ----+---/\/\----+------o ---> 0 to 10mA load
.. | |
.. R1 R2
.. 100 190 all resistors 1%
.. | | R1 = 10Rs
.. ,-------+ | R2 = 2R1 - Rs
.. | | |
.. | e e
.. | Q1 b --+-- b Q2 Q1 Q2 matched pnp
.. | c | c pnp Q3 Q4 matched npn
.. | | | | low-voltage high-gain
.. e +-----' |
.. b --- | ----------+
.. c Q5 +-----, |
.. | mps | | |
.. | A92 c Q3 | c Q4
.. | b --+-- b npn
.. | e e
.. | | |
.. | R3 R4 10k 1%
.. | |___________|
.. | |
.. \ Ib/2 | 270k 4.99k
.. / Ib '--/\/\---+--/\/\--- gnd
.. \ 100k 0.5W |
.. | 0.5W 5% \ 4.99k
.. | /
.. | 4.99k \
.. | ,--/\/\---, | 4.99k
.. | | __ | +--/\/\---, 0 to +5V
.. '---+--|- \ | 4.99k | __ | out, for
.. | >--+---/\/\--+--|- \ | 0 to 10mA
.. gnd ---|+_/ | >--+----
.. gnd ---|+_/

As before, the two opamps can be condensed into one,
with a difference-amplifier configuration.

 

[Michael]

Michael wrote:
Winfield wrote:
Michael wrote:
 Yes, the circuit suffers from errors from the Vbe dependence
 on transistor current.  Let's see.  At 10mA full-scale load
 current, we have 11mA through the 10 ohms and 2.2mA though
 the 50 ohms, with 0.2mA worth subtracted out by the second
 opamp.  The Q1 and Q2 transistors are operating at 2.2mA and
 1mA, for an offset error of about 20mV or -20%.
 At zero current load transistor Q1 would nominally operate at
 1mA 10/50 = 0.2mA, ignoring the Vbe issue, which adds an offset
 of about 40mV = +40% of full scale.  Hmm, doesn't look too good.
How did you calculate these values?  Wouldn't that value depend
on the Is of the transistor chosen?
No.  It's easy with two identical transistors, you can
take ratios and get a simple equation, dV = kT/q ln Ix/Iy.
See AoE page 91.  Figure 2.53 plots dV vs current ratios.
 [ snip ]
To be honest, I prefer the idea behind this circuit. I've always
preferred circuits that you can do without op-amps, and I really liked
how you connected the two bases together. I think it's because there's
this tendency these days just to connect black boxes, and I'm always
trying to avoid that tendency. It's unfortunate that it won't work
perfectly, though.
It's not hard to fix the circuit's Vbe-vs-Ic problem, if we
allow ourselves one more transistor, in order to force both
Ic's equal, as I should have done in the first place.  You
think about it, and I'll try to find time to post the answer
later today or tomorrow.

 The goals for our circuit are to present a 0-5V output
 at ground, for a 0-10mA current on a +250V rail, with a
 1MHz or greater bandwidth, to an accuracy of say 2%.

 Here's my original poor circuit, with very low gain and
 a 60mV input error required to get its full-scale output,
 With itshigherrors, this circuit met only two goals.

 To reduce the input error below say 1%, or 1mV, both of
 the input transistors must always operate at the same
 current, to within e^dV/25mV = 1.04, or 4% change for
 a full-scale output swing.  So we need more gain in the
 differential stage.  As before, it must operate from
 signals at the positive rail, and it must be pre-biased,
 to have more-than 1MHz bandwidth at zero input signal.

 We'll split our bias current into two equal parts, using
 new transistors Q3 Q4, to feed input transistors Q1 Q2.
 Except for the base current of Q5, thehigh-voltage PNP
 output transistor, Q1 and Q2 operate at equal collector
 currents of about 450uA each (assuming similar betas).
 The Q5 current ranges from 450uA, to insurehighspeed,
 to 1.45mA at full scale.  A 100k collector resistor holds
 its power dissipation down to 150mW.  [ snip drawing ]
 If Q5's beta is more than 100 (the datasheet plot says 150),
 then Q1's change in Ic is under 10uA, for 0 to full scale,
 or half of our 2% budget.  So this circuit should fix the
 problem.

 Oops!  That was positive, rather than negative
 feedback.  Sorry about that!  Here we go.

.   +250V        Rs 10.0               Iout
.    IN  ----+---/\/\----+------o ---> 0 to 10mA load
.            |           |
.           R1          R2
.           100         190        all resistors1%
.            |           |            R1= 10Rs
.    ,-------+           |            R2 = 2R1- Rs
.    |       |           |
.    |       e           e
.    |   Q1    b --+-- b    Q2     Q1 Q2 matched pnp
.    |       c     |     c  pnp    Q3 Q4 matched npn
.    |       |     |     |     low-voltagehigh-gain
.    e       +-----'     |
.      b --- | ----------+
.    c  Q5   +-----,     |
.    | mps   |     |     |
.    | A92   c  Q3 |     c  Q4
.    |         b --+-- b    npn
.    |       e           e
.    |       |           |
.    |      R3          R4 10k 1%
.    |       |___________|
.    |             |
.    \ Ib/2        |  270k       4.99k
.    /          Ib '--/\/\---+--/\/\--- gnd
.    \ 100k           0.5W   |
.    | 0.5W            5%    \ 4.99k
.    |                       /
.    |       4.99k           \
.    |   ,--/\/\---,         | 4.99k
.    |   |   __    |         +--/\/\---,  0 to +5V
.    '---+--|- \   |  4.99k  |   __    |  out, for
.           |   >--+---/\/\--+--|- \   |  0 to 10mA
.    gnd ---|+_/                |   >--+----
.                        gnd ---|+_/

 As before, the two opamps can be condensed into one,
 with a difference-amplifier configuration.

Hi Winfield - you tripped me up when you said it could be done with
one more transistor. I was trying to figure out a way to add a
transistor to your original circuit, instead of making a fairly
different one!

Anyways - I just ran through the numbers and I think I understand most
everything that is going on in your circuit, but I'm still confused
about a couple things:

First of all, R3 would be 10K, correct? I didn't see it marked, but
otherwise things wouldn't make sense.

Q5's operation is a bit strange to me. It really confused me until I
decided to just assume that the emitter currents in Q1 and Q2 were the
same and that Q5 would get the remainder of the current flowing
through R1. So that tells me that Q5 is acting as a current buffer, so
once again the MPSA92 is in the common base configuration. The common
base configuration really confuses me... Is there a section in AOE
that goes over it? I couldn't find anything about it. As far as I can
tell, however, the idea of the common base configuration is to keep
VBE large enough that the transistor attempts to push as much current
through the collector as possible. Is that at all accurate?

Lastly, could you please explain the feedback loop that you are using?
I must admit that I'm having trouble seeing why the circuit you posted
earlier today would not work while this one would.

Thanks!

-Michael

 

[Winfield]

To reduce the input error below say 1%, or 1mV, both of
the input transistors must always operate at the same
current, to within e^dV/25mV = 1.04, or 4% change for
a full-scale output swing. So we need more gain in the
differential stage. As before, it must operate from
signals at the positive rail, and it must be pre-biased,
to have more-than 1MHz bandwidth at zero input signal.
We'll split our bias current into two equal parts, using
new transistors Q3 Q4, to feed input transistors Q1 Q2.
Except for the base current of Q5, thehigh-voltage PNP
output transistor, Q1 and Q2 operate at equal collector
currents of about 450uA each (assuming similar betas).
The Q5 current ranges from 450uA, to insurehighspeed,
to 1.45mA at full scale. A 100k collector resistor holds
its power dissipation down to 150mW. [ snip drawing ]
If Q5's beta is more than 100 (the datasheet plot says 150),
then Q1's change in Ic is under 10uA, for 0 to full scale,
or half of our 2% budget. So this circuit should fix the
problem.

Oops! That was positive, rather than negative
feedback. Sorry about that! Here we go.

. +250V Rs 10.0 Iout
. IN ----+---/\/\----+------o ---> 0 to 10mA load
. | |
. R1 R2
. 100 190 all resistors 1%
. | | R1 = 10Rs
. ,-------+ | R2 = 2R1 - Rs
. | | |
. | e e
. | Q1 b --+-- b Q2 Q1 Q2 matched pnp
. | c | c pnp Q3 Q4 matched npn
. | | | | low-voltagehigh-gain
. e +-----' |
. b --- | ----------+
. c Q5 +-----, |
. | mps | | |
. | A92 c Q3 | c Q4
. | b --+-- b npn
. | e e
. | | |
. | R3 R4 10k 1%
. | |___________|
. | |
. \ Ib/2 | 270k 4.99k
. / Ib '--/\/\---+--/\/\--- gnd
. \ 100k 0.5W |
. | 0.5W 5% \ 4.99k
. | /
. | 4.99k \
. | ,--/\/\---, | 4.99k
. | | __ | +--/\/\---, 0 to +5V
. '---+--|- \ | 4.99k | __ | out, for
. | >--+---/\/\--+--|- \ | 0 to 10mA
. gnd ---|+_/ | >--+----
. gnd ---|+_/

As before, the two opamps can be condensed into one,
with a difference-amplifier configuration.

Hi Winfield - you tripped me up when you said it could be done with
one more transistor. I was trying to figure out a way to add a
transistor to your original circuit, instead of making a fairly
different one!

Anyways - I just ran through the numbers and I think I understand most
everything that is going on in your circuit, but I'm still confused
about a couple things:

First of all, R3 would be 10K, correct? I didn't see it marked, but
otherwise things wouldn't make sense.

Q5's operation is a bit strange to me. It really confused me until I
decided to just assume that the emitter currents in Q1 and Q2 were the
same and that Q5 would get the remainder of the current flowing
through R1. So that tells me that Q5 is acting as a current buffer, so
once again the MPSA92 is in the common base configuration. The common
base configuration really confuses me... Is there a section in AOE
that goes over it? I couldn't find anything about it. As far as I can
tell, however, the idea of the common base configuration is to keep
VBE large enough that the transistor attempts to push as much current
through the collector as possible. Is that at all accurate?

Well, imagine a current source feeding an emitter. If the
compliance voltage is high enough to create more than Vbe
across the base-emitter junction, it'll carry all of the
current-source's current. But we know the transistor only
needs 1/beta of that for the base; the rest, almost equal
to the original emitter current, comes from the collector.
That fraction, called alpha, is nearly equal to 1.

Lastly, could you please explain the feedback loop that you are using?
I must admit that I'm having trouble seeing why the circuit you posted
earlier today would not work while this one would.

One way: you can follow each element in the feedback path,
looking for an extra inverted element after adding them up.
Another way, imagine the current in the output transistor
Q5 is too high. Voltage across R1 too high. What happens
to the rest of the circuit, does it increase or decrease
Q5's current?

Imagine, if Q5's current is too high, the voltage across R1
is larger (more negative) than the R2 - Q2e node, and Q2's
emitter voltage is more negative than it should be, and Q2
takes more current from R2 than it should, and this pulls
the Q4-Q2 collector node toward Q2, tending to shutoff Q5,
and reverse the process. That's negative feedback. You
can see this quickly, by observing that every stage is a
follower, or noninverting -- except the Q2 inverting stage
that gives us negative feedback.

 

[Michael]

Winfield wrote:
 To reduce the input error below say 1%, or 1mV, both of
 the input transistors must always operate at the same
 current, to within e^dV/25mV = 1.04, or 4% change for
 a full-scale output swing.  So we need more gain in the
 differential stage.  As before, it must operate from
 signals at the positive rail, and it must be pre-biased,
 to have more-than 1MHz bandwidth at zero input signal.
 We'll split our bias current into two equal parts, using
 new transistors Q3 Q4, to feed input transistors Q1 Q2.
 Except for the base current of Q5, thehigh-voltage PNP
 output transistor, Q1 and Q2 operate at equal collector
 currents of about 450uA each (assuming similar betas).
 The Q5 current ranges from 450uA, to insurehighspeed,
 to 1.45mA at full scale.  A 100k collector resistor holds
 its power dissipation down to 150mW.  [ snip drawing ]
 If Q5's beta is more than 100 (the datasheet plot says 150),
 then Q1's change in Ic is under 10uA, for 0 to full scale,
 or half of our 2% budget.  So this circuit should fix the
 problem.
 Oops!  That was positive, rather than negative
 feedback.  Sorry about that!  Here we go.
.   +250V        Rs 10.0               Iout
.    IN  ----+---/\/\----+------o ---> 0 to 10mA load
.            |           |
.           R1          R2
.           100         190        all resistors 1%
.            |           |            R1 = 10Rs
.    ,-------+           |            R2 = 2R1 - Rs
.    |       |           |
.    |       e           e
.    |   Q1    b --+-- b    Q2     Q1 Q2 matched pnp
.    |       c     |     c  pnp    Q3 Q4 matchednpn
.    |       |     |     |     low-voltagehigh-gain
.    e       +-----'     |
.      b --- | ----------+
.    c  Q5   +-----,     |
.    | mps   |     |     |
.    | A92   c  Q3 |     c  Q4
.    |         b --+-- b    npn
.    |       e           e
.    |       |           |
.    |      R3          R4 10k 1%
.    |       |___________|
.    |             |
.    \ Ib/2        |  270k       4.99k
.    /          Ib '--/\/\---+--/\/\--- gnd
.    \ 100k           0.5W   |
.    | 0.5W            5%    \ 4.99k
.    |                       /
.    |       4.99k           \
.    |   ,--/\/\---,         | 4.99k
.    |   |   __    |         +--/\/\---,  0 to +5V
.    '---+--|- \   |  4.99k  |   __    |  out, for
.           |   >--+---/\/\--+--|- \   |  0 to 10mA
.    gnd ---|+_/                |   >--+----
.                        gnd ---|+_/
 As before, the two opamps can be condensed into one,
 with a difference-amplifier configuration.

Hi Winfield - you tripped me up when you said it could be done with
one more transistor. I was trying to figure out a way to add a
transistor to your original circuit, instead of making a fairly
different one!

Anyways - I just ran through the numbers and I think I understand most
everything that is going on in your circuit, but I'm still confused
about a couple things:

First of all, R3 would be 10K, correct? I didn't see it marked, but
otherwise things wouldn't make sense.

Q5's operation is a bit strange to me. It really confused me until I
decided to just assume that the emitter currents in Q1 and Q2 were the
same and that Q5 would get the remainder of the current flowing
through R1. So that tells me that Q5 is acting as a current buffer, so
once again the MPSA92 is in the common base configuration. The common
base configuration really confuses me... Is there a section in AOE
that goes over it? I couldn't find anything about it. As far as I can
tell, however, the idea of the common base configuration is to keep
VBE large enough that the transistor attempts to push as much current
through the collector as possible. Is that at all accurate?

 Well, imagine a current source feeding an emitter.  If the
 compliance voltage is high enough to create more than Vbe
 across the base-emitter junction, it'll carry all of the
 current-source's current.  But we know the transistor only
 needs 1/beta of that for the base; the rest, almost equal
 to the original emitter current, comes from the collector.
 That fraction, called alpha, is nearly equal to 1.

Lastly, could you please explain the feedback loop that you are using?
I must admit that I'm having trouble seeing why the circuit you posted
earlier today would not work while this one would.

 One way: you can follow each element in the feedback path,
 looking for an extra inverted element after adding them up.
 Another way, imagine the current in the output transistor
 Q5 is too high.  Voltage across R1 too high.  What happens
 to the rest of the circuit, does it increase or decrease
 Q5's current?

 Imagine, if Q5's current is too high, the voltage across R1
 is larger (more negative) than the R2 - Q2e node, and Q2's
 emitter voltage is more negative than it should be, and Q2
 takes more current from R2 than it should, and this pulls
 the Q4-Q2 collector node toward Q2, tending to shutoff Q5,
 and reverse the process.  That's negative feedback.  You
 can see this quickly, by observing that every stage is a
 follower, or noninverting -- except the Q2 inverting stage
 that gives us negative feedback.

So it looks to me like the base of Q5 can go from about .2V below it's
emitter to about 1V below it's emitter, assuming Q1-Q4 always stay
active. So then the feedback mechanism is in place to regulate that
and keep the proper amount of current flowing through Q5 by changing
Q1's VBE. Your description of the feedback mechanism makes sense, but
I'm still trying to figure out this common base configuration. So for
a common base amplifier you have to have some sort of feedback system
in place to keep VBE within the proper range, right? I mean without
feedback it seems like you could easily have VBE slightly too large
and have all the current flow from E to B.

I guess this all makes sense in a hand wavey way... but going from
understanding it to designing it... I still need alot of work there.

Thanks,

-Michael

 

[Winfield Hill]

So it looks to me like the base of Q5 can go from about .2V below it's
emitter to about 1V below it's emitter, assuming Q1-Q4 always stay
active. So then the feedback mechanism is in place to regulate that
and keep the proper amount of current flowing through Q5 by changing
Q1's VBE. Your description of the feedback mechanism makes sense, but
I'm still trying to figure out this common base configuration. So for
a common base amplifier you have to have some sort of feedback system
in place to keep VBE within the proper range, right? I mean without
feedback it seems like you could easily have VBE slightly too large
and have all the current flow from E to B.

Ebers-Moll tells us that Vbe relates not to Ie
directly, but instead to Ic. So what happens
in a common-base amplifier is that we present
a current Ie to the emitter, and the transistor
adjusts its own Vbe, so that Ic = Ie - Ic/beta,
or Ie minus the base current.

So to use the common-base configuration, all
we have to do is present the emitter current,
and collect the collector current.

But my emitter-coupled differential amplifier
works a little differently, as you can see.
Q1 sets up the base voltage for itself and Q2,
and forces Q2's emitter to follow it. In the
original "defective" version, I collected the
resulting collector current, ignoring the big
change in Vbe over the amplifier's range. In
the new version, we collect the current that's
left over, after the differential amplifier's
feedback loop responds to the forcing currents
from Q3 and Q4.

 

[Michael]

Ebers-Moll tells us that Vbe relates not to Ie
directly, but instead to Ic. So what happens
in a common-base amplifier is that we present
a current Ie to the emitter, and the transistor
adjusts its own Vbe, so that Ic = Ie - Ic/beta,
or Ie minus the base current.

So to use the common-base configuration, all
we have to do is present the emitter current,
and collect the collector current.

But you also have to present some way for VBE to change, right? You
can't have the base and the emitter held at specific voltages,
correct? Because otherwise, say you presented the transistor with a
VBE of .8V, it seems to me that all current would flow from B to E and
probably just bypass the collector.

But my emitter-coupled differential amplifier
works a little differently, as you can see.
Q1 sets up the base voltage for itself and Q2,
and forces Q2's emitter to follow it. In the
original "defective" version, I collected the
resulting collector current, ignoring the big
change in Vbe over the amplifier's range. In
the new version, we collect the current that's
left over, after the differential amplifier's
feedback loop responds to the forcing currents
from Q3 and Q4.

Why would Q2's emitter have to follow the base of Q2? Maybe that is
where I am getting confused... To me it looks like the base of Q5 is
not set at a specific voltage - more it looks like it can be anywhere
from about .2V below Q5's emitter to a couple VBEs below Q5's emitter.

By the way - do you have any idea what sort of frequency this circuit
could operate to? My goal is to measure and analyze the current
through an avalanche photodiode. It will have pulses of current though
it that should have rise and fall times in the single digit pico
seconds, and durations in the single digit nanoseconds. Ideally I'd
like to have this circuit's output follow that very closely. If there
is a set delay in the output that is fine as I can fix that elsewhere.
Do you think that a circuit like this would be a good fit for my
application?

Thanks, and sorry I'm being so thick skulled about this. Some days I
wonder if I'd be better off just sticking to soldering all day.

-Michael

 

[Winfield Hill]

But you also have to present some way for VBE to change, right?
You can't have the base and the emitter held at specific voltages,
correct? Because otherwise, say you presented the transistor with a
VBE of .8V, it seems to me that all current would flow from B to E
and probably just bypass the collector.


Why would Q2's emitter have to follow the base of Q2? Maybe that is
where I am getting confused... To me it looks like the base of Q5 is
not set at a specific voltage - more it looks like it can be anywhere
from about .2V below Q5's emitter to a couple VBEs below Q5's emitter.

I copied the drawing below, for reference. The
current through Q3 (which is equal to Q4 and is
half the bias current), goes through Q1 and sets
its Vbe voltage. Q2's base is tied to Q1's so,
assuming they have equal current, Q2's emitter
voltage has to equal Q1's emitter voltage. Once
we know that we can start evaluating the currents,
to see what they do. Q2's collector is especially
interesting because if Q1 and Q2 don't have the
same current, as programmed by Q3 and Q4, then
Q5's base changes to control the current Q5
takes from R1, and bring Q1 and Q2 in balance.


.. +250V Rs 10.0 Iout
.. IN ----+---/\/\----+------o ---> 0 to 10mA load
.. | |
.. R1 R2
.. 100 190 all resistors 1%
.. | | R1 = 10Rs
.. ,-------+ | R2 = 2R1 - Rs
.. | | |
.. | e e
.. | Q1 b --+-- b Q2 Q1 Q2 matched pnp
.. | c | c pnp Q3 Q4 matched npn
.. | | | | low-voltagehigh-gain
.. e +-----' |
.. b --- | ----------+
.. c Q5 +-----, |
.. | mps | | |
.. | A92 c Q3 | c Q4
.. | b --+-- b npn
.. | e e
.. | | |
.. | R3 R4 10k 1%
.. | |___________|
.. | |
.. \ Ib/2 | 270k 4.99k
.. / Ib '--/\/\---+--/\/\--- gnd
.. \ 100k 0.5W |
.. | 0.5W 5% \ 4.99k
.. | /
.. | 4.99k \
.. | ,--/\/\---, | 4.99k
.. | | __ | +--/\/\---, 0 to +5V
.. '---+--|- \ | 4.99k | __ | out, for
.. | >--+---/\/\--+--|- \ | 0 to 10mA
.. gnd ---|+_/ | >--+----
.. gnd ---|+_/

By the way - do you have any idea what sort of frequency this
circuit could operate to?

OK, I'm curious too. I'll run a spice model.

My goal is to measure and analyze the current through an
avalanche photodiode. It will have pulses of current though
it that should have rise and fall times in the single digit pico
seconds, and durations in the single digit nanoseconds. Ideally I'd
like to have this circuit's output follow that very closely. If there
is a set delay in the output that is fine as I can fix that elsewhere.
Do you think that a circuit like this would be a good fit for my
application?

No. I don't think you're allowed to let the APD
voltage change much, once you've found the magic
value. I allowed 100mV fs, which might be too
much. You could use a transimpedance amplifier,
sitting at 250V, but what you really want is a
Tektronix CT-1 AC Current Probe, like this one,
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=190175118289
You just run the wire through the little hole in
the 3rd photo, and you're all set! Measure away,
with your 2GHz 50-ohm scope! Ahem!

 

[Jim Thompson]

On Fri, 28 Dec 2007 13:54:02 -0800 (PST), Winfield Hill

[snip]

I copied the drawing below, for reference. The
current through Q3 (which is equal to Q4 and is
half the bias current), goes through Q1 and sets
its Vbe voltage. Q2's base is tied to Q1's so,
assuming they have equal current, Q2's emitter
voltage has to equal Q1's emitter voltage. Once
we know that we can start evaluating the currents,
to see what they do. Q2's collector is especially
interesting because if Q1 and Q2 don't have the
same current, as programmed by Q3 and Q4, then
Q5's base changes to control the current Q5
takes from R1, and bring Q1 and Q2 in balance.


. +250V Rs 10.0 Iout
. IN ----+---/\/\----+------o ---> 0 to 10mA load
. | |
. R1 R2
. 100 190 all resistors 1%
. | | R1 = 10Rs
. ,-------+ | R2 = 2R1 - Rs
. | | |
. | e e
. | Q1 b --+-- b Q2 Q1 Q2 matched pnp
. | c | c pnp Q3 Q4 matched npn
. | | | | low-voltagehigh-gain
. e +-----' |
. b --- | ----------+
. c Q5 +-----, |
. | mps | | |
. | A92 c Q3 | c Q4
. | b --+-- b npn
. | e e
. | | |
. | R3 R4 10k 1%
. | |___________|
. | |
. \ Ib/2 | 270k 4.99k
. / Ib '--/\/\---+--/\/\--- gnd
. \ 100k 0.5W |
. | 0.5W 5% \ 4.99k
. | /
. | 4.99k \
. | ,--/\/\---, | 4.99k
. | | __ | +--/\/\---, 0 to +5V
. '---+--|- \ | 4.99k | __ | out, for
. | >--+---/\/\--+--|- \ | 0 to 10mA
. gnd ---|+_/ | >--+----
. gnd ---|+_/

[snip]

Excellent! Pretty soon we can enroll you as an apprentice IC
designer... you put your finger right on it... balance... balance...
balance... match... match... match ;-)

...Jim Thompson

 

[Michael]

I copied the drawing below, for reference. The
current through Q3 (which is equal to Q4 and is
half the bias current), goes through Q1 and sets
its Vbe voltage. Q2's base is tied to Q1's so,
assuming they have equal current, Q2's emitter
voltage has to equal Q1's emitter voltage. Once
we know that we can start evaluating the currents,
to see what they do. Q2's collector is especially
interesting because if Q1 and Q2 don't have the
same current, as programmed by Q3 and Q4, then
Q5's base changes to control the current Q5
takes from R1, and bring Q1 and Q2 in balance.

. +250V Rs 10.0 Iout
. IN ----+---/\/\----+------o ---> 0 to 10mA load
. | |
. R1 R2
. 100 190 all resistors 1%
. | | R1 = 10Rs
. ,-------+ | R2 = 2R1 - Rs
. | | |
. | e e
. | Q1 b --+-- b Q2 Q1 Q2 matched pnp
. | c | c pnp Q3 Q4 matched npn
. | | | | low-voltagehigh-gain
. e +-----' |
. b --- | ----------+
. c Q5 +-----, |
. | mps | | |
. | A92 c Q3 | c Q4
. | b --+-- b npn
. | e e
. | | |
. | R3 R4 10k 1%
. | |___________|
. | |
. \ Ib/2 | 270k 4.99k
. / Ib '--/\/\---+--/\/\--- gnd
. \ 100k 0.5W |
. | 0.5W 5% \ 4.99k
. | /
. | 4.99k \
. | ,--/\/\---, | 4.99k
. | | __ | +--/\/\---, 0 to +5V
. '---+--|- \ | 4.99k | __ | out, for
. | >--+---/\/\--+--|- \ | 0 to 10mA
. gnd ---|+_/ | >--+----
. gnd ---|+_/


OK, I'm curious too. I'll run a spice model.

Could you send me the spice file? (or post it here - but not on ABSE
as I don't have access) I'd love to play around with it: viewing
circuits in spice really helps me to understand them.

No. I don't think you're allowed to let the APD
voltage change much, once you've found the magic
value. I allowed 100mV fs, which might be too
much.

Well, the current gain of the APD decreases as you decrease the
voltage across it. Take for example:
http://sales.hamamatsu.com/assets/applications/SSD/Characteristics_and_use_of_SI_APD.pdf
- figure 2-4 and 2-5 show gain vs. voltage. I guess at the higher
reverse voltages (where I will probably be running it) the gain is
quite sensitive to changes in that voltage. Then again, the main thing
I'm looking for is just an on/off pulse, so it's not the end of the
world if the gain fluctuates some. Just how much gain fluctuation is
allowable is just impossible to estimate at this point, though.

You could use a transimpedance amplifier,
sitting at 250V, but what you really want is a
Tektronix CT-1 AC Current Probe, like this one,
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=190175118289
You just run the wire through the little hole in
the 3rd photo, and you're all set! Measure away,
with your 2GHz 50-ohm scope! Ahem!

Ah phooie - here all I have is a 60GHz scope. Err, hmmmm... make that
MHz. My trusty old Tek 2215A, to be exact

So - allow me to go into my plan in a bit more detail: My plan is to
power everything off of a single 5V supply. The analog supply
(probably +-12V) will be generated with normal boost and inverting DC/
DC converters. The 250V supply will be generated with a non-isolated
flyback supply. It'll be held at something like 25V above the APD
supply. An op-amp driving a couple FETs will actively regulate the
voltage across the APD, with the idea of eliminating, or at least
reducing, the noise from the flyback supply. In theory, that could
keep the voltage across the APD at a constant, no matter what current
was flowing. In reality, of course, I'm not sure if it's possible to
get the bandwidth of that circuit high enough. The design I'm working
on looks alot like Figure 6.47 "High-voltage regulated supply" from
AOE, but I'm seeing bandwidth in possibly the single digit MHz region
at best. It still needs alot of work to be sure - I'm still at a
pretty early stage on it.

So the Tek current probe is just a current transformer, right? Are
there plain current transformers on the market with similar
performance? I'm aiming to keep parts cost for this entire project at
around the cost of a single one of those guys (~$500 new), as I'd like
to make a bunch of them. Also, what would the output of such a circuit
look like when looking at the current through my APD? I had thought
that current transformers were more for periodic signals, while the
signals through the APD will not be periodic.

To me what might make sense is to float a high speed instrumentation
amplifier with a ground at say 250-10V using a zener. Use that to
watch the voltage across a very small shunt, one that might produce
50mV or so at peak current. I don't believe a drop that small will
affect the APD in any significant manner. Run the output of the
instrumentation amp into an op-amp acting as a transconductance amp
that puts the current into say a current buffer like what you've done
with the MPSA92, then take that current and run it through a
transimpedance amp to generate a voltage.

Does that make sense at all? That's alot of black boxes, which always
scares me... Also, it seems most instrumentation amplifiers top out in
the single or double digit MHz region, which probably isn't enough.

The reason I wanted current sensing to be high side was so that I
could use the APD side of the shunt resistor as the feedback for my
high voltage regulator circuit. But if I could use a small enough
shunt, I think it'd be just fine for it to be low side, which would
make things easier I would think as I could just use a non-inverting
amplifier to get the voltage up to something reasonable. Maybe that
would be a more reasonable path to take?

I'm going to cut myself off now before my post becomes novel length.
Hopefully it isn't all completely insane.

Thanks!

-Michael