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

[Michael]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it. I then want to have a
ground referenced voltage that is the amplified difference across the
shunt with a range of about 0-5V. I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, 5V, and probably some sort of +-12V
analog (clean) supply. I would like to get 8b or more of resolution
from the ground referenced amplified shunt voltage signal. I would
also like this circuit to be fast. The faster the better, MHz
bandwidth at least. If this were a lower voltage I'd just do an op-amp
difference amplifier. Unfortunately, last I checked, Apex had the
market for op-amps that can handle such high voltages fairly well
cornered, and their op-amps are expensive, large, and not particularly
quick. High side current amplifiers seem to top out at about 100VDC.

So I'm trying out various ideas. My favorite right now is to put
voltage dividers on both sides of the shunt to drop the voltage down
maybe to 0-10V on both and then use an instrumentation amplifier to
amplify that voltage. For some odd reason this bothers me, but I can't
place my finger on just why.

Am I being paranoid and the instrumentation amp is a legitimate idea,
or is there a better way to do this?

Thanks!

-Michael

 

[Glenn Gundlach]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it. I then want to have a
ground referenced voltage that is the amplified difference across the
shunt with a range of about 0-5V. I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, 5V, and probably some sort of +-12V
analog (clean) supply. I would like to get 8b or more of resolution
from the ground referenced amplified shunt voltage signal. I would
also like this circuit to be fast. The faster the better, MHz
bandwidth at least. If this were a lower voltage I'd just do an op- amp
difference amplifier. Unfortunately, last I checked, Apex had the
market for op-amps that can handle such high voltages fairly well
cornered, and their op-amps are expensive, large, and not particularly
quick. High side current amplifiers seem to top out at about 100VDC.

So I'm trying out various ideas. My favorite right now is to put
voltage dividers on both sides of the shunt to drop the voltage down
maybe to 0-10V on both and then use an instrumentation amplifier to
amplify that voltage. For some odd reason this bothers me, but I can't
place my finger on just why.

Am I being paranoid and the instrumentation amp is a legitimate idea,
or is there a better way to do this?

Thanks!

-Michael

The resistor matching becomes nearly impossible an dthe opamp offsets
and DC performance is tough, too. Back in the '70s we were
retrofitting remote control and metering onto a 1956 RCA TV
transmitter. To monitor high side ( and in a transmitter, high side is
HIGH - several kV ) the manufacturer made a gadget that converted the
current to an AC voltage, transformer coupled it to the low side and
converted back to a level. Doesn't sound so good, huh ? It DID work
well enough to satisfy the FCC. I'm sure there are better ways now.

Something like this help? Unfortunately it is not recommended for new
designs.

http://focus.ti.com/lit/ds/symlink/3650.pdf

GG

 

[Winfield]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it. I then want to have a
ground referenced voltage that is the amplified difference across the
shunt with a range of about 0-5V. I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, 5V, and probably some sort of +-12V
analog (clean) supply. I would like to get 8b or more of resolution
from the ground referenced amplified shunt voltage signal. I would
also like this circuit to be fast. The faster the better, MHz
bandwidth at least. If this were a lower voltage I'd just do an op-amp
difference amplifier. Unfortunately, last I checked, Apex had the
market for op-amps that can handle such high voltages fairly well
cornered, and their op-amps are expensive, large, and not particularly
quick. High side current amplifiers seem to top out at about 100VDC.

I suggest that you consider the benefits of a bipolar-
transistor current-replicator approach, as in my ASCII
circuit below. (view in a fixed font, like Courier)

high-voltage high-bandwidth high-side current monitor

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 ---|+_/

The circuit drops 100mV across the sense resistor at
full current. It has a 5V fs output, as you requested.

The transistor arrangement is designed to operate Q1
and Q2 at similar currents for matching, and at low
voltages to keep junction heating low. They should be
high beta types to insure accuracy at low currents.

Q4 provides the power-dissipating 250-volt voltage drop
for the output current. High-voltage transistors have
reduced beta, so up to 2% of the signal current will be
steered away from the output, but you can add a second
mpsa92, as a Darlington, to capture all the current.

Normally circuits like this are run with bias currents,
Ia, that are very low compared to the measured current.
But your maximum current is only 10mA, and you want a
1MHz or more bandwidth, so I chose a Q2 bias of 1mA,
which means it adds 10% of fs to the measurement and
has to be subtracted from the output. What's more, I
set the reflected output current Ib = 1/5 Iout, to keep
the Q4 transistor power dissipation down (550mW max), so
we have to take away 4/5 of the Ia correction current
before the subtraction.

If you want to add an offset adjustment, you can use
your +/-12V supplies and a trimpot to get a current
into one of the opamp summing junctions.

There are ICs to accomplish much of what I've shown,
but that's not nearly as interesting this morning.

 

[Jan Panteltje]

On a sunny day (Sat, 8 Dec 2007 05:49:52 -0800 (PST)) it happened Winfield
<[email protected]>:


Jan Panteltje says: I see a direct crystal path from +250 to ground
via Q2, Q1, Q3,Q4, and op-amp.
This sort of guarantees big bang in case Q4 has a problem, possibly
taking any other logic on the same supply as 'opamp' with it.
I would like to see some series resistor in collector Q4 perhaps.

 

[John Larkin]

On a sunny day (Sat, 8 Dec 2007 05:49:52 -0800 (PST)) it happened Winfield
<[email protected]>:


Jan Panteltje says: I see a direct crystal path from +250 to ground
via Q2, Q1, Q3,Q4, and op-amp.
This sort of guarantees big bang in case Q4 has a problem, possibly
taking any other logic on the same supply as 'opamp' with it.
I would like to see some series resistor in collector Q4 perhaps.

Yeah, it's always a good policy to have no power-ground paths through
silicon, especially as the voltage goes up. Unless the transistors are
so big they can take down the power supply.

John

 

[John Larkin]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it. I then want to have a
ground referenced voltage that is the amplified difference across the
shunt with a range of about 0-5V. I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, 5V, and probably some sort of +-12V
analog (clean) supply. I would like to get 8b or more of resolution
from the ground referenced amplified shunt voltage signal. I would
also like this circuit to be fast. The faster the better, MHz
bandwidth at least. If this were a lower voltage I'd just do an op-amp
difference amplifier. Unfortunately, last I checked, Apex had the
market for op-amps that can handle such high voltages fairly well
cornered, and their op-amps are expensive, large, and not particularly
quick. High side current amplifiers seem to top out at about 100VDC.

So I'm trying out various ideas. My favorite right now is to put
voltage dividers on both sides of the shunt to drop the voltage down
maybe to 0-10V on both and then use an instrumentation amplifier to
amplify that voltage. For some odd reason this bothers me, but I can't
place my finger on just why.

Am I being paranoid and the instrumentation amp is a legitimate idea,
or is there a better way to do this?

Thanks!

-Michael

Could you get away with a Hall sensor? The MHz requirement makes
things tricky.

If you use a cheap dc-dc converter to float a, say, +-15 volt supply
on the bus, you could use a good opamp (OP-27, LT1028) to scale up the
shunt voltage to 10 volts maybe, then you may be able to drive a
grounded diffamp at decent accuracy. Or do Win's current pumper, but
with power you can do a very precise closed-loop current source at MHz
bandwidth.

PWM with fiber optics would be fun, too.

Classic DCCTs (DC current transformers) like the Danfysik things, are
phenomenally accurate and would be ideal here, but the cheapest ones
are something like $400.

There are some cheaper open-loop or closed-loop Hall sensors, LEM or
somebody.

John

 

[John Larkin]

I suggest that you consider the benefits of a bipolar-
transistor current-replicator approach, as in my ASCII
circuit below. (view in a fixed font, like Courier)

high-voltage high-bandwidth high-side current monitor

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 ---|+_/

The circuit drops 100mV across the sense resistor at
full current. It has a 5V fs output, as you requested.

The transistor arrangement is designed to operate Q1
and Q2 at similar currents for matching, and at low
voltages to keep junction heating low. They should be
high beta types to insure accuracy at low currents.

Q4 provides the power-dissipating 250-volt voltage drop
for the output current. High-voltage transistors have
reduced beta, so up to 2% of the signal current will be
steered away from the output, but you can add a second
mpsa92, as a Darlington, to capture all the current.

Normally circuits like this are run with bias currents,
Ia, that are very low compared to the measured current.
But your maximum current is only 10mA, and you want a
1MHz or more bandwidth, so I chose a Q2 bias of 1mA,
which means it adds 10% of fs to the measurement and
has to be subtracted from the output. What's more, I
set the reflected output current Ib = 1/5 Iout, to keep
the Q4 transistor power dissipation down (550mW max), so
we have to take away 4/5 of the Ia correction current
before the subtraction.

If you want to add an offset adjustment, you can use
your +/-12V supplies and a trimpot to get a current
into one of the opamp summing junctions.

There are ICs to accomplish much of what I've shown,
but that's not nearly as interesting this morning.

You could hang a zener on the bus with a big resistor to ground, and
derive a -5 volt supply up there. Then use a r-r opamp. 9, 10 parts
maybe to deliver a grounded output voltage.

John

 

[Fred Bloggs]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it..... I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, ...

Stop right there: a "noisy" 250V supply, all of which is across the
load, and you want a sense drop in the "low" mv range?

And all these "I want" , "I would like" 's,

Go away...

 

[Jim Thompson]

Stop right there: a "noisy" 250V supply, all of which is across the
load, and you want a sense drop in the "low" mv range?

And all these "I want" , "I would like" 's,

Go away...

Scrooge is back ;-)

Cranky! Cranky! Cranky! Cranky! Cranky! Cranky! Cranky!

...Jim Thompson

 

[John Larkin]

Stop right there: a "noisy" 250V supply, all of which is across the
load, and you want a sense drop in the "low" mv range?

And all these "I want" , "I would like" 's,

Go away...

Hell, this is just along the borderline between "hard" and "fun."

The MHz+ bandwidth requirement makes it interesting; otherwise, it
would be easy.

John

 

[John Larkin]

Scrooge is back ;-)

Cranky! Cranky! Cranky! Cranky! Cranky! Cranky! Cranky!

...Jim Thompson

But not creative. Goes along.

John

 

[Winfield Hill]

You could hang a zener on the bus with a big resistor to ground,
and derive a -5 volt supply up there. Then use a r-r opamp.
9, 10 parts maybe to deliver a grounded output voltage.

Yes, and thanks to a lower offset voltage, more
accurate as well. Here's the circuit I used in
my RIS-496 +/-250-volt amplifier, adjusted for
the O.P.'s specs, with Jan's extra resistor:

10.0 +250V Iout
------+---+---/\/\----+----o ---> 0 to 10mA load
| | |
/ '----, |
\ 200 | |
/ 1% +------|--+---,
| | | | |
+--------|----, | | | 7.5V zener
| _|_ | | _|_/ |
Q1 e / -|--' | /_\ _|_
mpsa92 b --< | | | --- 0.1uF
PNP c \__+|----' | |
| | LT1783 | |
| +---------+---'
\ |
/ \ 330k
\ 100k / 0.5W
| \
| |
| gnd
| __
+-----|+ \
| | >--+--- 0 to +5.0V for
\ ,-|-_/ | 0 to 10mA load
/ |________|
\
/ 10.0k 1%
|
gnd

It does have fewer parts, 10 vs 15. The LT1783
RRIO rail-rail 300uA opamp has f_T = 1.5MHz, so
the O.P. might get his desired bandwidth.

I don't know why it's less appealing to me. I
designed both circuits, so that can't be it...

 

[Rich Grise]

Hi - I want to put a high side shunt resistor on a grounded load (~10ma
peak) that has about 250VDC across it. ....
So I'm trying out various ideas...

Have you thought of just an optoisolator? I'd heard that they're
terribly nonlinear, but I actually sat at a bench once and
measured one - a 4N something-or-other, and when I saw the
graph, I almost fell out of my chair - the CTR was linear within
eyeball percent.

And there are other opto tricks, like use a dual, with one for
feedback; use a V/F converter, optoisolate it, then use an
F/V converter; the latter two would need some kind of power
supply, of course, which could be snitched from the 250V with
a zener and resistor.

And just sticking the opto's LED in series with the load would
also give you free overload protection! ;-)

Good Luck!
Rich

 

[Fred Bartoli]

Winfield Hill a écrit :

Yes, and thanks to a lower offset voltage, more
accurate as well. Here's the circuit I used in
my RIS-496 +/-250-volt amplifier, adjusted for
the O.P.'s specs, with Jan's extra resistor:

10.0 +250V Iout
------+---+---/\/\----+----o ---> 0 to 10mA load
| | |
/ '----, |
\ 200 | |
/ 1% +------|--+---,
| | | | |
+--------|----, | | | 7.5V zener
| _|_ | | _|_/ |
Q1 e / -|--' | /_\ _|_
mpsa92 b --< | | | --- 0.1uF
PNP c \__+|----' | |
| | LT1783 | |
| +---------+---'
\ |
/ \ 330k
\ 100k / 0.5W
| \
| |
| gnd
| __
+-----|+ \
| | >--+--- 0 to +5.0V for
\ ,-|-_/ | 0 to 10mA load
/ |________|
\
/ 10.0k 1%
|
gnd

It does have fewer parts, 10 vs 15. The LT1783
RRIO rail-rail 300uA opamp has f_T = 1.5MHz, so
the O.P. might get his desired bandwidth.

I don't know why it's less appealing to me. I
designed both circuits, so that can't be it...

Because you need a small bypass cap across the 100K series resistor
otherwise you can say goodbye to your MHz BW

 

[Winfield Hill]

Winfield Hill a écrit :








Because you need a small bypass cap across the 100K series
resistor otherwise you can say goodbye to your MHz BW

Yep, right, 11 parts.

 

[John Larkin]

Yes, and thanks to a lower offset voltage, more
accurate as well. Here's the circuit I used in
my RIS-496 +/-250-volt amplifier, adjusted for
the O.P.'s specs, with Jan's extra resistor:

10.0 +250V Iout
------+---+---/\/\----+----o ---> 0 to 10mA load
| | |
/ '----, |
\ 200 | |
/ 1% +------|--+---,
| | | | |
+--------|----, | | | 7.5V zener
| _|_ | | _|_/ |
Q1 e / -|--' | /_\ _|_
mpsa92 b --< | | | --- 0.1uF
PNP c \__+|----' | |
| | LT1783 | |
| +---------+---'
\ |
/ \ 330k
\ 100k / 0.5W
| \
| |
| gnd
| __
+-----|+ \
| | >--+--- 0 to +5.0V for
\ ,-|-_/ | 0 to 10mA load
/ |________|
\
/ 10.0k 1%
|
gnd

9 parts exactly! 10 if you include the bottom amp.

It does have fewer parts, 10 vs 15. The LT1783
RRIO rail-rail 300uA opamp has f_T = 1.5MHz, so
the O.P. might get his desired bandwidth.

Not through that 100K resistor. Dang, 11 parts!

I don't know why it's less appealing to me. I
designed both circuits, so that can't be it...

You like to design circuits, whereas the momentum these days is to
connect boxes.

John

 

[John Fields]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it. I then want to have a
ground referenced voltage that is the amplified difference across the
shunt with a range of about 0-5V. I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, 5V, and probably some sort of +-12V
analog (clean) supply. I would like to get 8b or more of resolution
from the ground referenced amplified shunt voltage signal. I would
also like this circuit to be fast. The faster the better, MHz
bandwidth at least. If this were a lower voltage I'd just do an op-amp
difference amplifier. Unfortunately, last I checked, Apex had the
market for op-amps that can handle such high voltages fairly well
cornered, and their op-amps are expensive, large, and not particularly
quick. High side current amplifiers seem to top out at about 100VDC.

So I'm trying out various ideas. My favorite right now is to put
voltage dividers on both sides of the shunt to drop the voltage down
maybe to 0-10V on both and then use an instrumentation amplifier to
amplify that voltage. For some odd reason this bothers me, but I can't
place my finger on just why.

Am I being paranoid and the instrumentation amp is a legitimate idea,
or is there a better way to do this?

---
This oughta work: (View in Courier)

.. FWB REG
.. +----+ +-----+
..20AC>-+ +--|~ +|---+---| |--+------------+------>Vcc
.. P||S | | |+ +--+--+ | |
.. R||E | | [BFC] | | |
.. I||C | | | | +--|----+ |
..20AC>-+ +--|~ -|-+-+------+ | | | | ADC
.. +----+ | | Vcc [RFB] +---+---+
..250DC>-----------+-+-[Rs]--+---+-|-\ | | Vcc |
.. | | | >--+---|IN OUT1|-->ADOUT1
.. +---------|---+-|+/ . .
.. | | GND | OUT8|-->ADOUT8
.. | | | | GND |
.. | +--+ +---+---+
.. [LOAD] | |
.. | +------------+
.. |
..250GND>--------------------+


Typical ADC output circuitry:


.. Vcc
.. |
.. [R]
.. |
.. | OPTO
.. +-+----+
.. | A C|--> To logic
.. | |
.. | K E|--> To logic ground
.. +-+----+
.. |
.. D
..ADOUTn>--G
.. S
.. |
..250DC>-----+

Note that everything except the load is running with 250VCD as its
common, so if you need to talk to the ADC you'll need to do it
through optocouplers unless you can use the same common for your
micro. Pretty dangerous if you don't know what you're doing!

No offense intended.

 

[Phil Endecott]

Hi - I want to put a high side shunt resistor on a grounded load
(~10ma peak) that has about 250VDC across it. I then want to have a
ground referenced voltage that is the amplified difference across the
shunt with a range of about 0-5V. I would like the voltage across the
shunt to be in the low mV range. Power supplies available to me will
be the 250V, which will be noisy, 5V, and probably some sort of +-12V
analog (clean) supply. I would like to get 8b or more of resolution
from the ground referenced amplified shunt voltage signal. I would
also like this circuit to be fast. The faster the better, MHz
bandwidth at least. If this were a lower voltage I'd just do an op-amp
difference amplifier. Unfortunately, last I checked, Apex had the
market for op-amps that can handle such high voltages fairly well
cornered, and their op-amps are expensive, large, and not particularly
quick. High side current amplifiers seem to top out at about 100VDC.

So I'm trying out various ideas. My favorite right now is to put
voltage dividers on both sides of the shunt to drop the voltage down
maybe to 0-10V on both and then use an instrumentation amplifier to
amplify that voltage.

I built myself something like this for scoping mains currents (order of
10-100mA) into power supplies, and it seems to work OK. In my case I
have a 4R7 series resistor and a pair of 18k:330k potential dividers,
into an AD620 instrumentation amplifier with a +/- 9v supply. I'm not
really sure what the frequency response is, because I don't have a
better measure to compare it with. (High frequency components are
probably also messed up by the fact that my -9v comes from a
switched-capacitor voltage doubler!) The box also has a scaled-down
voltage output and I feed both into my scope on "multiply" mode with a
whole 50Hz cycle visible. It has way more frequency response than is
needed for this, but whether it gets anywhere near a MHz I don't know.

I've put a picture of my contraption here:
http://chezphil.org/tmp/mains_probe.jpeg

For some odd reason this bothers me, but I can't
place my finger on just why.

Try it out and see if the reason jumps out at you.

I think that the best solution would involve a battery, an ADC, some
fibre-optic, and either a DAC into a scope or directly capturing the
digital signal on your PC. I have a very vague feeling that I may have
seen a circuit like this in a applications note or something like that.


Cheers,

Phil.

 

[Winfield Hill]

9 parts exactly! 10 if you include the bottom amp.


Not through that 100K resistor. Dang, 11 parts!


You like to design circuits, whereas the momentum
these days is to connect boxes.

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 ---|+_/

 

[John Larkin]

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.

One more resistor if you're willing to run a bit of current through
the shunt. Two more if not.

Actually, at MHz speed, to keep the bus noise from coupling through
capacitances, one might decide to throw a differential current down at
ground, which also fixes the zero problem. More parts on the high
side, ballpark 12, and a few more near ground.

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.
Chopamp.




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 ---|+_/

Won't the 1M squirt in an error proportional to bus voltage? Are you
trying to compensate the collector resistance of Q1?

You could take the standing-current offset bias thingie from the
zener, or better yet use a bandgap.

John