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

On Dec 23, 7:30 pm, Winfield <winfieldh...@xxxxxxxxx> wrote:
On Dec 23, Michael wrote:





On Dec 9, Winfield Hill wrote:
John Larkin wrote:
Winfield Hill wrote:
John Larkin wrote:

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

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.

 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 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.

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

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.

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
.

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