I want to make a circuit with an output impedance of 50 ohms.

I am dealing with negative photomultiplier tube pulses that in 50 ohm load look like in the picture below:

enter image description here

I want to use the pulses to trigger the one shot sub-circuit described here which uses positive signals:

enter image description here

I understand that I must use a simple sub-circuit with an inverting fast operational amplifier.

Is it possible to make the op-amp sub circuit have an \$ Z_{out} \$ = 50 ohm so it can drive the one shot sub-circuit?

  • 1
    \$\begingroup\$ Andy's advice is spot on for Zo, however you can modify the one-shot to accept negative input pulses \$\endgroup\$
    – sstobbe
    Commented Apr 15, 2019 at 19:00
  • \$\begingroup\$ @sstobbe hello and thank you for commenting , such a modification would that would be very good . Do you have any suggestion how yo modify it ? Best regards. \$\endgroup\$
    – Chris Sy
    Commented Apr 16, 2019 at 2:15
  • \$\begingroup\$ Hello! I am really interested in this question since I am facing the exact same issue. How did you end up solving it? We're you able to finally detect the PMT pulses? \$\endgroup\$ Commented Mar 24, 2023 at 14:34

1 Answer 1


Is it possible to make the op-amp sub circuit have an Zout = 50 ohm so it can drive the one shot sub-circuit?

A basic op-amp circuit is usually one where its natural output impedance is close to zero ohms but, this is only true at low frequencies. For instance, a "normal" op-amp circuit might start to have significant output impedance as frequency rises above (say) 1 MHz. This is dictated by the data sheet i.e. faster op-amps will maintain a lower output impedance for higher frequencies but there will always be a tipping-point and you (as the designer) have to decide how to tackle this.

For your one-shot circuit I see that the amplitude required at the input can be a few tens of milli volts and this helps a lot because you can use a potential divider on the output of the op-amp to convert a naturally high voltage (like a pulse of 2 volts p-p for instance) to a signal level 20 mV p-p.

This requires an attenuation of 100:1 so, if you used 5000 ohms and 50 ohms to give an approximate attenuation of 100:1 you get a natural output impedance of about 50 ohms irrespective of what the op-amp circuit might be struggling to deliver.

  • \$\begingroup\$ Hello and thank you for answering, i don't seem to understand you mean i should connect the output of the opamp to a 5k and 50 ohm resistors in series and connect the point above the 50 ohm to the one shot input ? ( it seems i can't use images in comments to show you what i understand , sorry about that) \$\endgroup\$
    – Chris Sy
    Commented Apr 15, 2019 at 9:36
  • 2
    \$\begingroup\$ Yes, that's what I'm suggesting - your op-amp should be able to deliver a couple of volts pulse to the attenuator and the attenuator will deliver 20 mV p-p to your one-shot circuit. You would probably get-away with a 4k7 resistor directly feeding the 49.9 ohms on the one-shot circuit. It won't be critical if wire lengths are short. \$\endgroup\$
    – Andy aka
    Commented Apr 15, 2019 at 10:02

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