I have designed a peak detector to capture the peaks of high speed pulses. This is the circuit:Peak detector circuit

This is the waveform obtained on simulation: enter image description here

Waveform when input peak is about 2.5 V:

enter image description here Orange is PR3, output of peak detector. Blue is PR4, output of Op- Amp U1. Red is PR1, input pulse. I use Comp_1 to discharge the cap C2, after the pulse has ended. This is done to capture peak of all the pulses (regardless of the value).

Could someone suggest how I can fix this issue?

  • \$\begingroup\$ @Trevor, it doesn't help. On the contrary, the value stored during pulse is further lower than the actual peak. \$\endgroup\$ – Ash Oct 30 '17 at 20:17
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    \$\begingroup\$ You're powering the op-amp from VDC5V0. Presumably this is 5.0 V. How close to the positive rail is LT1809 able to drive its output? Does the circuit work if the input peak is only at 2.5 V? \$\endgroup\$ – The Photon Oct 30 '17 at 20:27
  • \$\begingroup\$ @ThePhoton, Yes VDC5V0 is 5V. I tested the op-amp as a unity gain buffer, it was able to follow the input without issue. With a 2.63 V peak, there is a drop initially ( Vpeak= 2.54V). However due to bias current, the peak increases to 2.68 V before the pulse ends. \$\endgroup\$ – Ash Oct 30 '17 at 20:36
  • \$\begingroup\$ When you did the buffer test, what was the load on the output? (Try with a load that will give a similar current to what will be needed in your final circuit) \$\endgroup\$ – The Photon Oct 30 '17 at 20:37
  • \$\begingroup\$ I just ran the sim again, the load is 220 pF cap, with series resistor 33 ohm. Feedback taken at ouput of op-amp. Resistor is used to avoid settling issue. The output follows the input. \$\endgroup\$ – Ash Oct 30 '17 at 20:45

For high speed pulses, I would avoid amplifiers with fancy things in their feedback path.

Go back to the basic principles of how a diode or BJT work. A diode or emitter follower driving a capacitor is a peak detector. You can use a FET to reset the capacitor voltage to 0 to be ready for the next peak, or a resistor that bleeds off the voltage with a long enough time constant to not interfere with whatever you are trying to do.


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