Problem statement

in one of my design i need to stretch narrow 10 ns pulses coming from photo diode after TIA stage, if i have to measure the pulse amplitude i have to stretch it at least by a factor of 100, in this search i have found a approach in which the circuit was using a capacitor after its feedback path, but i have not understood how exactly the circuit works, but when i tried to replicate it, no pulse stretch was happening, here below is the circuit and kindly suggest few modifications to the circuit or suggest an approach for the analog pulse stretch

circuit enter image description here Opamp is a high slewrate 3000V/us and high BW of 100M as BW = 0.35/risetime and slewrate = 2* pi * BW * Vpp, so for my 10ns input the opamp would do the job.

pulse output is just zero !

enter image description here

EDIT1: primarily i forgot to mention that my 10ns pulses will repeat at a rate of 1Hz-50Khz, so just after some small modifications, i came with below circuit where i increased the gain and obtained pretty decent results, kindly provide your valuable comments

enter image description here


enter image description here PS: i was using TINA from TI as simulator in above case

  • 1
    \$\begingroup\$ A signal of 1nsec requires a gain bandwidth product of 1GHz, well beyond the capability of your amplifier. Note that slew rate and GBW are not directly related. The slew rate shows the fastest output available, not what the internal gain stage can handle. \$\endgroup\$ – Peter Smith Apr 4 '16 at 10:15
  • \$\begingroup\$ @PeterSmith find the update please \$\endgroup\$ – kakeh Apr 18 '16 at 3:11

You are trying to charge a 330 pF capacitor up very quickly. Going by information from your other question this needs to be done in about 1ns. Given your pulse maybe be (say) 1V in amplitude you can work out how much current the op-amp has to supply to charge the capacitor up....

Q=CV or...

dQ/dt = C dV/dt (note that rate of change of charge equals current)

So current = 330pF * 1V per nanosecond = 330 mA

You are not going to find a fast enough op-amp that can supply this amount of current with any degree of accuracy.

You are going to be looking at something around 10pF rather than > 100 pF and you are going to need a much better op-amp as has been mentioned in the other answer.

  • \$\begingroup\$ please find the updated modification and its results \$\endgroup\$ – kakeh Apr 18 '16 at 3:12
  • \$\begingroup\$ 90mA can be supplied by the op-amp to charge a 500 pF capacitor. To charge to 1V will take 5.6 ns. If you can live with this then that's fine but remember if the pulse is (say) 3 volts it'll take 3 times longer. Also remember that a simulation may not perfectly represent reality. Things like maximum output current may not be simulated in the model. \$\endgroup\$ – Andy aka Apr 18 '16 at 7:26

The gain bandwidth product of the LM6172 is insufficient to handle a 1 nsec pulse; note that GBW and Slew rate are not directly related. See this application note.

For this you could try something like the LTC6409 or the LMH5401 which have a GBW sufficient for this application.

Other solutions exist, but it is highly unlikely a simple breadboard circuit would operate properly, and layout for such a circuit will require significant attention to detail.

  • \$\begingroup\$ can you please suggest the other solutions also, it would be a great help \$\endgroup\$ – kakeh Apr 5 '16 at 2:58
  • \$\begingroup\$ kindly note that my pulse is 10ns not 1ns, sorry if you have misunderstood from my question \$\endgroup\$ – kakeh Apr 5 '16 at 3:01
  • \$\begingroup\$ please find the updated modification and results \$\endgroup\$ – kakeh Apr 18 '16 at 3:13
  • \$\begingroup\$ seems like both are FDAs, if my next stage is comparator and not ADC can i just ignore one of the outputs ? \$\endgroup\$ – kakeh Apr 26 '16 at 7:32

Perhaps send the pulse into a low pass filter. 50 ohms and ... 1nF or so (if I punched the numbers right.) Then amplify that signal with your slower opamp. Or design the TIA to give longer pulses.

  • \$\begingroup\$ can you please throw some light on design of TIA to give longer pulses,seems pretty interesting, how is that possible ? \$\endgroup\$ – kakeh Apr 5 '16 at 2:59
  • \$\begingroup\$ Well lower bandwidth, to make the pulses longer will also reduce the voltage. Do you have a schematic for the TIA? Typically increasing the feedback resistor will decrease the BW. \$\endgroup\$ – George Herold Apr 5 '16 at 16:30
  • \$\begingroup\$ currently my TIA is based on OPA656 with a 12M feedback resistor for gain and 4pF feedback cap to reduce ringing effects \$\endgroup\$ – kakeh Apr 6 '16 at 2:53
  • \$\begingroup\$ @kakeh, Ahh, that is a 500 MHz opamp...What's the photodiode capacitance? I don't see anyway you are getting 1 ns pulses from that. (Got a 'scope shot of the output?) (12 Meg and 4 pf is ~ 50 us time constant!) You can use an led to measure the BW. \$\endgroup\$ – George Herold Apr 6 '16 at 14:02
  • \$\begingroup\$ photo diode capacitance is around 12pF, I am using the feedback capacitance 4pF in parallel with resistor Rf to reduce the parasitic effects of amplifier, cap is not placed after the opamp stage \$\endgroup\$ – kakeh Apr 7 '16 at 2:36

Even if the original poster has solved this problem long ago, this sort of situation does come up frequently so here is my answer, over five years later...

This is similar to problem SLAC had when designing the "Z zero Factory" in the 1980s. Very narrow pulses would arrive from beam position sensors. Their amplitudes had to be digitized for use in steering the beams. No beam intersections, no Nobel Prize! I spend about a year working with the EEs and techs calibrating the electronics, so I've seen this before.

The pulse were first put through a passive low pass filter, specifically a Gaussian filter that would produce an approximately Gaussian-shaped (iirc) pulse with a peak occurring several nanoseconds after the input pulse. Pure LC circuitry, with no noise, slew rates or anything else to screw with the measurements. A sample/hold was triggered at the expected time of that peak. The delay of the Gaussian peak relative to the original pulse gave time for the trigger circuitry to work. The S/H used a diode bridge powered by, as I recall, four transistors arranged in an H-bridge, like a motor controller. The pulse edges of the S/H did not have to be anywhere as sharp as the edges of the original input pulse, though sharp by any ordinary standards.

We did not try to make diode-based peak detector as in the diagram you show. The forward bias of the diode would make calibration at lower beam intensities (smaller pulses) difficult. You overcome forward bias with an op amp, a standard textbook circuit perfect for such things in audio and sub-MHz applications, but for accurate measurement, forget it. Gain-bandwidth product, slew rate, stray capacitance, blah!

Good results are obtained with passive filter pre-processing, then using digitization techniques that are still quite fast but able to handle pulses on a scale of tens of nanoseconds.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.