2
\$\begingroup\$

I have the following situation: I have a signal with periodic, short (~2ns), ~1.5V pulses spaced ~10-20ns apart. I would like to detect when the pulses disappear, i.e. when there is no pulse for ~1us or so.

I am currently planning on using a first comparator to detect the pulses, then a low-pass filter to followed by a second comparator to keep the output on as long as the pulses arrive frequently enough. Something like the following:

enter image description here

Here, V3 is the pulse source, V1 is the threshold for the "pulse-detection" comparator, and also for the output of the low-pass filter. (I understand that the threshold for U2 might need to be different from the one for U1, let's just assume that the thresholds work out)

I tried to simulate this in KiCADs spice simulator, but ran into errors regarding singularities in the system, which makes me a bit worried.

As I understand it, the first comparator should just charge the capacitor C6 through R5, and drain it again in between the pulses (so it should work, even if there is no actual load attached to the RC filter). Will this circuit work as designed?

\$\endgroup\$
2
  • \$\begingroup\$ Seeing as the response time of the LM393 is measured in microseconds and you are giving it single-digit nanosecond pulses, I think that might be what is giving the simulator a fit. \$\endgroup\$
    – vir
    Commented Aug 11, 2023 at 21:17
  • \$\begingroup\$ @vir Thanks, that might indeed be a problem - I just assumed that since I don't need the detection speed to be that high, I might get away with the slow speed of the comparators... That being said, I have removed the model number from the image now, since I'd like to ask the question regarding the general circuit, not a particular choice of components \$\endgroup\$
    – Lukas Lang
    Commented Aug 11, 2023 at 21:22

1 Answer 1

3
\$\begingroup\$

The integrated voltage across C6 in your schematic is not related to the input voltage level. So you need another, probably lower reference voltage for U2A.

May I suggest a simple BJT solution?

schematic

simulate this circuit – Schematic created using CircuitLab

If you use an RF transistor with FT >= 1 GHz as Q1 you can integrate the pulses in C1.

Q2 can be a slower standard type. It is not a proper comparator, but provides the required signal.
The simulator needs some time to work it out.

\$\endgroup\$
4
  • \$\begingroup\$ A classic missing-pulse problem and solution. I would suggest moving R3 to the left of R2, in direct parallel with C1. this should increase the independence between the resistor value for the detection window and the resistor value for the output transistor base drive. \$\endgroup\$
    – AnalogKid
    Commented Aug 12, 2023 at 0:51
  • \$\begingroup\$ @AnalogKid Yes, I agree, I started the schematic with R3 on the left side as you suggest, but I need a voltage divider for the base drive anyway. The result was better with R3 on the right side, at least in the simulation. I may have used bad values in the first approach. I'm not happy with the input circuit at these frequencies, but the OP didn't specify a source impedance. \$\endgroup\$
    – Jens
    Commented Aug 12, 2023 at 1:12
  • \$\begingroup\$ Thanks! So you are effectively suggesting the "same" structure for the circuit, but with simple transistors instead of the comparators? Can you explain what R1 is for exactly? What I forgot to mention in the question is that the entire thing is supposed to "monitor" a signal that is transmitted to another device over a 50Ohm transmission line (with the receiver having the proper termination), which is why I need a high-input impedance for this circuit. But as I understand it, that's still true for your circuit, even with R1? \$\endgroup\$
    – Lukas Lang
    Commented Aug 12, 2023 at 8:56
  • \$\begingroup\$ @LukasLang The 1.5 V pulse amplitude, R1 and the Vbe voltage drop define the peak base current of around 16 mA. The current gain of Q1 at these frequencies is not much above 2 and so R1 defines the charge current of C1. You need another input stage if you want to monitor a transmission line. This is worth a question of its own if the impact on the signal must be low. Even a high speed comparator has a relevant input capacitance and your T connector will produce losses. \$\endgroup\$
    – Jens
    Commented Aug 12, 2023 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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