0
\$\begingroup\$

I have an inverting integrator

enter image description here

that integrates the input signal (green) into the blue signal.

enter image description here

If I do nothing, the integrator output will go down and down and down... I don't want this. I want to reset my integrator back to ~9V when the second green pulse comes.

With some knowledge in digital circuit, I already produce the "switch signal" (green narrow pulses) as shown below.

enter image description here

Finally, I need to build a switch circuit that resets the integrator (basically C1) every time it sees the green narrow pulse. How could I build this switch circuit?

My Attempt

I tried using an npn BJT who is on when V_{BE}>0 and V_{BC}>0. However, this is not working, maybe because my green pulse is only 1V (from digital circuit) and the integrator output is as high as 9V. There is no way whereby V_{BE}>0 and V_{BC}>0. Also, I feel this way, the switch circuit is somewhat "disturbing" the integrator.

\$\endgroup\$
1
\$\begingroup\$

Funny but it seems like a very similar question has been asked just recently, and a piece of a circuit I've used before may help you too. This would be my approach.

Integrator with auto reset

You would have to set the voltage divider resistors at the (+) input of the left OP amp so that the input pulse would cross the threshold. If you think the pulse will be less than perfect, you might also want to experiment with an high added resistance to offer some hysteresis. When the pulse goes high, the output of that first OP am will quickly switch low, and the output of the integrator will begin to rise slowly over time. When the pulse goes low, however, its output will quickly switch high, and the output of the second OP AMP will drop almost immediately, like a RESET. This is due to the much lower resistance (and hence shorter) integration timing for the RESET case. You can reverse the action to behave like your version by switching the twodiode resistors, putting the voltage divider on the (-) input of the first OP AMP, and feeding your input to its (+) input through an additional resistor. The important thing is that now you'll have a RESET after each pulse event. If you were sampling the output of the second OP AMP, and saving the highest result recorded every time the pulse went low, you will a have a good measure of the pulse duration. In addition, I'd argue that resetting AFTER the event is always better than resetting at the start of the event. The reason is that any reset is likely to take a short but still non-zero amount of time. So resetting AFTER the accumulation ensures that any reset time does NOT corrupt your pulse duration measurement.

\$\endgroup\$
  • \$\begingroup\$ Thanks a lot. The interesting thing is the signal that I need is actually from the - pin of the right op amp, instead of its output pin. This is weird. Also, the reason why I wanted to reset at every pulse's start instead of the end is that I wish to show the value on LED for a period of time. What you said totally makes sense. But seems that I need another small circuit to latch the peak so as to be shown for a period of time on LED. \$\endgroup\$ – Sibbs Gambling Oct 16 '14 at 3:02
  • \$\begingroup\$ What if I insist discharging the integrator at the start of a hump? How should I modify yours design? Thanks! \$\endgroup\$ – Sibbs Gambling Oct 16 '14 at 5:01
  • 1
    \$\begingroup\$ Just remember that you can't reset in zero time. So if you insisted in resetting at the start of the pulse, you would first have to decide how much time and precision you're willing to sacrifice. Wouldn't it be better to add maybe a simple one gate DELAY circuit in the path between left OP-AMP out and the top diode? Then, between the time the pulse ends and the time delay expires, OP-AMP-2 will HOLD its output, giving you time to "READ" it. I'm new to this forum and am not sure how to add an additional circuit to a comment, but maybe what I've suggested already will help? \$\endgroup\$ – Randy Oct 16 '14 at 14:03
1
\$\begingroup\$

First things first, you haven't shown the full circuit. The full circuit of a resettable integrator has a resistor on the input like this: -

enter image description here

Without the input resistor the capacitor will charge at a rate determined by the output impedance of the op-amp and not in a controlled manner unless you are injecting a current into the input but I can see you are not because your green traces are indicating against a scale of voltage on your graph.

Note that there is also shown a switch across the capacitor to discharge it or "reset" the integrator and this is the key to perhaps what you want to achieve. If this suits your requirements, try using an analogue ("analog" for west of the Atlantic ocean) switch - it contains FETs that can be controlled digitally to short out the capacitor very quickly.

Here is also a useful tutorial on integrators.

\$\endgroup\$

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.