OpAmp Integrator with fast current pulse input

op amp integrator

I'm trying to determine if an opamp, such as LM124, can integrate the current of a very fast pulse. I am simulating the circuit in PSpice. The problem is that the voltage proportional to the integral is appearing at the input rather than the output. Should this be expected? I've been told that such a voltage should not appear at the input because of the virtual ground (the positive input is grounded).

For my input current pulse, the voltage across the capacitor should rise 75mV.
The current pulse is a triangle:
t=0, i=0
t=3ps, i=10mA
t=15ps, i=0

fast integrator circuit

joker12 said:

I'm trying to determine if an opamp, such as LM124, can integrate the current of a very fast pulse. I am simulating the circuit in PSpice. The problem is that the voltage proportional to the integral is appearing at the input rather than the output. Should this be expected? I've been told that such a voltage should not appear at the input because of the virtual ground (the positive input is grounded).
For my input current pulse, the voltage across the capacitor should rise 75mV.
The current pulse is a triangle:
t=0, i=0
t=3ps, i=10mA
t=15ps, i=0

Click to expand...

In your circuit, the positive input is NOT grounded (contrary to your description)
Nevertheless, in principle the circuit could work also for +1V at the + input.
But why this dc bias ?
However, the 1 pf cap is too small if compared with input/output caps of the opamp model (and with parasitic caps of the real circuit). Therefore the strange behaviour with a weird output voltage.
Use 10 or 100 pF instead - with the consequence of lower output voltage.
Are 75 mV required ?

op amp integrator circuit

joker12,
I believe the problem is that the LM124 is not nearly fast enough to respond to such a fast pulse. Your input current pulse is 10^-2 Amperes. This current must be matched by the current through the feedback (integrating) capacitor. The current through the capcitor is i = C dv/dt. Or dv/dt = 10^-2/10^-12 = 10^10 volts/second. I dont believe that any op-amp can provide this slew rate. The result is, during the pulse period, the op amp output remains at virtual ground. Essentially you have a passive RC integrator with the output appearing at the op amp (-) input. You might try simulating the circuit with an ideal op amp. i believe this circuit will behave as you expect.
Regards,
Kral

lm324 integrator

It seems somewhat - excuse me - ridiculous, to use a LM324 integrator for ps current pulses. Generally, you can expect the LM324 output to give a correct charge measurement, as long as no excessive input current is caused at the inverting input. Applying 10 mA current however simply forces the input substrate diodes into conduction, thus the charge measurement is lost.

There is a very simple means to avoid this kind of dynamic overload. Connect a passive integrator in front of the active OP integrator. A pf capacitor to ground would be sufficient. In a real circuit, it's present anyway. For a more precise charge amplifier operation, a faster OP should be used, as said. But basic cahrge amplifier operation is possible even with presented circuit. If you are intending to use an antique OP, an µA741 would be more stylish to my opinion.

opamp integrator

I think there will be a 1V bias when connected to the test subject.

Yes I could use a 10pF because I don't need 75mV.

Yes, I thought about the slew rate and bandwidth not being good enough. Which opamp would you suggest?

I started taking the simulation out to the miliseconds, and the output voltage actually does reflect the integration or charge after the initial spike in the pico-seconds. The charge might decay to quickly though. If I used a larger feedback resistance it decays slower. I try using no feedback resistance and PSpice has an error.

op-amp integrator

Yes, without a feedback resistor, you need another reset method for the charge amplifier. However, it doesn't make much sense to use a high ohmic resistance with a bipolar amplifier, because the drift and particularly current noise are unsuitable high.

So if you need to extent the charge amplifier operation to a long time constant, you should use a FET amplifier.

op amp integrator

Thanks. By FET amplifier, do you mean something like a common-source amp?

I think I'm going to forget about the active integrator. My simulations were producing some ok results, but I don't think any opamp is nearly fast enough for this type of pulse.

I'm now looking at a passive integrator, which was basically be a capacitor connected to ground. Ground can supply the opposite charge readily. Then we want to use a voltage follower configured opamp (or something else, suggestions?) to buffer the voltage for measurement. Still, the opamp will take time (slew rate). The problem is, even with the single current pulse, the voltage on the capacitor starts to take off towards the rail voltage after about 100ns. I don't know why that would happen. Do you call this drifting?

I also tried an instrumentation amp in simulation and the input voltage did the same thing.

I just need some current to voltage device to measure this very fast single pulse of current. Do you have any suggestions?

pulse integrator circuit

I primarly meant a FET OP. But I aggree, that a feedback integrator isn't necessarily needed by the application.

Unfortunately, you didn't really specify your problem. At a simulation level, drift is more or less meaningless, because you can simply assume ideal parts. So as the more interesting question, what's the real world problem, that you are trying to simulate?

How about offset currents in the real application? What defines the measuring time respectively integrator time constant? Depending on the application, it can be microseconds up to seconds. I prefer suggestions for real world problems or at least realistic simulation set-ups.

integrator operational amplifier input current

A transistor's gate oxide layer is going to have an ion beam incident upon it, which will produce the current pulse. I want to know the charge from the current.

We need to maintain an acurate charge representing voltage for about 1 microsecond or maybe less.

fast op amp

Thank you for the clarification, it's a classical charge amplifier applications, I think. 1 us is comfortable short. The timeconstant requirements depend also on the ability of the connected measurement electronics, e.g. an oscilloscope, to detect the pulse peak and determine it's magnitude.

For a charge amplifier or passive integrator with time constants in the us range, the amplifier selection isn't very critical I think. An amplifier with some MHz bandwidth and low input current should be O.K. I don't see a general drift problem with this setup. Of ourse, also a passive integrator must have a bias resistor respectively a finite time constant.

As an active integrator advantage, the pulse height doesn't depend on the connected input capacitance.

fast integrator

Ok, I have a new design I'm going with, but I think I need a 1nF low-esl capacitor for the current pulse. Where can I buy that? I looked in Digikey and Mouser. Digikey doesn't have any in stock and they require a purchase of 100-400.