Simple Integrator Circuit

[ThermalRunaway]

Hi guys,

I'm busy doing some assignment work and I've come across a troublesome question regarding an Integrator circuit based around an Op-Amp. I understand what an integrator is and what it does, but the question that has been put to me seems a little ridiculous - unless of course I'm misunderstanding something.

The question is as follows. A circuit has been provided as part of the question where a 741-OpAmp is setup as an integrator. There's a 2uf capacitor for the feedback, and a 500K resistor connected to it's inverting input. A square wave running at 1Khz with a 1:3 mark-space ratio is applied to the input, so one complete cycle of the input signal takes 1ms and it's only on for a 3rd of the time that it's off. The amplitude is +4V for the on-time, 0V for the off time.

The author of the question wants me to sketch a graph of what would happen at the output for the first 5ms, assuming it's 0V to begin with.

And here's where my confusion begins. Obviously the output is going to decrease linearly at a rate which depends upon the amplitude of the signal (+4V) and the time constant of the CR circuit. But the time constant for his circuit is 1s, and it's only at +4V for 1/3 of a millisecond each cycle. So to me, not a great deal will happen at the output during 5ms!!!

I've tried simulating the circuit on multisim but it doesn't seem to like it - the output falls immediately the the bottom supply rail (-15V) and stays there.

The author of the question also wants me to calculate how long it will take for the output to reach -2V and I'm confused about that too!!!

If anyone can offer any constructive advice I'd appreciate it, and in the meantime I'll do some googling and some scratching of head.

Brian

 

[on1aag]

Desintegrator.

Hi Brian,

After 5 mS the output voltage will reach -5 mV, that's 1 mV / mS.
Do I have to calculate the time for it to reach - 2 volts ?

on1aag.

 

[Cabwood]

When the input is +4V the output voltage drops by 4V/s. When the input is zero the output remains constant.

For each cycle, the output drops linearly for 250us, and then stays where it is for 750us. The drop during each cycle is 250us x 4V/s = 1mV

So during the first 5ms the output does the above 5 times. After 5ms the output voltage will be 5mV lower than it was at the start of that period.

If for each cycle of 1ms the output drops 1mV, and assuming an initial output of 0V, it will take 2000 cycles to reach -2V. That's 2s.

 

[ThermalRunaway]

Thankyou both for your replies. I had actually sat down and worked all that out for myself and, thankfully, the graph I've drawn agrees with you cabwood!

Thanks again for the advice, appreciated.

Brian