# 741 Integrator Circuit

#### BrianH

Joined Mar 21, 2007
43
Hi guys,

I've come across a troublesome assignment question regarding a simple integrator circuit and, whilst viewing some information regarding Integrators on the AllAboutCircuits website I came across this forum. I haven't managed to find any information which would clear up my misunderstanding of the assignment question so I thought I'd give you lot a try - perhaps something will immediately jump out at you which I hadn't considered.

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 in multisim but it doesn't seem to like it - the output falls immediately to 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

#### BrianH

Joined Mar 21, 2007
43
Well, I've sat down with this, scratched my head for a bit, and I've managed to draw my graph. The output of the integrator drops by 1mV for a 3rd of a millisecond, then stays constant for the remaining 2/3 of a millisecond. This continues with each cycle so that by 5mS the output has dropped to -5mV.

I actually asked this same question on another forum and the answer I received there agreed with what I've done here so hopefully I'm right!

Thanks anyway guys

Brian

#### hgmjr

Joined Jan 28, 2005
9,027
It would be very helpful if you could provide an attachment showing the circuit and the sketch you have developed.

hgmjr

#### BrianH

Joined Mar 21, 2007
43
Well for my part my original misunderstanding is completely cleared up so I don't need any further advice, however it might be a good idea to post that information incase anyone else is interested at a later time.

I'll need to do that this evening so please bare with me.

Brian