# Operational Amplifier

Discussion in 'Homework Help' started by KyleJones91, Jan 6, 2016.

1. ### KyleJones91 Thread Starter New Member

Jan 6, 2016
3
0
Hi All,

Im new here so please go easy

I have got a piece of work to do regarding op amps (integrator and differentiator circuits)

I have to use my own figures to show how the op amp works

I understand that to work out voltage out = RC dvin/dt

Say for example my Capacitor is 4.7mf and I have 35V input aswell as a 1kohm resistor, I got 1000*0.0047 35v/??

How do I work out the time, what has the time got to do with it?

Kyle

2. ### artmaster547 Member

Jan 6, 2016
89
1
If you look at the formula you can see that its voltage differentiated with respect to time, and not just a division, the capacitor in this scenario the voltage across the capacitor is effected by time, the resistance and capacitance in theory are fixed values as you have stated in the calculation. If you have a value for V out you could work out the time by integrating everything with respect to time.

3. ### KyleJones91 Thread Starter New Member

Jan 6, 2016
3
0
Thanks very much artmaster547.

I have been told to use my own values, so if I (make up) a voltage out, then I can work out my time across the capacitor?

So say I have an output of V

Am I right in thinking 1000*0.0047 35v/10 seconds = 4.7*3.5 = 16.45v output?

4. ### artmaster547 Member

Jan 6, 2016
89
1
Yeah that's right, and as you said if you make up a value for the output voltage you can switch it around and workout time, when you integrate with respect to time you would get either two solutions:
i.e. if you have two different times t1 and t2:
then you get Vout*t2-Vout*t1=RCVin
or usual what you have is from 0 to time t (it is most likely this equation):
Vout*t-Vout*0=RCVin
Which is of course:
Vout*t=RC*Vin
and you rearrange but it's important you don't treat dV/dt as a fraction but in most questions it ends up that way