# OP-AMP circuit confusion

Discussion in 'Homework Help' started by andy24691, Feb 16, 2012.

1. ### andy24691 Thread Starter Member

Nov 25, 2010
42
0
I would like to know whether this circuit can be used to increase the output voltage across the capacitor by momentarily connecting the base of the PNP to the -3V terminal. I'm unsure how to calculate the current through the transistor when it saturates, if it has negligible resistance when turned on does that not mean a large current will flood through and exceed the the collector current rating. Do I need a current limiting resistor? I would like to know what the OP-AMP is trying to do when the inverting input is connected to the 10V.

cheers

2. ### jimkeith Active Member

Oct 26, 2011
539
99
Yes, you need a resistor in the emitter of the transistor to limit the current.
When the base of the transistor is connected to -3V, the current through the resistor will be approx (3V-0.6V)V/R, assuming that the op amp input node is 0V.
This will cause the output of the op amp to integrate in the positive direction until it reaches saturation, than the input node will eventually discharge to the emitter voltage.

3. ### andy24691 Thread Starter Member

Nov 25, 2010
42
0
Could you tell me how you derived the emitter current, these things aren't obvious to me and I need them explained. I've been trying to work it out on a piece of paper and I'm at a bit of a loss I'm afriad.

cheers

4. ### jimkeith Active Member

Oct 26, 2011
539
99
Too simple to see...
You are starting with -3V on the base
the pnp emitter follower has the emitter approx 0.6V higher than the base, so Ve = (-3 + 0.6)
I = E/R, so I = (-3 + 0.6V)/R
(the sign is different than the previous post because it now indicates direction of current)
For practical limits, I should not exceed about 10mA, because that is what the op amp can easily source.

5. ### andy24691 Thread Starter Member

Nov 25, 2010
42
0
I think I'm getting confused because the emitter is connected to 0V, yet is 0.6V above the base voltage and it can't be both. But if a resistor is connected in between the inverting input of the op-amp and the emitter then the two are not directly connected and the 2.4V is dropped across the resistor. Do I need some sort of resistor to limit the base current also, the voltage to the base is from the output of a multiplexer. Here's a diagram showing whats going on, my lecturer said you should be able to get -3V at the base of the transistor with a single resistor in between the comparator and multiplexer though this didn't seem to work in the lab.

*The resistor connected to ground in the voltage divider should be 47K not 22k

Last edited: Feb 17, 2012