Op Amps

[JasonMcG]

Hi

I need help with this problem and how to go about doing it.

An inverting amplifier of closed loop gain of -15 is designed with an assumption that the op amp is ideal. Then the inverting amplifier is connected to a source 10 mV peak sinusoidal signal whose source resistance is 100 kΩ.
i) If R1 of the inverting amplifier is 10 k, draw the equivalent circuit of the input of the inverting amplifier with the source connected and determine the input voltage that appears at the input of the amplifier. What has happened to the source signal? [6]
ii) Draw the circuit diagram of the inverting amplifier with the source signal connected and carry out a full analysis to determine the closed loop voltage gain (Vo/Vs) of the inverting amplifier system. [8]
iii) Explain what can be done to remove the effects in part i) and part ii) on the inverting amplifier. Draw the resulting circuit diagram. [5]

Jason

 

[cowboybob]

What have you done thus far?

Do you have a SIM program?

 

[JasonMcG]

for i) I calculated Rf to be 150k. I said -15=-(Rf/R1).
I then said i=V/R = 10mV/100k = 10nA.
I used this 10nA from the source and multiplied it by the the 10k (R1) resistor to get 1mA that is at the input of the op amp.
Dunno if I did anything correct there or not?

 

[JimB]

for i) I calculated Rf to be 150k. I said -15=-(Rf/R1).

That is OK.

I then said i=V/R = 10mV/100k = 10nA

.
Why?
Dont worry, it is just wrong anyway!

I used this 10nA from the source and multiplied it by the the 10k (R1) resistor to get 1mA that is at the input of the op amp.

Ugh?
Now you are well away with the fairies.

To get you back on track, you need to redraw the circuit as I have attached it here.

Calculate the input voltage to the amplifer part.
Remember that there is a virtual short circuit between the + and - inputs of the op amp.
You should now see that there is a voltage divider, which is dividing the 10mV from the ideal source. ( The resistance 100k ohm, of the source is effectively in series with the source).

Now you should be able to calculate the output voltage from the amplifier when it is connected to this voltage source.

JimB

 

[JasonMcG]

I then said i=V/R = 10mV/100k = 10nA
Why?
Dont worry, it is just wrong anyway!

Is this incorrect?

 

[JasonMcG]

Since there is a virtual short, everything flows from Rf to R1 to the source, therefore should the amplifier input voltage not be 10mV since the 100k resisitor is in series with the source?

 

[JimB]

JimB said: ↑
I then said i=V/R = 10mV/100k = 10nA
Why?
Dont worry, it is just wrong anyway!
Is this incorrect?

The current flowing in the input circuit is
I = 10mV/(100k + 10k)
But why you need to know this I am not sure.

Since there is a virtual short, everything flows from Rf to R1 to the source, therefore should the amplifier input voltage not be 10mV since the 100k resisitor is in series with the source?

Just look at the input circuit, ther is a loop from the voltage source (10mV), through the source resistance (100k), through the Ri (10k), through the op-amp input terminals (0k), and back to the voltage source.

The 100k and 10k in series form a voltage divider.
The input to the op-amp is the voltage developed across the 10k resistor.

does that make it clearer?

JimB

 

[JasonMcG]

So the input voltage is therefore calculated to be 0.909mV?

 

[KeepItSimpleStupid]

Jim drew the circuit in post #4, so now analyzing it should be relatively easy with known tools. You know that the voltage at the inverting input is zero. It's determined by two currents. You also know that they are negatives of each other.