# Operational Amplifier Circuit - common Vin to both inputs

Discussion in 'Homework Help' started by Fraser_Integration, Dec 12, 2009.

1. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Hi AAC. About that time again for me to bother you with what is no doubt an easy question, it's got me stumped.

If you see the attached diagram, the question is what is the gain of the circuit when the switch is open and closed. Open is easy, as the non-inv input forms a virtual earth and the gain is just -Rf/Ri: lovely jubbly.

However when the switch is closed, I enter a world of pain.

Obviously some of the initial current, well the larger portion of it, will flow down to the 27k resistor, but I have no idea how this relates to gain, so far I have only dealt with voltage differences being amplified.

To me, the non-inverting input will be at Vi as there are no resistive elements between the input and the terminal, but there is surely no way that the inverting input could be at Vi owing to the 56k resistor.

I have the answer sheet giving +1 gain, but would appreciate some advice.

Many thanks,
Fraser

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2. ### cumesoftware Senior Member

Apr 27, 2007
1,330
14
I thing I might have it figured. The gain is 1, yes, because of the negative feedback. When the switch is closed V+ = Vi. Since we have negative feedback, V- = V+ (V- tends to follow V+, due to Vo). Therefore, theoretically V- = Vi. Since V- = Vi, there is no current flowing through the first 56K resistor. As so, there is no current flowing through the second 56K resistor (because the feedback resistor network makes a voltage divider between Vo and Vi, or lets call it a voltage "averager" LOL). Since there is no current flowing through this second resistor, the voltage drop across it will be also 0 and Vo = V- = Vi.

Since Vo = Vi, the gain is 1.

Last edited: Dec 12, 2009
3. ### kdillinger Active Member

Jul 26, 2009
141
3
Remember the golden rules of op-amps. The inverting node will be equal to the non-inverting node.

So when the switch is closed Vin will be measured at the non-inverting input. Negative feedback will slew the inverting input to the same potential; i.e., Vi.

Since the input is Vi, the inverting input is Vi, then the output will be equal to Vi.

Think of it this way, the op-amp will drive the error toward 0V (the difference between the inverting and non-inverting pins). In order to do this the output must slew positive or negative to force the inverting node to equal the non-inverting node.

Since the non-inverting node is Vi then the negative feedback of the op-amp will slew the output voltage to drive the error voltage to 0. Well, since the non-inverting node is Vi in order to drive the error voltage to 0 then the inverting input will be forced to Vi.

Since the input is Vi and the inverting input is equal to Vi then no current will flow thru the input resistor. Because of the op-amps high input impedance (GΩ or TΩ) any current thru the input resistor will also flow through the feedback resistor. However...

Since we just found out that no current flows thru the input resistor then no current will flow thru the feedback resistor and thus no voltage drop.

Ergo Vout = Vin.

4. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
Ah damn, seems so simple now. Thank you two for that.

5. ### Fraser_Integration Thread Starter Member

Nov 28, 2009
142
6
I think it just seemed illogical to me that a voltage could "travel" over the resistor without actually dropping any potential.

6. ### hgmjr Moderator

Jan 28, 2005
9,030
215
It happens that the voltages at each end of the resistor are always equal to each other. This results in zero volts dropped across the resistor at all times. Ohm's Law therefore calculates zero current.

hgmjr