# Virtual short concept

1. Feb 20, 2008

### raybuzz

hi everyone,
I am having a tough time convincing myself about the virtual short concept in opamps,
As i understand from texts, on the application of a voltage on one of the input terminals of opamp, the voltage at the other input terminal is also forced to match the voltage on the first input terminal. How does this happen? How come this is valid irrespective of the different configurations in ehich the opamp is used?

2. Feb 20, 2008

### dionysian

First think about what the opamp does. It creates a "Large" differential gain A(V2- V1) = Vo. For the ideal opamp A is considered infinate. In real life its real big around 10000 or 100000. Think what would happen if there was a small diffrence between V2 and V1: (V2- V1) = .01V for example the output voltage would be 100 that is if A was 10000. Now if you hook the output terminal back to V2 with a resistor in between, there will be a negtive voltage going from the output to V2. this negative voltage will bring down the voltage at V2 untill V2 = V1. But notice that once V2 = V1 there will be no differential gain because: A(V2 - V1) = A*0=0. So at this point the opamp will be in equalibrium and V2 wil be "for all practical perposes" equal to V1. The same reasoning applies if V2 < V1 except there would be positive voltage from the output t V2 which would increase V2 untill it equaled V1 and thus V2 = V1... Virtual short!

It's kind of hard to imagine it because it happens so damn fast but think about the opamp as device that voilently reacts to any diffrence in voltage between its input terminals. When used with feedback it basically makes sure that its input terminals are equal by QUICKLY lowering or raising the voltage of one of the terminals untill they are both equal. And once they are equal the opamp is happy and doesn't need to voilently apply voltage back to the input terminals.

Is this clear? Let me know if its not.

Last edited: Feb 20, 2008
3. Feb 20, 2008

### ranger

Negative feedback.

With negative feedback, the op-amp will try to drive its output voltage to whatever level necessary so that the differential voltage between V+ and V- is practically zero.
This is in essence one of the two golden rules of op-amps.

4. Feb 21, 2008

### raybuzz

Why does the opamp reach equlibrium on the differential voltage reaching 0? can you explain in terms of the internal structure of opamps, being made of controlled voltage or controlled current sources?

5. Feb 21, 2008

### dionysian

well the differential voltage doesn't actualy reach "0". That is just the ideal. In reality it will be really small and thus there will be very little feedback to change the differential voltage.

Since V2 - V1 = very very small V0 = very very very small since V0 = (V2 - V1)/A. And there will be only a small amount of feedback trying the change V2 so you can say its in equalibrium.

As for the internal structure i think someone else is going to have to explain this....

6. Feb 21, 2008

### chroot

Staff Emeritus
The op-amp is configured in such a way that it amplifies the difference between its two inputs. This is usually done with two transistors, each driven by one input, like this:

(Image courtesy of Wikipedia, http://en.wikipedia.org/wiki/Differential_amplifier)

The circuit is called a differential pair. It is a transimpedance amplifier, which means it accepts inputs in the form of voltages and produces outputs in the form of currents.

At the bottom, you'll see a current source. This is called the "tail" of the amplifier, and the current source is often called a "tail current source." This current source pulls a constant current, which is necessarily the sum of the currents through each of the bottom transistors.

(The transistors on top form a current mirror, the only point of which is to create a single-ended output.)

When you apply the same voltage to each input, each transistor conducts the same amount of current. The circuit is balanced, and the output current will be zero. When you apply different inputs, the tail current flows more through one transistor than the other. When you apply very largely different inputs, all of the tail current flows through one transistor, while the other transistor is turned off.

If you connect the output of this circuit back to its inputs via a feedback network, the circuit will actively drive current into (or out from) the network until its inputs are balanced again.

- Warren

7. Feb 21, 2008

### ranger

An extremely good book on the subject is the classic by Jiri Dostal (1993). Its one of the best op-amp books that I've read, though it is very rigorous and takes a lot of competence to read. It carries a very hefty price tag. You should check out your school's library for a copy.