operational amplifier

[belal elkady]

why the input impedance of op-amp is very high , and the out put impedance in very small ?

 

[Miguel Gaspar]

It is desig goal of the manufacturers that is because it permits a modular design in amplification. So the input current is very low about nano Amps for a BJT input and pA for MOSFET.

 

[belal elkady]

It is desig goal of the manufacturers that is because it permits a modular design in amplification. So the input current is very low about nano Amps for a BJT input and pA for MOSFET.

so why the o/p impedance very small ?

 

[Miguel Gaspar]

The uotput impedance is very small for the same reason.

 

[ckshivaram]

Input impedance is comparable to putting a resistor in in parallel with a circuit. The higher the value, the closer it becomes to acting like it doesn't exist. If you put a low input impedance device on a circuit, its the same as putting something more resembling a short in your circuit. For output impedance, its the same as putting a resistor is series. The lower the output impedance, the less effect it has on the total resistance of the circuit.
Why Does an Op Amp Need a High Input Impedance and a Low Output Impedance?
http://focus.ti.com/lit/an/slaa068a/slaa068a.pdf

 

[ninju]

An op-amp is not supposed to take any input current into the device through the input terminals and besides, it is designed as some kind of a voltage controlled voltage source. Since it is voltage controlled,we connect a voltage source or a battery at the input. Any voltage source will have a voltage source (Vs) in series with a source resistance (Rs). When this is connected to an op-amp, the Rin (input impedance) comes in series with the source voltage and its resistance. The amplifier takes in only the drop across Rin. The drop across Rin is Rin*Vs/(Rin+Rs). Here, if the Rin is not very large, only a part of the applied input voltage will drop across Rin. So, Rin has to be very very large, ideally infinite for it to take in most of the input voltage. You can observe this in low grade multimetres which when measuring voltage tend to consume almost half the voltage being measured.

In case of Rout (the output impedance), the same reason applies. Rout is in series with a resistance RL (The load resistance). Let the gain of the op-amp be A. So, the voltage at the output will be A*Vs. The output voltage will be read out from the load resistance. So, we need the entire voltage to drop across it. So, here, the voltage drop across RL is A*Vs*RL/(RL+Ro). Here, we'd prefer Ro to be 0 ideally.

Hope it was clear.

 

[LvW]

Yes, ninju gave you a good explanation.
Again in short:
* An opamp has the task to amplify (and to provide at its output) a voltage. Thus, it shall function (ideally) as a voltage controlled voltage source (VCVS)
* There are other integrated amplifier types like VCCS (voltage-controlled current source, known as OTA) or CCVS (Current-feedback amplifier) or CCCS (current conveyor).