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To obtain maximum "voltage" transfer from one stage to another.
Imagine you have a battery with a high series resistance (i.e. high input impedance) and hook a light bulb with it. You'll realise that it's dimmer than w/o it. That's because there is a voltage drop across the series resistance which is considered "wasted".
Similarly, you'll want high input impedance so that you can obtain maximum voltage across your DUT provided by the source.
hi all,
think that you have a signal source Vs which has an output impedance Ro and you want to amplify the signal by an operational amp. which has an input impedance
Rin then at the input of the opamp Vin=Vs*Rin/(Rin+Ro)
if Rin<<Ro signal is attenuated very much at the input of the opamp
For driving small resistive loads like speakers the voltage division as above limits
your gain so you minimize the Rout
CMOS opamps need not to have small Rout because they generally drive capacitive loads
You can think simply as voltage division. When using high input impedance, all the voltage from voltage division go to amplifier. In similar at the otuput low output impedance causes most of the voltage appears on the load.
If you have current amplifier you need low input impedance ana high outpu impedance, you can think it with current division
A high input impedance makes your opamp take
only the input voltage from its source and not any
current. (Imagine your input source as an ideal source
in series with a resistance which is the internal resistance
of the source; your opamp input impedance adds to this
resistance and causes a voltage drop between the input
source and the actual opamp input, if your opamp has
low input resistance).
Similarly, if your opamp has high output impedance,
a part of the opamp output swing is wasted as a drop
in the output resistance (again imgine the output as a
ideal source in series with a resistance).
high impedance at input is needed so as to pass whole of the upcoming voltage to the input of opamp.
low impedance at o/p is needed so as to pass whole of the o/p voltage to next stage.
You want your op am to transfer voltage, then you need to make the input source more like a "voltage" source to your op am, and your op am also more like a "voltage" source to the load. When the op am input impedance is large (compared to the input source impedance), the input source will be more like a voltage source. The same reason requires the op am have a low output impedance (compared to the load).
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