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# Amplifier input and output impedance - Why R1//R2 and why Rc//RL?

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#### atferrari

##### Well-Known Member
After some time searching, found **broken link removed** explaining input and output impedance of an amplifier.

There is a doubt (I suspect it is related to Thevenin or Norton):

Why R1 is considered in parallel with R2 and why Rc is considered in parallel with RL? After all the current going into RL left already Rc.

Starting from the circuit schematic, as usually drawn here:

Something which is not shown is the decoupling capacitor from +12v to 0v.
The purpose of the decoupling capacitor is to remove any signals from the 12v line, in other words, it creates a very low impedance (short circuit) at signal frequencies.

The version of the circuit which you show, can be described as the "AC Equivalent Circuit", it is showing the circuit from the point of view of an AC signal.
The DC bias conditions are not shown.

So, consider R2 and Rc, each resistor has one end connected to the +12v line, but, the 12v line is short circuit to the 0v line as far as AC signals are concerned, and so that put R1 and R2 in parallel, and Rc is in parallel with RL (via C2).

Does that make sense?

JimB

So, consider R2 and Rc, each resistor has one end connected to the +12v line, but, the 12v line is short circuit to the 0v line as far as AC signals are concerned, and so that put R1 and R2 in parallel, and Rc is in parallel with RL (via C2).

Does that make sense?

JimB

First time I get the complete story, Jim. Sure it makes sense. Gracias for replying.

BTW, for anyone reading this later, the R2 in bold should be R1.

BTW, for anyone reading this later, the R2 in bold should be R1.
Well spotted!
JimB

The input impedance of the transistor is not calculated.

The input impedance of the transistor is not calculated.

And how relevant is it for the case, AG?

And how relevant is it for the case, AG?
The input impedance of the transistor is much lower than the bias resistors. The article shows ho to calculate its 2500 ohms value.

It so happens that in a DC circuit, the impedance of a potential divider is just impedance of the two resistors in parallel.

The maths of proving it is a bit tedious, and I'm sure it's on the web somewhere, but the result is easy to remember.

In the transistor circuit, I agree that the transistor impedance is likely to be smaller and much more difficult to calculate.

It so happens that in a DC circuit, the impedance of a potential divider is just impedance of the two resistors in parallel.

The maths of proving it is a bit tedious, and I'm sure it's on the web somewhere, but the result is easy to remember.

In the transistor circuit, I agree that the transistor impedance is likely to be smaller and much more difficult to calculate.

It took me so long time to hear of something that intrigued me every time I run across input impedance of a BJT amplifier. Have to search and check with LTSpice. Gracias.

The article in the first post explains how easy it is to calculate the input impedance of a common-emitter transistor.
Simply calculate the base voltage to give the emitter voltage and emitter and collector current, then use the simple formula of (25/IE) times the beta..

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