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Impedance relation between 2 Bjts


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If there are two BJT small signal amplifiers (common Emitter) cascade connected using a suitable decoupling cap
The first preamp has an o/p imeadance Zo, and the second has an i/p impedance Zin
Now we have Zo and Zin and a cap betwwen them
My question is: Is the connection series or parallel? That, is the o/p voltage divided or the o/p current?

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Hi Hero999
At first I like to thank you on the fast of response


I mean here is if the output impedance of the first stage = 10k and the input impedance of the second phase = 10k
Are these two impedances connected in series or parallel?
thanks alot

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There is no decoupling capacitor. Instead there are coupling capacitors. C3 couples the transistors together. C5 is the input coupling capacitor and C4 is the output coupling capacitor.

Instead of wrongly guessing the impedances, calculate that the output impedance of the first transistor is 4.7k ohms and the input impedance of the 2nd transistor is 2.41k ohms. They are in parallel so the first transistor has a total collector load of 10k//2.41k= 1.94k ohms so its gain is 0.41 times the gain when it has no load.

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Thanks audioguru
I am very happy with your presence and I hope to be in good health and happy in your life.

I have read in a book "The art of electroncs" page 66, the following

I take it most likely the first stage is a source of signal with internal resistance (Zout1), and the second stage with input resistance of (Zin2),,,, what would you say?

Note: I miss you here too long and I had to register on many forums and I put the same question to ensure the receipt of a convincing answer from one of them.

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  • 2 weeks later...
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Guest remonx6

Impedance relation between 2 Bjts  use this formula regarding this issue
Zout and Zin form a voltage divider. The actual input voltage seen by the second amplifer is is calcluated from that, i.e. V = Vin * Zin / (Zin + Zout)

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