3 Transistor Amplifier - Analysis

Thread Starter

elec_eng_55

Joined May 13, 2018
214
I am trying to analyze a 3 transistor amplifier in order to understand how it was designed.

I cannot figure out how to compute the voltage on Q3 base.

Is R4 some type of feedback resistor? How was this value computed?

upload_2018-8-31_0-47-38.png
(Schematic added by moderator -please post schematic for those who do not have LTSPICE, such as cell phone users)
 

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Jony130

Joined Feb 17, 2009
5,145
I cannot figure out how to compute the voltage on Q3 base.
This voltage will be equal to Vb = Vcc - Vbe3 ≈ Vcc - 0.65V

Is R4 some type of feedback resistor? How was this value computed?
Yes, it is a negative feedback resistor but at the same time, it provides a path for a Q1 quiescent current.
And designing is "easy" because you as a designer decide about the current and the voltage across this resistor. So, you can use the Ohm's law.
 

Thread Starter

elec_eng_55

Joined May 13, 2018
214
This voltage will be equal to Vb = Vcc - Vbe3 ≈ Vcc - 0.65V


Yes, it is a negative feedback resistor but at the same time, it provides a path for a Q1 quiescent current.
And designing is "easy" because you as a designer decide about the current and the voltage across this resistor. So, you can use the Ohm's law.
How would you breakdown this amplifier in order to compute gain, input impedance, output impedance etc. of the various
stages?
 

Jony130

Joined Feb 17, 2009
5,145
How would you breakdown this amplifier in order to compute gain, input impedance, output impedance etc. of the various
stages?
First, let us try to find the DC operation point. To simplified the calculation I will ignore the base current (β = ∞). Later on, if you want you can include the beta value in your hand calculations.

The voltage tha the base Q1 is:

Vb1 ≈ Vcc * R7/(R6+R7) ≈ 10V and the Q1 emiter voltage is Ve1 ≈ (Vb1 - Vbe1) ≈ 9.4V.

We also see that the voltage drop across R1 resistor is equal to Vbe2, therefore the Q1 collector current will be around Ic1 ≈ 0.6V/10k = 60μA (at this stage I ignore Q2 base current).

And this current (Ic1) must also flow through R4 hence the voltage drop across R4 is VR4 ≈ R4*Ic1 ≈ 1.32V.

So, the voltage at Q2's emitter is

Ve2 ≈ Ve1 - VR4 ≈ 8V and the Q2's emitter current is around Ic2 ≈ 8mA and the Q3 collector current is Ic3 ≈ 8.6V/68kΩ ≈ 126μA.

And this is the end of the first iteration. We could continued and used this values in the next iteration and include the transistors current gain.
But I do not have time for this right now, you can try it yourself if you want.

And to find the AC parameters try to read this post where I a long time ago analysis a very similar circuit.
https://forum.allaboutcircuits.com/threads/schematic-analysis-for-voltage-drops.85763/page-3#post-616893

In short the voltage gain will be equal to R4/R5 within reasonable limits. And Rin ≈ R6||R7
 
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