DC amplifier analysis help.

[logar0]

Hello everyone. I am a beginner with electronics. I came across with the following DC transistor amplifier. It is supposed to have a gain of (R2+R3)/R2. The simulation seems to confirm that.



I am trying to solve the circuit, but I can't find a valid solution. My equations are sort in relation to the unknown values. Any help?

[logar0]

Are my assumptions and calculations done correctly?

[LvW]

Are my assumptions and calculations done correctly?

Formally (from the mathematical point of view) your result seem to be "correct". However, from the practical point of view your basic assumption (Vin-0.7=Vin) is rather optimistic. This means: You have a signal output for a very small differential input voltage (near zero) across the BE-path o f the 1st transistor. With other words: Infinite open-loop gain. And therefore, your result equals the gain of the classical non-inverting opamp configuration.
The correct result should contain the finite gain of the two-transistor amplifier - together with the feedback resistors [general form: A=Ao/(1-LG)]  with LG=Loop gain.

[logar0]

Thanks you!

[XFDDesign]

I'm actually unsure if this circuit will work correctly. IR2 will be the sum of Ie1 and IC2. IC2 will be a gained up version if ib2, which is a function of the current in ir1 and ic1 (ic1 = ir1 + ib2). It is wholly possible that the emitter voltage seen on Q1 will be much less than 0.7V due to the current contribution from IC2. The absence of values really makes it hard to give a definitive answer.

[LvW]

I'm actually unsure if this circuit will work correctly. IR2 will be the sum of Ie1 and IC2. IC2 will be a gained up version if ib2, which is a function of the current in ir1 and ic1 (ic1 = ir1 + ib2). It is wholly possible that the emitter voltage seen on Q1 will be much less than 0.7V due to the current contribution from IC2. The absence of values really makes it hard to give a definitive answer.

I think, the shown circuit works as a dc amplifier only in case the voltage Vin (as shown in the diagram) is larger than a certain lower limit (to allow app. 0.6 volts across the BE junction).
EDIT:  I have simulated the circuit for some selected resistor values - it works with a gain that is - as expected - slightly below (1+R3/R2).

[XFDDesign]

My concern was based on there being enough Vbe drop to have Q1 "on enough." What really saves it is the feedback design.

Analysis can be simplified by shorting out R3, in which case the whole thing resolves to a simple feedback design. Being at DC, we can cheat and assume that Beta is "large enough" not to worry too much about, in which case attention can be paid to R1 and Q2.
(Vcc-R1Ic1)=Vt ln (Ic2/Is)

Eventually this falls out to the nice "clean" form of Is*exp[(Vcc-R1Ic1)/0.0258] = Ic2

Let R1 be small, and you have Ic1 having a strong contribution to the total currents into R2. Let R1 be large, and the majority of the currents into R2 are supplied from Q2. The clever part of the circuit is that the feedback drops the required base current needed by Q1, and makes the input look much higher in impedance as an input.

Given the feedback in use, devices are both kept "on" but simply using the 0.7V drops is not going to be sufficient. The answers here are going to rely on the use of the currents and the device gains, of which Ic2 will depend on R1. If R1 is not a component of the total gain, something is wrong.


Edit:
I decided to run this out through lunch.My final equation is:
Vo = (Vin - 0.7) + (Bf/(Bf+1))*( (Vin-0.7)/R2 - 0.7/R1 )*R3

SPICE confirms this for values: R1=23k, R2=1.3k, R3=729 (Test 1) and R1=10k, R2=5.3k and R3=760.3 (Test 2)

The catch is, the "gain" isn't flat for Vin, but rather it is only the value at your "target" Vin set point.

In the above cases, G=1 at Vin=Vo=2 (Test 1) and Vin=Vo=6 (test 2).

For a gain of 2, at 2.0Vin, R1=10k, R2=5.3k, R3=15.559k, Vo=3.983V so the equation checks out.