Question: DC analysis of a transistor amplifier?

[samy555]

I could not complete the analysis of this circuit.

This is what I managed to do:

IB1 = (VCC - VBE1)/R3 = (3 - 0.7)/100K = 23uA

If β = 100, then IC1 = 2.3mA

VCE1 = VCC – IC1 * R1 = 3 – (2.3m* 2.2K) = - 2.06 V (Error)

The Voltage between points A & B (VAB) = 0.7 volt

So, IC1 = VAB/ R1 = 0.7/2.2K = 0.318 mA

Then, VCE1 = 3-0.7 =2.3 V

I stoppedhere. I wish I couldcalculate IC2 and VCE2 to complete the picture.

Thank you.

 

[LvW]

Yes - it is an interesting circuit - sometimes called "current starving" circuit.
The value of Ic1 sounds reasonable. However, it does NOT go entirely through R1. In contrary, the majority of Ic1 is identical to the base current Ib2 of the pnp (Q2).
Hence, continue the calculation with I(R1)=0.7/R1 and Ic1=I(R1)+Ib2.

 

[Ratch]

I could not complete the analysis of this circuit.

This is what I managed to do:

IB1 = (VCC - VBE1)/R3 = (3 - 0.7)/100K = 23uA

If β = 100, then IC1 = 2.3mA

Yes, base current is an indicator and somewhat proportional to the collector current.

VCE1 = VCC – IC1 * R1 = 3 – (2.3m* 2.2K) = - 2.06 V (Error)

No, that is not correct. IC1 is the sum of the current existing in R1 AND the base current of IC2.

The Voltage between points A & B (VAB) = 0.7 volt

Yes, that is correct.

So, IC1 = VAB/ R1 = 0.7/2.2K = 0.318 mA

Wrong. IC1 is the value stated above.

Then, VCE1 = 3-0.7 =2.3 V

Yes, the emitter-base diode of IC2 holds the voltage of R1 at around 0.7 volts.

I stoppedhere. I wish I couldcalculate IC2 and VCE2 to complete the picture.

IC1 is 2.3 mA. 0.318 mA comes from R1, so 2.3 mA minus -0.318 mA = 1.95 mA comes from the base current of IC2. Multiply this current by 100 and you get IC2=198.2 mA. You can take it from there. You should notice that IC2 saturates, and cannot supply198.2 mA. It seems to me that a simple diode across R1 would not change current present in IC1.

Ratch

 

[samy555]

Thank you very very much Ratch