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[SOLVED] Homework - voltage amplifier analysis

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Jun 15, 2011
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Hello, I'm an electronic engineering student. I'm trying to solve this exercise but
it's too difficult for me:


I do not understand elementary stages and how they work :oops:
Help me please

I think:
Q3 is a current generator (transistor connected diode like)
Q2 is a common base amplifier because the base is connected to ground for the signal
Q1 is a common base amplifier because the base is connected to ground. It takes the input signal from Q2 collector.
Q4+Q5 looks like a Sziklai Pair.
Is this correct? Maybe Q1+Q2+Q3 form a canonical amplifier + active load and I don't realize that!
How those stages work togheter?

The homework requires to find the continuous value of VI that corresponds to Vo=0.
But, supposing all transistor on active region:
VC3=-Vcc+VBE3=5-0.7=4.3 V
VB2= VC3=4.3 V
VI = VC3+VBE2=4.3+0.7 = 5 V????

I'm confused.. can you help me please? Sorry for my english :oops:
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VC3=-Vcc+VBE3=5-0.7=4.3 V
- I can't imagine, that VC3 is positive. Please reconsider.
- Apparently you're required to calculate the exact voltages and currents expected at each transistor stage, not just put in default 0.7V VBE. That's why Is, Vt and beta are given. The operation points should be calculated backwards output to input.
Yes Q1 and Q2 are common base amplifiers.

Yes Q4 and Q5 are a Sziklai pair.

Pulling the input low will (I believe) have the effect of turning on Q1 and Q2. This will allow the PNP transistor to turn on. This will cause Q5 to conduct.

Pulling the input hi will turn off all transistors except Q3 (I think).

You want to input the correct V so that Q5 will be at the same resistance as R3. This means causing Q4 to turn on just a little bit.
This means causing Q4 to turn on just a little bit.
Yes, in a qualitative view. In other words, the transistors are operated in their linear operation area. The problem asks for applying the Shockley equation to calculate the exact operation points.
Thank you all, now it's more clear! I'm starting calculating the current IR3, equal to IE5 and so on, perfect!
Maybe I will ask you something later if I will get into some trouble :grin:
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