# Calculating currents in transistor of common-emitter amp

#### Davrum

Joined May 14, 2015
8
Hi all. In black is the assignment question and in blue my gropings in the dark towards the answers.

The value I have for (ii) Vce = 7.62V seems too high, although I'm not sure if that's just because I'm comparing it to other examples. On the other hand it's between 0.3 and 25V, so it's in the right range for linear operation, right?

Am I correct in putting that 7.7V across V2 according to the information given in the asking of (i)? C2 doesn't change the voltage does it?

Part (iii) is where I'm most unsure (50% unsure about (i), 65% unsure about (ii) 94% unsure about (iii))

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#### Jony130

Joined Feb 17, 2009
5,212
You do not need to know R1 and R2 values.
Also notice that Vc - voltage between collector and ground is 7.7V So from this we can find Ic quiet easy. Ic = (Vcc - Vc)/Rc = (12.7V - 7.7V)/5kΩ = 1mA
From this we can find Ie, Ib. Also notice that Vce is a voltage difference between collector and emitter voltage. And this is why your Vce is wrong.
As for maximum undistorted output voltage. The minimum voltage occurs when BJT is start to saturation. And the max. when is start to cut-off.

#### shteii01

Joined Feb 19, 2010
4,644
I think Part 2 has an error.
Vs=Vrl+Vce+Vre
Vce=Vs-Vrl-Vre=Vs-IcRl-IcRe
However, as Jonny pointed out Vce+Vre=7.7 volts so you don't need to go all the way to the source.
Vce+Vre=7.7 V
Vce+IcRe=7.7 V
Vce=7.7-IcRe=7.7-(1 mA*3.3 kOhm)

#### Davrum

Joined May 14, 2015
8
Thanks guys, I think I over-complicated it trying to do some kind of small-signal analysis thing and got in over my head.

The distortion thing really threw me because although we've covered it in class briefly, it wasn't enough to get an understanding of it to the point I could get my head around part 3. Your explanation make a ton of sense Jony, especially with how it makes me consider the characteristic curves.

#### Davrum

Joined May 14, 2015
8
I think Part 2 has an error.
Vs=Vrl+Vce+Vre
Vce=Vs-Vrl-Vre=Vs-IcRl-IcRe
However, as Jonny pointed out Vce+Vre=7.7 volts so you don't need to go all the way to the source.
Vce+Vre=7.7 V
Vce+IcRe=7.7 V
Vce=7.7-IcRe=7.7-(1 mA*3.3 kOhm)
This makes a lot more sense than what I had before thanks shteii01, although it seems like we should consider Ie rather than Ic as the current through Re, so that the last two lines would be:

Vce+IeRe=7.7 V
Vce=7.7-IeRe=7.7-(1.007 mA*3.3 kOhm)

Or am I missing something (again)?

#### shteii01

Joined Feb 19, 2010
4,644
This makes a lot more sense than what I had before thanks shteii01, although it seems like we should consider Ie rather than Ic as the current through Re, so that the last two lines would be:

Vce+IeRe=7.7 V
Vce=7.7-IeRe=7.7-(1.007 mA*3.3 kOhm)

Or am I missing something (again)?
Ie=Ib+Ic
Ib is much smaller than Ic. The common way of doing the math fast and loose is to ignore Ib. Therefore, Ie=Ic.