# Common-Emitter Amplifier (Collector Voltage)

1. Jun 1, 2013

### roam

Hi,

My question is regarding an example of a common-emitter transistor amplifier shown in my electronics book. Here is the diagram of this particular amp operating from a DC power supply:

[Broken]​

My book says: "the transistor has β = 99, with no signal applied the voltage at the base of the transistor is measured to be 6 volts, so the dc voltages at the emitter and collector of the transistor are Ve=5.3 V, Vc=7 V respectively."

So I don't understand how you can have 7 volts at the collector in this particular circuit.

My book always makes the assumption that b-c voltage drop is 0.7. So we have

$I_e = \frac{V_e=6-0.7}{5.3 \ k\Omega} = 1 \ mA$​

And $I_e \approx I_c$, therefore $V_c=I_cR_c = 5 \ V$! So is the book wrong, or is there something wrong with my calculation?

We know that $V_{cb} = 12-6 =6 V$. Hence Vc can't be greater than 6. I'm very confused and I think it might be a typo. Any help would be greatly appreciated.

Last edited by a moderator: May 6, 2017
2. Jun 1, 2013

### gnurf

If IC = 1mA then the voltage drop across RC is 5V. Your book is correct.

3. Jun 1, 2013

### The Electrician

Why do you think that Vc can't be greater than the base voltage?

4. Jun 1, 2013

### technician

I agree with The Electrician, you are wrong to state that Vc-b =12-6=6V

5. Jun 1, 2013

### jim hardy

What polarity?

double check that statement. i'd have expected them to say that about voltage b-e not b-c.

6. Jun 2, 2013

### roam

My mistake, I meant to say b-e voltage drop is 0.7 volts. That is why we had $V_e =6-0.7= 5.3 \ V$.

Last edited: Jun 2, 2013