Common-Emitter Amplifier (Collector Voltage)

[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:

​

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 V_e=5.3 V, V_c=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 V_c can't be greater than 6. I'm very confused and I think it might be a typo. Any help would be greatly appreciated.

 

[gnurf]

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

 

[The Electrician]

Hence V_c can't be greater than 6.

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

 

[technician]

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

 

[jim hardy]

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

What polarity?


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

 

[roam]

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

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