Common-Emitter Amplifier (Collector Voltage)

  • Thread starter roam
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  • #1
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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. :confused:

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.
 
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  • #2
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If IC = 1mA then the voltage drop across RC is 5V. Your book is correct.
 
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  • #3
The Electrician
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Hence Vc can't be greater than 6.

Why do you think that Vc can't be greater than the base voltage?
 
  • #4
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I agree with The Electrician, you are wrong to state that Vc-b =12-6=6V
 
  • #5
jim hardy
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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.
 
  • #6
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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##.
 
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