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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:

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

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

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

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 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.
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