# Common Emitter BJT amplifier

Discussion in 'Homework Help' started by Petrucciowns, Jul 12, 2009.

1. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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Last edited: Jul 12, 2009
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I think it means AC output voltage - so it relates to the overall gain of the circuit and the AC input voltage.

3. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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Now that I look at it more closely I see that you're right. It's the voltage of the input resistor x the gain. I just cant figure out how to find that voltage. How is the input current found?

Last edited: Jul 12, 2009
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You can use the voltage divider equation to calculate the AC base voltage.

You are given Zin=3.12k - so you have a voltage divider comprising the source [10mV], the 600 ohm resistor and the input resistance [3.12k].

This will enable you to obtain the AC voltage at the base.

You are also given the gain relative to the base input as 60.2 so with Vb (AC) known form above calcs and Gain known you should be able to obtain Vout (AC).

5. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
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Sorry for being a little slow ,but what would the equation look at. I tried 3.12k /3.12k + 600 The answer comes out with the same number just 100 times bigger so I know I'm not doing it right.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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OK - let's go through the calcs.

Given source voltage = 10mV

Vbase (AC) = 3120/(3120+600)*10 mV = (3120/3720)*10 = 8.39 mV

Gain = AvTR = 60.2 (per your data)

Vout (AC) = Gain * Vbase (AC) = 60.2 * 8.39 = 504.9 mV

Which is close enough to 505mV

7. ### Petrucciowns Thread Starter Active Member

Jun 14, 2009
62
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Awesome! Thank you very much, I was forgetting to include the source in my calculation for Vbase.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You're welcome - glad to help.