# Basic Small Sig BJT Amplifier

Discussion in 'The Projects Forum' started by XwyhyX, Nov 18, 2012.

1. ### XwyhyX Thread Starter New Member

May 20, 2012
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0
I need a voltage gain of at least 100 for a preamp which will receive an input of 100mV. 2 or 3 stages is fine. The DC supply is 12V. Is this possible with these given conditions?

I have only tried cascading identical common emitter circuits but already there's clipping seen on the signal, how could I fix this so that I can connect another stage or would it be possible to reach a voltage gain of 100 using 2 stages? (I also think 12V does not provide enough swing space? Idk)

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2. ### tubeguy Well-Known Member

Nov 3, 2012
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You've actually demonstrated you can easily achieve a gain of 100 with two stages.
Because R5 is bypassed with a large cap; basically R3/R4 determines gain.

And gain doesn't add, it multiplies. Try increasing R4, and R10, see what happens.

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3. ### crutschow Expert

Mar 14, 2008
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If the input is 100pp then the minimum required power supply would need to be greater than 100mV*100 = 10V to avoid clipping so 12V can work. But for that, the output stage collector would need to be carefully biased at the 1/2 the supply voltage.

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4. ### tubeguy Well-Known Member

Nov 3, 2012
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Also, Your 2nd stage is driving a load which is R12 - 1k.
The collector resistor R9 is 2.7k. This forms a voltage divider of 2700/1000. So roughly 1/3 of the max available voltage will been seen at the output.
Edit: Output voltage would be about 2.7v p-p

If my math is correct C3 impedance is less than 100 ohms at normal audio frequencies so doesn't play a big part.

You could increase R12 to 50k for experimental purposes to increase the output.

Last edited: Nov 18, 2012
5. ### XwyhyX Thread Starter New Member

May 20, 2012
4
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Oh ok so all i need is a little fix here right? Whew

6. ### XwyhyX Thread Starter New Member

May 20, 2012
4
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Thanks! I was looking for that, how to know if the DC supply is enough so that I can stop thinking about it. Setting the collector voltage at 1/2 of 12V which is 6V would mean centering the Q point? Am I right? So I have to manipulate the circuit still to have a collector voltage of 6V?

7. ### XwyhyX Thread Starter New Member

May 20, 2012
4
0
I have my load constant as 1K. Not being able to modify it has already gave me headaches. haha

8. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
You may be able to get 10v p-p in the simulation but it would likely be high distortion. Ok for experimenting, but not good for a real-world circuit.
You would need to increase the power supply voltage.

Also, the first stage cannot be connected directly to the 2nd.
It makes it impossible to get the 2nd stage biased properly.
Put a coupling cap between stage 1 and 2. C1 and C3 are in the circuit for that reason - they pass the AC signal, but block DC.

9. ### tubeguy Well-Known Member

Nov 3, 2012
1,157
197
Well... Unfortunately, the circuit shown just cannot drive a 1K load to the levels you require.
Edit: An output buffer circuit is required which can provide the needed current drive capability.

Do you really need to use transistors?

What will this circuit be used for?

Last edited: Nov 19, 2012