# small signal BJT amplifier

Discussion in 'The Projects Forum' started by princessERI, Feb 19, 2013.

1. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
Can somehow help me with design this amplifier?
It should have a gain of 5 yet whenever I made DC analysis of the circuit it always show a clipped signal output. I already try changing the value of the I1 still the graph won't change as I change the value of the resistors.
However, if I change the current without changing the values of the resistor, I can get a full signal output but the gain will be too large.

I also want to know how to get the current gain.
Thanks for any help.

(I attach the picture of the circuit and the graph of the output signal)

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2. ### #12 Expert

Nov 30, 2010
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9,678
You need a resistor in the emitter circuit. The transistor is running as hard as it can go if you don't suppress the gain at the emitter.

3. ### BC107C New Member

Apr 3, 2011
14
1
The gain for the circuit, as drawn is
G=gm*Rc; where gm= the BJT small( small !) signal gain.
gm=40*Ic. ( ma/V)
Therefore at about 1 mA you will get G=5.This is valid for small signal analisys.
400mV rms at the input is not small signal, hence what you see on the scope is the large input signal response.
A small signal would be max 10 mV rms.
At 1mA in emmiter the samll signal gain is 5.

For 400 mV input the schematic needs to be adjusted so the transitor will still work in a small signal input range( Delta Vbe<=25mV)

4. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
Actually, that 400 mV is the input of mp3 player and we are asked to design a small signal audio amplifier.

5. ### crutschow Expert

Mar 14, 2008
21,391
6,127
1mA bias current through the 130 ohm collector resistor gives a DC voltage drop of 130mV, thus the collector is biased very close to the +5V (+4.87V) and the maximum output signal with out clipping is 130mV peak. For a gain of 5 and a 400mVrms input the pp output would be 0.4*1.4*2*5 = 5.6V thus you need a larger plus voltage supply, say 10V.

Then you want to bias the collector in the middle of the supply voltage at about 5V so you would need a 5KΩ collector resistor with 1mA bias current. To get a gain of 5, add a 1kΩ resistor in series with C2 (giving a theoretical gain of 5k/1k = 5). The negative feedback from the emitter resistor determines the gain and also reduces the distortion of the amp. Because of the transistor base-emitter resistance, the added resistor needs to be slightly smaller. A value of 967Ω gave me a gain of 5 in my LTSpice simulation.

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6. ### #12 Expert

Nov 30, 2010
18,076
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So many ways to say the same thing!
I hope the Princess will find one that fits her level of education and the terms she is familiar with.

7. ### crutschow Expert

Mar 14, 2008
21,391
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I wasn't aware that we all said the same thing.

8. ### #12 Expert

Nov 30, 2010
18,076
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You're right. BC107C did not mention the need for an emitter resistor.

9. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
Thanks for helping me with that design
But how do I compute current gains?
It is such a tricky stuff to me.

The circuit I attached is my first stage so I can have a voltage gain but the output current is too small and the speaker needs 0.25 A (8 ohm, 0.5 W)

10. ### #12 Expert

Nov 30, 2010
18,076
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That's what the second stage is for, right?

11. ### crutschow Expert

Mar 14, 2008
21,391
6,127
Current gain of what?

You could use a single emitter follower to give current gain (equal to the transistor beta or Hfe) but such a Class A stage has low efficiency.

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12. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
yes, and I do not really get the idea of computing the current gain

13. ### princessERI Thread Starter New Member

Feb 18, 2013
11
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Then how can I make the efficiency go higher?
All of my designs have output much less than the desired one whenever I simulate.

14. ### crutschow Expert

Mar 14, 2008
21,391
6,127
Post the diagrams of your circuits that have "less than the desired output". Otherwise I'm just guessing as to the problem.

The current gain is simply the output signal current of the amplifier divided by the input signal current.

15. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
I can't find my other designs that have a full output signal with less voltage and most of my designs are always clipped especially the bottom half.
I found alternative biasing techniques in the internet and I do not know where I always get wrong especially with the clipping part
But this is the latest one I made though I wanted to bias the emitter of the follower with a current source (ended up being another failure) for easy connection of a potentiometer.

I made the stages direct coupled and with a capacitor but I just can get my desired output and waveform

In the second diagram, the values of I are just randomly assumed and the two last stages just made a square wave and when connected to the first stage they made a full wave but not symmetrical to a sine wave. Furthermore, the output voltage is much less than the input about micro volts.

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16. ### crutschow Expert

Mar 14, 2008
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As you have determined, "random assumed" values will likely not work unless you happen to lucky at the lottery.

It would seem that you need to understand more about circuit design and transistor biasing than I can readily do on this forum. I suggest you find some tutorials on that (here or here are examples), or just build a simple amp using a audio power IC of which there are many.

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17. ### patricktoday Member

Feb 12, 2013
157
42
Looking at your second circuit, why don't you try a standard class AB output stage after your second transistor? That can supply much more current to your speaker without pulling down your signal voltage much.

http://www.soundonsound.com/sos/jun06/articles/loudandlight.htm

Choose some transistors that can handle a few watts and probably 1k or less for the resistors. You can try class A, AB or B; there are tons of variations on these you can find, but the main purpose is to take a signal which has been amplified to your desired voltage and output it with a lot more power, without affecting the voltage much.

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18. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
By the way, what is the allowable range for the common emitter input resistance

19. ### princessERI Thread Starter New Member

Feb 18, 2013
11
0
Can someone help understand the reason why the signal clipped in that way?
I used an ideal current source and the waveform was good but when I replaced it with the equivalent transistors, the signal clipped and the output was reduced.
(the first stage went well as I see the waveform when it is not connected with the second stage

___________________________________________________________________
Thanks for any help. I really appreciate any comments as it helps in designing for our project

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20. ### patricktoday Member

Feb 12, 2013
157
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When you create a current source with transistors, the voltage signal can only fall as low as your Vbe value or about .7 volts above the emitter voltage. You could try raising your bias point a bit.

Did you verify you're getting the desired current out of Q3 and Q6 as expected?