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the input voltage is from mp3 player.

Now, why is that when the input voltage is around 5 mv pk, there are no clipping issues but if we change the input voltage to around 400 mv pk(which is common in every mp3 player including mobile phones), there is severe clipping as described below in the picture.

enter image description here

here, we can see there are no clipping issues

but here, we start to inject the mp3 player as the voltage input (400mV)

enter image description here

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2 Answers

up vote 1 down vote accepted

The R1-R2 voltage divider biases the base of Q1 at about 1.17 V.

A 0.4 V excursion below that puts it at 0.77V, which would normally keep it in forward active mode. But the 100 uF capacitor is preventing the emitter voltage from changing, keeping that node at about 0.4 V. That means you only have 0.3 V across Q1's b-e junction at the peak of your input signal, which will pretty effectively put the BJT in cutoff mode.

Having the BJT in cutoff mode will put the Q1 collector at about 10 V, not sensitive to the exact level of the base voltage. And that flat top at 10 V (after a shift going through the C2(?) capacitor, is what you're seeing as clipping.

It might be slightly clearer what's going on if you remove C2 and just directly connect your virtual oscilloscope to the Q1 collector. Then you'll see the actual level where clipping occurs is about 10 V, making it slightly easier to diagnose the problem. (I'm assuming the virtual oscilloscope has a "perfect" input and doesn't cause any loading on the circuit)

You can probably reduce or eliminate the clipping by removing Cb. This will allow the circuit to behave like an "emitter-degenerate" amplifier at the frequency you're running it, reducing the gain as suggested by Anindo, and increasing the input impedance as a side benefit. You would still expect to see clipping if you increased the input amplitude to about 0.5 V (1 V peak-peak).

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The amplifier is going into cutoff saturation with the higher input signal. Also, the input biasing is off by a bit so the clipping is asymmetrical.

Actions:

  • Reduce the gain of the amplifier to keep the output within the linear part of its curve.

  • Change the DC biasing of the input signal to center it on the midpoint of the linear part of the output curve.

To do this empirically, use a potentiometer instead of one of the two fixed resistors, to figure out what resistor ratio puts the null point closes to the middle of the output linear curve portion.

Thanks to @ThePhoton for pointing out my incorrect use of "saturation".

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If the problem was putting the BJT into saturation, you'd expect the clipping to happen when the input voltage is high. Instead, he's seeing clipping when the input is low. –  The Photon Nov 22 '12 at 6:55
    
@ThePhoton Isn't it flattening out when the input approaches lower limit? Maybe I'm misreading it, my mistake. –  Anindo Ghosh Nov 22 '12 at 7:12
    
@AnindoGhosh Not correct - the transistor goes into cutoff but the amplifier "black box" at that time goes to saturation. You shoudl type "the amplifier is going into saturation" or "the transistor is going into cutoff". when transistor does not conduct, output of amplifier is saturated to max it can go. Although the Actions is still correct. –  ExcitingProjects Nov 22 '12 at 7:26
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@ExcitingProjects I see merit in both viewpoints, yours and ThePhoton's. –  Anindo Ghosh Nov 22 '12 at 7:33
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