Output waveform of emitter amplifier

[mathman44]

Homework Statement

What causes the output distortion seen in the above diagram? Output on right, input on left.

The Attempt at a Solution

This isn't homework, just something I can't figure out, playing with a simulator. The gain here is dependent on the input voltage... but why is the output shaped the way it is?

 

[mathman44]

Input is 50mV, that bit got cut out.

 

[vk6kro]

Looks like you are overdriving it. I can produce the same effect here.

Try 10 mV p-p input.

Also check that the collector voltage is about half the supply voltage with no input signal.

 

[mathman44]

What exactly is overdriving and what are the consequences of it? I don't quite understand, because V_in is quite small, so how could it overdrive the transistor?

 

[vk6kro]

Overdriving is having enough input to produce a distorted output. Your Vin is small, but not small enough.

That is a high gain amplifier (about 170 with a 2N2222 as the transistor) and transistors are inherently non linear, so you cannot get high gain and undistorted, high output at the same time.

If you removed the emitter bypass capacitor, you could get more output, but the gain would drop a lot. You could use up to 0.5 volt peak to drive it and still get good linear output, and more of it.

Try it. Drop your input to 5mV peak and you should get undistorted output.

 

[mmmboh]

I think he's interested in why the wave looks like that in the first place, not just how to fix it (or am I wrong?)..I think the wave is distorted on top, because the value of the capacitor is so much greater than the value of the resistor, relatively.

 

[mathman44]

Overdriving is having enough input to produce a distorted output. Your Vin is small, but not small enough.

That is a high gain amplifier (about 170 with a 2N2222 as the transistor) and transistors are inherently non linear, so you cannot get high gain and undistorted, high output at the same time.

If you removed the emitter bypass capacitor, you could get more output, but the gain would drop a lot. You could use up to 0.5 volt peak to drive it and still get good linear output, and more of it.

Try it. Drop your input to 5mV peak and you should get undistorted output.

Ok, so why is it that the distortion occurs only around this input range, and not when it's more or less? I just don't understand why the output is so sensitive only around this range of input voltage.

 

[vk6kro]

It isn't. As you increase the drive, the distortion gets worse. The reason it starts at this input level is that this produces an output swing that isn't directly proportional to the input.

Up to this level, the output only swings in a region where the output depends directly on the input.

Real transistor behaviour is shown on charts supplied by manufacturers. If you draw a load line on one of these curves, you can see where it goes through areas where the lines are unevenly spaced and this tells you that you will get distortion if you drive the transistor into these areas.

Why don't you want to try driving it with less signal? You can't do anything about what is inside a transistor, but you can learn ways to make them useful.

 

[vk6kro]

This is a curve showing base current vs base-emitter voltage for a 2N2222

[PLAIN]http://dl.dropbox.com/u/4222062/Base%20emitter%20curve.PNG

The input is a voltage as shown across the X axis, but the output depends on the input current, shown here on the y axis.

You can see that if the bias point was set to 650mV and the input swung from 600 to 700 mV, the negative going current swing would be 3.4 uA but the positive going swing would be 22 uA.

So, an undistorted voltage at the input will result in a distorted output at any input level but it gets really bad at high drive levels

 

[mathman44]

Great, thanks for the clarification!

 

[skeptic2]

I think there's another effect that is involved also. Remove the 15 uF cap and look at V(Q1:c) again. Most of the distortion is gone.

I think what is happening is that the base-emitter junction along with the gain of the transistor is acting like a diode detector. The 15 uF capacitor can charge a lot faster through the transistor than it can discharge through the 1K resistor so it tends to bias the transistor off very slightly, causing it to operate class AB. This also explains why when the signal is reduced, the distortion is too.

The critical parameter is that the maximum slope of the signal at the emitter must never exceed the minimum discharge slope of the capacitor.

 

[vk6kro]

You may be right, but you could consider the following.

It is normal for an amplifier with an unbypassed emitter resistor to produce a very clean output. You also get a high input impedance and low gain, but clean output. Look at the gain, it should be about 7 where it was 170 before, with the capacitor.

The distortion is the same whether there is a bypassed emitter resistor there or not. Try it.

The emitter bypass capacitor cannot discharge via the emitter resistor because this has the same voltage on it as the capacitor, until you remove the power.

If the base emitter juction was getting reverse biased, you would be able to see it on your simulator. Can you?
Just show both emitter and base waveforms and see if they cross over.
If it did get reverse biased wouldn't that stop any collector current flowing?

In the following diagrams, LT Spice predicts the effect of non-linearity of the base emitter junction on the collector voltage and it agrees perfectly with the distortion that the poster was asking about. He has now removed the pictures.

[PLAIN]http://dl.dropbox.com/u/4222062/transistor%20distortion.PNG

 

[skeptic2]

You may be right, but you could consider the following.
It is normal for an amplifier with an unbypassed emitter resistor to produce a very clean output. You also get a high input impedance and low gain, but clean output. Look at the gain, it should be about 7 where it was 170 before, with the capacitor.

The distortion is the same whether there is a bypassed emitter resistor there or not. Try it.

I'm not sure I understand the difference between a clean output and an undistorted output. When I take off the 15 uF cap I get much less distortion.

The emitter bypass capacitor cannot discharge via the emitter resistor because this has the same voltage on it as the capacitor, until you remove the power.

Not quite. When the signal is rising, the capacitor is charging through the transistor. As the base voltage falls, the charged capacitor must discharge through the resistor and if the input voltage falls faster than the capacitor can discharge, the Vbe is reduced and the transistor approaches cuttoff.

If the base emitter juction was getting reverse biased, you would be able to see it on your simulator. Can you?
Just show both emitter and base waveforms and see if they cross over.
If it did get reverse biased wouldn't that stop any collector current flowing?

No I do not see the base emitter junction being reverse biased and I've been careful not to say that the transistor is going into cuttoff, only that it approaches it. For low distortion amplifiers it is usually better to use more stages at lower gain than fewer at higher gain. In amplifiers for audiofiles, bypass capacitors are often eliminated not only because of variation of gain by frequency but also because they introduce phase distortion between low frequencies and high frequencies.

In the following diagrams, LT Spice predicts the effect of non-linearity of the base emitter junction on the collector voltage and it agrees perfectly with the distortion that the poster was asking about. He has now removed the pictures.

[PLAIN]http://dl.dropbox.com/u/4222062/transistor%20distortion.PNG[/QUOTE]

 

[vk6kro]

I understand what you are saying but I can't find any evidence of it here.

When I increase the drive from 0.005 volts peak to 0.05 volts peak, the emitter voltage actually decreases from 0.636 volts to 0.592 volts.