Nonlinear distortion

Discussion in 'Wireless & RF Design' started by PRS, Mar 12, 2010.

1. PRS Thread Starter Well-Known Member

Aug 24, 2008
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35
I made a CE amp from first order design considerations. It has a gain of 10 volts per volt. A triangle wave input maintains its shape until the frequency is increased beyond a certain point. Then it gets distorted. What causes this distortion and how can I fix it?

2. hgmjr Moderator

Jan 28, 2005
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Post your schematic and one of our members can see if there is anything obvious.

hgmjr

3. t_n_k AAC Fanatic!

Mar 6, 2009
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Also give an indication of the nature of the distortion occurring and the frequency at which it appears.

If the frequency is high enough you may see the Miller effect causing the triangular wave to become more "sinusoidal". The higher order harmonics in the input wave would be more severely attenuated than the fundamental frequency.

It would be also likely that with a capacitively coupled CE amplifier one would see low frequency side distortion effects.

4. PRS Thread Starter Well-Known Member

Aug 24, 2008
989
35

I attached the schematic. I appologize for the sloppy drawing, but it's the best I could do using PAINT and a mouse.

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5. PRS Thread Starter Well-Known Member

Aug 24, 2008
989
35
A triangle wave becomes ever more like a sine wave in that the top and bottom get rounded and the diagonals get warped at around 400 kHz. Below that the triangle shape is retained.

I posted the schematic just above this. The signal maintains its shape until the output of the CE stage.

Now that sounds plausible and that would explain it. The output capacitance would be in parallel with the output and load resistances and consitute a LP filter. Is that it? If so, then every CE amp has the same problem at some frequency just before its -3dB point. Unfortunately, I didn't record its BW. It worked at 455 kHz, and that was all I was concerned with at the time. I'm tempted to build it on a copper clad board and retest it just to get to the bottom of this.

Could you elaborate on this? I did not use capacitive coupling with the CE stage, but the concept is interesting.

Apr 5, 2008
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Hello,

A triangular signal consists of odd harmonics.
If the band with of the amplifier is to low the output will not amplify the higher harmonics.
This will lead to a signal that looks more and more like a sinewave (the base frequency of the triangle ).

Greetings,
Bertus

7. PRS Thread Starter Well-Known Member

Aug 24, 2008
989
35
Yes, this is exactly what it turned out to be. Tnk explained it the same way. I rebuilt the circuit, this time on copper clad board and tested it. The BW is 1.05 Mhz, but it starts rolling off ever so slowly at 350 kHz. The higher harmonics got attenuated and the triangle rounded itself into a near sine wave at 455 kHz.

What Tnk said about low frequency distortion is also true. A triangle starts distorting at 50 Hz and gets worse as I decrease the frequency. I would conjecture this is due to the effect of the coupling capacitors acting as HP filters and wiping out the lower harmonics first. ???

8. skeptic Active Member

Mar 7, 2010
51
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What is the purpose of the 33 uF capacitor? How did you arrive at it's value as well as the values of the input and output capacitors?

9. PRS Thread Starter Well-Known Member

Aug 24, 2008
989
35
When I first made this circuit it was for 455kHz. The formula for the -3dB rolloff or rollon of an RC filter is $f = \frac{1}{2piRC}\$. This applies to both high pass and low pass filters.

In the case of the coupling capacitors (at the input and output) R is the resistance the capacitor is looking at and you pick C such that your desired f lies within the passband.

Originally, this circuit was for a 455kHz amp, but when I made it universal so to speak, I wanted it to pass lower frequencies. The critical capacitor in the case of low frequencies was the capacitor at the emitter of the CE amp. It is a high pass filter and the formula given above requires a big capacitor due to the small R at that point. More if you want.

10. skeptic Active Member

Mar 7, 2010
51
9
Thanks for your reply. The thing that seems the most obvious to me is the 0.3 V drop from base to emitter of the second transistor. If we look at the 10 K resistor from the collector to 15 V and calculate the current it works out to 0.9 mA. If you multiply the 680 and the 1.5 K resistors in the emitter circuit by 0.9 mA and add them together you get only 1.962 V. Allowing for resistor tolerances it allows for about a 0.6 V base to emitter voltage drop. So why are you seeing only a 0.3 V drop?

It seems to me the 33 uF capacitor is charging through the 680 resistor and discharging through the 1.5 K resistor with each cycle. That puts the junction between the 680 and 1.5 K resistors at a higher voltage than it would be without the capacitor. This higher voltage tends to bias the transistor into cutoff. If the emitter is only 0.3 V below the base it means that only the positive peaks of the waveform are being amplified. Does the distortion change if you remove that capacitor?

Naturally your gain will drop when you remove the capacitor. To get the gain back you may want to increase the collector resistor from 10 K to 33 K.

11. PRS Thread Starter Well-Known Member

Aug 24, 2008
989
35
Hi skeptic! Thanks for your your insiteful criticism. You're right about the inconsistent DC voltage values. As I hinted above, this circuit was breadboarded at first and I tweaked some values without recording the changes in voltages. The 680 ohm resistor, for example was found experimentally to give a gain of exactly 10v/v when all was said and done.

I hooked the board back up and found the voltage from the source of the stage 1 into the emitter of stage 2 is 2.8 volts. The voltage at the emitter of stage 2 is 2.2 volts and the voltage at the collector of stage 2 is actually 4.6 volts and the voltage at the emitter of stage 3 is 4.0 volts.

I can tell you positively after I go look.... Since it is now built on copper clad I had to cut a lead of the 33uF cap. Gain got reduced without improving the shape of a triangle.

I think the fix lies in using a faster transistor. But thanks for your input. Your attentiveness makes me feel like I'm not just wasting my time drawing a circuit with Paint! LOL. Thanks again.