Asymmetrical Clipping in Modified JLH Output Stage - diyAudio
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Old 7th May 2013, 02:18 PM   #1
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Join Date: Jan 2007
Default Asymmetrical Clipping in Modified JLH Output Stage

I have attached the circuit of a JLH-style amplifier I built for headphone use. It uses an opamp and a +/-18 volt power supply to eliminate the DC-blocking capacitor at the output. The NPN driver transistor of the original was replaced with a PNP emitter follower configuration - the original design would require the output of the opamp to venture too close to the negative rail to be stable.

With no load but a Zobel network, the amplifier is totally clean to within a volt or so of the rails. As soon as a load is added, however, the positive-going signal clips at about half the supply voltage. This is largely independent of the output stage bias current.

From memory, the feedback resistor is 100k, the resistor to ground is 10k, and the emitter resistor varied between 4.7k and 10k (depending on bias current being tested).

Does anyone have any ideas as to why this topology would clip only on the positive-going signal, and only when the load is present? Any suggestions would be greatly appreciated.

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Old 8th May 2013, 04:27 AM   #2
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Join Date: Nov 2007
Location: Dallas
The output transistor bases draw more current as the load draws current to GND.
That pull-up resistor may need to be a smaller value, and probably the midpoint
of it bootstrapped by a capacitor to the output.

You got plenty of voltage across that resistor to keep bias current flowing on the
downswing, so you are only suffering on the upswing. Bootstrap will help.

Last edited by kenpeter; 8th May 2013 at 04:33 AM.
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Old 8th May 2013, 02:24 PM   #3
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Join Date: Jan 2007
I accidentally deleted my original post and had to retype it. In doing so, I forgot to mention that I also tried replacing the emitter resistor with a simple constant current source (another PNP transistor with a base resistor connected to ground) and it suffered from the same issue - flat-topping of the positive-going signal at eight or nine volts. My understanding is that a bootstrap is essentially a different way of approaching the same problem. 8\
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