Krill - The Next Generation

[Steve Dunlap]

I would like to start this thread off by looking at three different biasing methods. The first is class B, followed by class A and finally my method which has come to be known as the Krill. I will be looking at the voltage across one of the emitter resistors to determine if that transistor is conducting or not. I will be doing this over a number of posts to keep the size of any one post reasonable.

I don’t expect to change certain minds with this, but those of you that believed me earlier may find it interesting. As many of you know, I can no longer run tests in the physical realm, so this has all been done in simulation. Remember, this is only a simulation. Your reality may vary.

First up is the Douglass Self “Blameless” class B amp. The schematic I used is from here:

http://www.dself.dsl.pipex.com/ampins/dipa/dipa.htm

All simulations were run at 1 kHz.

This is at two watts output. No surprises here, the output transistor cuts off.

[Steve Dunlap]

Next up is the same amp biased into class A at 1.77A per output or 131W dissipation at idle. I used the 37V supplies shown on his schematic. The output is 50W into 8 ohms. As you can see, the transistor is still conducting with 146UV across the 0R1 emitter resistor. This is 1.46ma of current which is safely above the 5UA (max) emitter or collector current specified for the output transistors I use. I did not check this specification on the transistors he uses.

[Steve Dunlap]

Next is the same amp and conditions, but with a 4 ohm load. The output switches off as expected. There also appears to be a slight deformation of the first half of the wave form.

[Steve Dunlap]

And now, the Krill. This is at 50 watts into 8 ohms. You will notice that this amp is of the inverting persuasion. There is no reason for this other than this was what I was working on at the time I decided to do this. I used 0R22 emitter resistors and 34V supplies. I should also state that this amp uses 4 pair of output transistors. Again, it is what I was working on. This is at 129ma bias current for a dissipation at idle of 35.1W of dissipation. The bias was set for minimum distortion at 50W output into 8 ohms at 1 KHz. The wave form flattens on bottom, but you can see that there is still almost 96UV across the emitter resistor. To me, this says the transistor is still on.

[Steve Dunlap]

Much of the flattening is due to scale. Here is the “flat” portion when I zoom in.

[Steve Dunlap]

So what happens when I switch the load to 4 ohms? There is still 2UV across the emitter resistor. Does this mean the transistor has switched off or not? That is still 9UA of current which is above the 5UA cut off value. How on does a transistor need to be to be on?

[andy_c]

Hi Steve,

I thought we finally all agreed what was going on at the end of that big brouhaha? Here's my quick version.

1) On a given half-cycle of the load current, one driver transistor turns off completely, and very fast (due to the capacitor between the driver bases).

2) This leads to a reverse-biased Vbe of that driver, which, because there's no resistor connecting the driver emitters together, leads to zero base current in the output device that the driver is connected to (the one that would ordinarily turn off quickly).

3) The reverse-biased driver Vbe also leads to the voltage difference between the bases of the output devices not being constant, but getting larger when the output stage is sourcing or sinking lots of current.

4) The zero base current of one of the output devices means that the stored charge cannot be swept out of the base of the output device by the driver.

5) This stored charge manifests itself as a slowly decreasing collector current in the output device that normally turns off in a conventional class AB circuit. It will eventually turn off, given time. Try it for 20 Hz for example. Of course, since it's an exponential decay, it will never be identically zero, but it will get arbitrarily close.

One interesting thing would be to show the driver Vbe's in this situation, with polarity chosen so that Vbe of the NPN and Veb of the PNP are shown (both positive in the active region). The other interesting thing is to show the base current in each output device going all the way to zero on alternating half cycles.

[GK]

Alternatively, you could just remove the driver bias resistor from the plain double EF (with standard Vbe multiplier bias) for the same behaviour.

[andy_c]

I agree .

[Steve Dunlap]

5) This stored charge manifests itself as a slowly decreasing collector current in the output device that normally turns off in a conventional class AB circuit. It will eventually turn off, given time. Try it for 20 Hz for example. Of course, since it's an exponential decay, it will never be identically zero, but it will get arbitrarily close.

Here you go, 20Hz. Nothing changed but the frequency. There is now more current flowing in the "off" output.