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Old 15th October 2003, 10:20 PM   #1
Steven is offline Steven  Netherlands
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Default Spreading the heat in class A

Just an idea that popped up during some browsing through all the class A threads, showing BIG amplifiers and BIG heatsinks to get SOME power for your speakers.
It is about a push-pull output stage using emitter followers, where only the top half of the output circuit is shown, using an NPN transistor. The other half can be a PNP mirror of this.

What if we keep the output transistor (Q1) in class A at high current but limit the voltage across the transistor? This keeps the dissipation limited, so less transistors needed in parallel and a smaller heatsink. But of course we want reasonable voltage swing and thus need a power supply with a less limited voltage. If we now insert a power resistor (R1) in between the collector of the output transistor and the supply line, this resistor can dissipate most of the power that would otherwise be dissipated in the output transistor. Because a resistor is allowed to be much hotter than a transistor, in this way more power can be dissipated in a smaller volume. Do not mount the resistor on the same heatsink as the transistor, though.
But as soon as the base of Q1 goes up and Q1 needs to deliver current to the load, Vce of Q1 may become too small and Q1 would saturate. Here Q2 comes into action. It is a kind of cascode for Q1 and is driven from the output via D1. Current beyond the bias current of Q1 will be delivered via Q2. Probably even part of the bias current of Q1 is already delivered by Q2 and part by R1. Q2 keeps Vce1 at a voltage of Vz-Vbe2. Current source I keeps the zener diode biased and supplies the base current for Q2.

I have not started making calculations (nor simulations) for this setup to determine where the optimum is. I think we should strive for a situation where most of the dissipation is in R1 and some in Q1 and Q2 for most of the time, normally when signal levels are low.

This circuit may also improve linearity of the class A amp because Vce1 is kept fairly constant for all signal levels.
This circuit may also be interesting for class AB amplifiers hat drive capacitive loads. For capacitive loads the output current is already large for small output voltages, making the dissipation in Q1 large (large Vce, large Ic). Insertion of R1 will then lower Vc1. For class AB amplifiers Ic1-bias will be low, so the voltage across R1 will be low with no signal. In order not to reverse bias Q2 it will be required to insert a diode in the base of Q2, otherwise emitter-base zenering will occur.

I don't know if this circuit is new. I did a quick search, but could not find the right keywords for a usefully limited answer. I just post it on the forum to see what happens.

Steven
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Old 16th October 2003, 12:41 AM   #2
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Old news to some, but Bob Carver patented the idea of using three different +/- supply rails and switches in only the amount of voltage needed.

Your idea is interesting. I wonder if anyone has tried it.


JF
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Old 16th October 2003, 01:27 AM   #3
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Old as the hills. ;-) Has been tried before, maybe patented by Nelson Pass.
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Old 16th October 2003, 06:08 AM   #4
Steven is offline Steven  Netherlands
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Quote:
Originally posted by johnferrier
Bob Carver patented the idea of using three different +/- supply rails and switches in only the amount of voltage needed.
I know, that is 3-stage class G. But then you get voltage jumps on the collector of Q1, which show up as a kind of crossover distortion at the output for each transistion to another power rail. With 3 positive and 3 negative supply rails you get 5 transitions that may generate crossover distortion. The purpose of class G is to make the amplifier more efficient. My circuit is not more efficient but only moves part of the heat to a resistor that is more capable of dissipating than a transistor. Also the circuit does not have any transition 'areas' that could introduce distortion.

Quote:
Originally posted by john curl
Has been tried before, maybe patented by Nelson Pass.
Any references, you know of?

Steven
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Old 16th October 2003, 10:04 AM   #5
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Quote:
Originally posted by Steven
Also the circuit does not have any transition 'areas' that could introduce distortion.
Steven

would you get any distortion as Q2 starts to kick in?
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Old 16th October 2003, 11:44 AM   #6
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Quote:
Originally posted by Steven
Any references, you know of?
U.S. Patent #5,343,166, which can be viewed by doing a patent number search at www.uspto.gov.
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Old 16th October 2003, 11:47 AM   #7
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Default Re: Spreading the heat in class A

Quote:
Originally posted by Steven
Just an idea that popped up during some browsing through all the class A threads, showing BIG amplifiers and BIG heatsinks to get SOME power for your speakers.
It is about a push-pull output stage using emitter followers, where only the top half of the output circuit is shown, using an NPN transistor. The other half can be a PNP mirror of this.

What if we keep the output transistor (Q1) in class A at high current but limit the voltage across the transistor? This keeps the dissipation limited, so less transistors needed in parallel and a smaller heatsink. But of course we want reasonable voltage swing and thus need a power supply with a less limited voltage. If we now insert a power resistor (R1) in between the collector of the output transistor and the supply line, this resistor can dissipate most of the power that would otherwise be dissipated in the output transistor. Because a resistor is allowed to be much hotter than a transistor, in this way more power can be dissipated in a smaller volume. Do not mount the resistor on the same heatsink as the transistor, though.
But as soon as the base of Q1 goes up and Q1 needs to deliver current to the load, Vce of Q1 may become too small and Q1 would saturate. Here Q2 comes into action. It is a kind of cascode for Q1 and is driven from the output via D1. Current beyond the bias current of Q1 will be delivered via Q2. Probably even part of the bias current of Q1 is already delivered by Q2 and part by R1. Q2 keeps Vce1 at a voltage of Vz-Vbe2. Current source I keeps the zener diode biased and supplies the base current for Q2.

I have not started making calculations (nor simulations) for this setup to determine where the optimum is. I think we should strive for a situation where most of the dissipation is in R1 and some in Q1 and Q2 for most of the time, normally when signal levels are low.

This circuit may also improve linearity of the class A amp because Vce1 is kept fairly constant for all signal levels.
This circuit may also be interesting for class AB amplifiers hat drive capacitive loads. For capacitive loads the output current is already large for small output voltages, making the dissipation in Q1 large (large Vce, large Ic). Insertion of R1 will then lower Vc1. For class AB amplifiers Ic1-bias will be low, so the voltage across R1 will be low with no signal. In order not to reverse bias Q2 it will be required to insert a diode in the base of Q2, otherwise emitter-base zenering will occur.

I don't know if this circuit is new. I did a quick search, but could not find the right keywords for a usefully limited answer. I just post it on the forum to see what happens.

Steven
Steven,

I have done a design like that for Audio Amateur in the 80's, but without the resistor. The main advantage of my design was that the output transistors could be selected for linearity and high bandwidth, while the "regulator' transistor can be selected for robustness and high power dissipation. The result is that the supply for the output transistors shows signal-related ripple, but that can be taken care of by having a good PSRR in the main amp channel. Worked quite well.

Jan Didden
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Old 16th October 2003, 04:09 PM   #8
Steven is offline Steven  Netherlands
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Default Re: Re: Spreading the heat in class A

Quote:
Originally posted by millwood
would you get any distortion as Q2 starts to kick in?
No need to kick in for Q2. It can be conducting all the time (in class A). Both R1 and Q2 take care of the current through Q1.

Quote:
Originally posted by Joe Berry
U.S. Patent #5,343,166, which can be viewed by doing a patent number search at www.uspto.gov.
I found this patent too this morning (in Europe), but had no time to answer. Still thanks.
This patent describes something else, although on first sight it has some similarities. Patent #5,343,166 from Nelson Pass describes a floating current source powered from an additional (low voltage, high current) power supply that sources a current through Q1 in order to keep Q1 in class A and to minimize the current through Q2 because that would cause a lot of dissipation in Q2 which runs on a higher voltage. Purpose of this circuit is then to get a higher efficiency. Q1 and Q2 are not mentioned in the patent, but are the equivalents in my diagram. Numbering in the patent is completely different. The patent does not use resistors, as in my circuit R1. The only reference to a resistor in this sense is in column 3 at the bottom: "Alternatively, the high source impedance of the current source could be provided by a resistor." But what is meant in the patent is that you can have a floating power supply with a series resistor to increase the impedance across Q1. The impedance should be high because the voltage across Q1 is fixed by Q2 and the zenerdidode, and these two should not interfere.

As already said, the purpose of my circuit is not to get a higher efficiency but only to move part of the dissipation to a resistor. I do not use an additional floating power supply.

Quote:
Originally posted by janneman
I have done a design like that for Audio Amateur in the 80's, but without the resistor. The main advantage of my design was that the output transistors could be selected for linearity and high bandwidth, while the "regulator' transistor can be selected for robustness and high power dissipation. The result is that the supply for the output transistors shows signal-related ripple, but that can be taken care of by having a good PSRR in the main amp channel. Worked quite well.
This is more in line with the original patent of Nelson Pass 4,107,619 that uses a cascode on the output transistors to keep the Vce of each output transistor nearly constant. There is no resistor R1 that takes over part of the dissipation. The advantage of improved linearity is still part of my circuit, since I use the cascode too.

Steven
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Old 18th October 2003, 07:29 PM   #9
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Quote:
As already said, the purpose of my circuit is not to get a higher efficiency but only to move part of the dissipation to a resistor. I do not use an additional floating power supply.
Using multiple parallel output devices does also spread the load.
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Old 24th October 2003, 01:45 AM   #10
PRR is offline PRR  United States
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The idea of blowing heat in a sturdy resistor instead of a fragile transistor seems to make sense. It does work fine in a voltage regulator when you have to get rid of excess voltage.

Modern silicon is shockingly tough. I have seen transistors melt the solder on their legs yet keep working. Probably not for long, but it was like that for 20 minutes before I wondered why the soldered-in transistor was flopping around. You can't run resistors really-really hot and expect long life either. I don't think the saving in space will be all that large. (The savings in cost, surplus power resistors versus oversize heatsinks, may indeed be substantial for one-off construction.)

But I say, at risk of being proved wrong, that "this won't work".

Draw the full schematic, top and bottom. As you raise the input positive, you are dragging less current through the top resistor, but more current through the bottom resistor. WHat you have really done is added a couple of power resistors in parallel with the load. Although the resistors drop a few volts less than the load, the bootstrap drops the same amount, so supply voltage must be higher and total dissipation is higher for the same power.

I can't see a way it won't switch under any condition. I have not plotted every point nor variant topologies, but I don't think it can be done. If it doesn't switch going up, then it will switch going down. The switching spike may be small, but there will be some bleed-through, and roughly comparable to a Class AB design.

If you can show and defend a set of values that does not switch, I'll be very surprised.

If we take the situation where resistor drops essentially the whole idle power, the value of the resistors is about 16Ω for an 8Ω load, so the effective load is 16||16||8= 4Ω. To even stay Class A, we have to double the idle current just to drag these resistors around. But for the resistors to carry that new doubled idle current, they have to be 8Ω, total load 8||8||8= 2.6*Omega;, and now we have to triple idle current to stay Class A. I have not resolved this paradox, but I think all the solutions lead to higher total power dissipation, except the case where the resistor current is so small that it really does nothing.

So I think all this does is add dummy resistors and heat. And I think it won't actually reduce transistor dissipation at all.

If you can show and defend a set of values with total dissipation even as low as the conventional Class A topology, you can "Nya-Nya!! Dummy!!!" me.

I don't mean to discourage you. I'd be real interested to see a system like this with reduction of transistor dissipation. I've certainly been wrong enough before. But I've run through a lot of such schemes, thinking I was "on to something", and eventually found the fallacy. (One recent one troubled me for 20 years.) My gut says this one won't work either. Prove me wrong.
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