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Old 8th February 2004, 08:00 PM   #1
mikelm is offline mikelm  England
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Smile The sound of two transistors switching

Having concluded that the only way forward for me was a new CD player, yesterday, after visiting a local Hi Fi show, I felt inspired to do a tweak or two. Well who would have thought just two cheapo resistors could have such a huge effect.

I simply biased my LM 6182 cd723 o/p device into class A using two 2K resistors to the neg rail. I put them after the RC filter to avoid connecting the RF hash to the supply.

Judging from what I heard at the show I've just added a grand or to to the value of my player. It now sounds huge, relaxed and much cleaner in the mid and treble. I think I might try a new clock next that will replace the original one and feed the DAC.

I don't know if biasing op amps into class A would be quite as dramatic 1) in other cct's... or 2) if the amp following was not also class A, but I would stronly recommend that anyone using an op amp in any circiut to give this simple mod a try.

If you have this chip in a cd723 you're in for a real treat

mike
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Old 8th February 2004, 08:33 PM   #2
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How did you bias the opamp into Class A operation?
Looking at the schematic for the LM6182, I can only see how you can create a d.c. offset by adding resistors to the negative rail.
Am I missing something?
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Old 8th February 2004, 09:15 PM   #3
subwo1 is offline subwo1  United States
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I saw that idea here from mikelm and thought that, yes, I can see the benefit of trying such a thing. In the past, I would not have thought much of such an idea, but my research into class D circuits has expanded my insight.

The resistors connect from the output of the chip to the negative supply rail of the chip to force it to keep supplying current from the positive rail so that the output transistors of the chip will not crossover completely. It is not very inefficient for such a low power circuit, or, rather, the losses are insignificant for the CD player.

It makes sense that in low-level circuits that any crossover distortion is amplified when the power amp drives the speaker, and what was a small amount in proportion to the total is still in the same proportion, but crossover distortion is so very egregious that even small amounts may have a negative subjective effect.
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Old 8th February 2004, 09:18 PM   #4
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Quote:
Originally posted by Frank Berry
How did you bias the opamp into Class A operation?
Looking at the schematic for the LM6182, I can only see how you can create a d.c. offset by adding resistors to the negative rail.
Am I missing something?

I think it is more of an urban legend than anything else. the added resistor is more like another load to the output stage, except that it draws more from the positive side of the push-pull transistors. The switching is still there, assuming that you are not overloading the output stage.

Now, if you were to inject a small DC voltage to the signal such that you use only the one side of the push-pull, you are indeed eliminating the switching.

See your input is 20mvpp, and you add a +10mv DC signal to it. Now the combine signal swings from 0mv to 20mv peak. after the 10x opamp, it goes from 0mv to 200mv peak, all on one side of the opamp.

If you are really concerned about cross-over distortion, you can add a larger DC offset to the orignal signal, see 20mv, and your output will swing from 10x (20mv - 20mv/2) to 10x (20mv + 20mv / 2). and you are definitely in class A.

But I am not sure if that does anything to the sound.
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Old 8th February 2004, 09:25 PM   #5
mlloyd1 is offline mlloyd1  United States
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I'm a little surprise you would sense such a big change - doesn't this device idle at about 10mA per op amp before this change?

Maybe I'm thinking about another part?

Ah well, if you like it, so be it :-)

mlloyd1
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Old 8th February 2004, 09:29 PM   #6
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Quote:
Originally posted by subwo1
The resistors connect from the output of the chip to the negative supply rail of the chip to force it to keep supplying current from the positive rail so that the output transistors of the chip will not crossover completely.
it will cross-over completely. except in this case, the upper transistor in the PP stage "pushes" out more current, and the lower transistor in the PP stage "pulls" in less current, when they are conducting during their respective cycles. Both are conducting exactly 180 degrees, as they did without the added resistor.

otherwise, you would have created DC offset at the output.

To me, the only way you get a class B amp to work in Class A is if you inject intentionally a DC offset (at the input) so that only one of the two transistors in the PP stage conducts during 360 degrees. You can then use a cap to filter out the DC offset (at output).
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Old 8th February 2004, 09:33 PM   #7
mikelm is offline mikelm  England
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here is the cct

because the load is much higher resistance that the 2Kohm resistor to the neg rail whatever happens dc will always flow through only one of the chip o/p tranistors to the neg. The signal will be modulated onto this dc - class A

probably it's more elegant to use a current source - might sound better

mike
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Old 8th February 2004, 09:39 PM   #8
subwo1 is offline subwo1  United States
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Yes, I think a current source would be even better.
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Old 8th February 2004, 09:41 PM   #9
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Quote:
Originally posted by millwood
To me, the only way you get a class B amp to work in Class A is if you inject intentionally a DC offset (at the input) so that only one of the two transistors in the PP stage conducts during 360 degrees.
I simulated a push-pull stage (2n5551+2n5401, no feedback and no bias) driven by a 2vpp 1Khz signal. THD at 46%. Adding a 2v DC offset to the input signal gets THD at 0.5% from the same circuitry.

Quote:
Originally posted by mikelm
because the load is much higher resistance that the 2Kohm resistor to the neg rail whatever happens dc will always flow through only one of the chip o/p tranistors to the neg. The signal will be modulated onto this dc - class A
all it does, Mike, is that the upper transistor will work harder and the lower transistor will work less harder. Both will still switch on and off at exactly the same points.

Quote:
Originally posted by mikelm
probably it's more elegant to use a current source - might sound better

mike
a current source wouldn't do a whole lot of anything for you.
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Old 8th February 2004, 09:49 PM   #10
subwo1 is offline subwo1  United States
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all it does, Mike, is that the upper transistor will work harder and the lower transistor will work less harder. Both will still switch on and off at exactly the same points.
I would agree that its effect would not be class A if the output is having to drive a load through ground or a point between the power supply rails beyond a certain point. But if the amplitude of the output is able to be kept low enough, the class A operation could hold up.
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