Does V+ == -(V-) have to hold? - diyAudio
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Old 4th April 2006, 05:22 PM   #1
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Default Does V+ == -(V-) have to hold?

I was thinking that although most people do +/- say 30VDC, that maybe the only reason that its done that way is because thats how the trasformer+rectifier output it. But looking at schematics I was thinking that really all that should matter is the difference between V+ and V-, at least for a simple GC setup like Brians. Am I wrong?
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Old 4th April 2006, 06:02 PM   #2
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It will clip asymmetrically and at lower power.
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Old 4th April 2006, 06:04 PM   #3
DcibeL is offline DcibeL  Canada
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As long as your common point is half way between the rails, I think you are okay. If common (ground) is not half way between the rails, then you will se a DC offset at the output, and will be forced to use a huge capacitor at the output to block that DC.
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Old 4th April 2006, 06:38 PM   #4
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Quote:
Originally posted by DcibeL
As long as your common point is half way between the rails, I think you are okay. If common (ground) is not half way between the rails, then you will se a DC offset at the output, and will be forced to use a huge capacitor at the output to block that DC.
Wrong. the supplies can be assymetric, and the output will still quiess at 0 volts (as long as the minimum voltage for operation is maintained (at both rails))
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Old 4th April 2006, 09:24 PM   #5
DcibeL is offline DcibeL  Canada
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Hmm..I'll take your word about this for now, but I will have to test this when I get the chance.
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Old 4th April 2006, 09:32 PM   #6
Bazukaz is offline Bazukaz  Lithuania
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I think hitsware is right.

If both output and input is referenced to one ground point, output should sit at ~0 V dc , even if this "ground" is not exactly half between the rails.
Try a simulator and you will see.
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Old 4th April 2006, 10:34 PM   #7
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Ok, lets work this out properly. Say you have an (ideal) chip amp with a gain of 20, and rails at +20V and -10V, and the signal is coming in as a 1V peak to peak sinewave. The input has no bias, i.e. it oscillates around the zero point, so the output signal will swing fine to the +20, but when it needs to swing negative, it will only be able to get to -10V before the wave is clipped.

To be able to swing rail to rail with no clipping, you need to bias the input up to a point where it is halfway between the rails, and reduce the gain so the amp doesn't clip, so in this case, you need to add an offset voltage to the input signal, and reduce the gain to 15. This bias voltage carries right through the amplifier, producing an output offset of +5V as well, so to avoid blowing up your speakers, you then need to remove this, which is why you would need a DC blocking cap on the output.

Does that make sense?
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Old 4th April 2006, 11:21 PM   #8
DcibeL is offline DcibeL  Canada
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Yes, the bias voltage is required at the input in order for output to swing rail to rail. I must have had a brain fart this morning, I was trying to think back to when I learned single supply op-amps in school. I guess I am already loosing my memory in my old age .
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Old 4th April 2006, 11:51 PM   #9
cpemma is offline cpemma  United Kingdom
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Quote:
Originally posted by DcibeL
...I was trying to think back to when I learned single supply op-amps in school. I guess I am already losing my memory in my old age .
Maybe you're not ready for the knackers yard just yet - some opamps aimed at single-supply applications (eg, LM324, LM358) will swing down to ground but only up to 1.5-2V short of the positive rail, so there may be a point in setting their virtual ground a volt below midway to get the maximum swing.

Not hi-fi though.
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Old 5th April 2006, 12:46 AM   #10
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Default V+==V-

Pinkmouse is very correct, for the amplifier to swing symetrically rail to rail the center has to be the center, see the point in center???
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