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babadono

Amp Bridging

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Do we have any amp gurus left here on the forum? I am having a hard time understanding what I am measuring and would like to know the math involved to explain it.

I am going to build a hypex class D amp that is bridgeable into 8 ohms using two of their amp modules. It will be capable of 400W into 8 ohms. This works out to 56.56 Vrms across the 8 ohms. That in turn works out to 160V peak to peak. Hypex sells and recommends a power supply that outputs +/- 65 volts for these amp modules. So I asked them how will the output will be able to go 160 Vpp when there is only 130v(+/- 65) available. They said basically don't worry you will get double the voltage from the power supply when bridging.

So I brought one of my Crown D75 amps to my bench. I put it in bridge mode and loaded it with 8 ohms. And I get 112W across 8 ohms which is the same total wattage I get if running in 2 channel with 4 ohm loads. Ok so cool I get that. But when the amp is bridged and outputting 112w into 8 ohms the power supply is loaded down to +/- 25 volts so 50 volts total. But to put 112W across 8 ohms the peak to peak output voltage is 85 volts. How does this happen with 50 volts to work with? What is the mathematical relationship? I know it ain't magic. Anybody know the math?

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Well first 56.6 x 2 is not 160V it's 113.2V. 

 

Bridging lets one amp carry the + and the other the - of the wave form, so you get a swing of 2x the rail voltage. 

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6 minutes ago, JohnA said:

Well first 56.6 x 2 is not 160V it's 113.2V. 

Yes but 56.6 Vrms = approx 160Vpp.   Vrms x 2.83(2 x square root of 2) = Vpp.  No?

 

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14 minutes ago, JohnA said:

so you get a swing of 2x the rail voltage. 

That is what Hypex said too. But offered no math to back it up.

So that would imply 260 volts available instead of 130. Which would mean the amp is capable of over 1000 W. when bridged.  And that is not true. So there has to be a more mathematical and true explanation.

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How many watts can the amps put into a 4 ohm load when run single-ended?  I wonder if there's a current limiting issue.

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The Crown D75 puts out half of 112W into both of the stereo outputs when run 2 channel into 2 4 ohm loads. So 2 x 56 w. The power supply is then down to +/-25 volts just like when bridged into 8 ohms.

The Hypex amps? I don't know not working with them yet.

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Your Crown D75 is a perfect example of current limiting.  It's 40 watts into an 8 ohm load, but 55 watts into a 4 ohm load.  This tells me that it's limited to 3.7 amps per channel, not the necessary 4.5 amps required for 80 watts.  Your loaded +- 25 volt supply equates to 35.35 rms volts bridged, more that the 30 volts required for 110 watts into an 8 ohm load.

 

If the doubling of the voltage when bridged concept is still confusing, consider this.  When driving a speaker single-ended, when the + side of the speaker has +25 volts, the - side has 0 volts, a 25 volt potential difference.  When bridged, the + side of the speaker has +25 volts, the - side has -25 volts, a 50 volt potential difference.  

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You've got the two DC rails at +/- X volts.  When a waveform is heading toward one rail in "normal" use you can end up with only X volts maximum across the load at any given instant.  When bridging, you're inverting the signal to one of the amp channels and driving them simultaneously to the single load, so when a waveform is heading to one one rail, it's also heading to the other rail at the same time on the other lead, so you can end up with 2X volts maximum across the load at any given instant.  Did you not understand that?

 

So on only 1 of the two peaks in "peak-to-peak" you've got 2x the rail voltage, and on the inverse peak it's another 2x.  Ultimately you wind up with 4x a single rail voltage for your available peak-to-peak output.

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5 hours ago, babadono said:

when the amp is bridged and outputting 112w into 8 ohms the power supply is loaded down to +/- 25 volts so 50 volts total. But to put 112W across 8 ohms the peak to peak output voltage is 85 volts. How does this happen with 50 volts to work with? What is the mathematical relationship?

 

Like I said above, what you've got in that case is +/- 50 volts (25 volts each in equal and opposite directions simultaneously), or 100 volts to work with.

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So the +/- 25 volt supply can provide a potential of 50 volts positive for the amp that is driving the signal positive and a 50 volt potential that is negative to the amp that is driving the signal negative at the same time? And the 15 volt loss from the theoretical 100volts down to the actual 85 is just the loss across the output transistors?

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I don't know about loss across transistors...  I do know that when bridged, the load appears to the amp as half what it would be un-bridged.  

 

Just build it the way they suggest.  You'll be satisfied, I'm sure.

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You remind of someone I miss very much around here.  

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12 hours ago, babadono said:

So the +/- 25 volt supply can provide a potential of 50 volts positive for the amp that is driving the signal positive and a 50 volt potential that is negative to the amp that is driving the signal negative at the same time? And the 15 volt loss from the theoretical 100volts down to the actual 85 is just the loss across the output transistors?

 

Sorry,  but can't seem to let this go.  The 100 volt comparison is misleading.  It's true than bridging two amps with a +- 25 volt power supply can deliver the same voltage across a load as one single-ended amplifier with a +- 50 power supply,  but it's not the same circuit.  Again, each of the bridged amps (+- 25) can apply a 17.7 volt rms signal to it's side of the load, with the second amp's signal inverted.  The single-ended amplifier (+-50) can apply a 35.4 volt rms signal to one side of the load and will apply zero volts to the other side. 

 

Notice I said "can" apply.  The D75 has a voltage gain of around 20, so a .5  volt rms input signal will be amplified to 10 volt rms to drive one side of the load.  10 volts rms equals +- 14.414 volts.  Since the power supply can deliver up to +-  25 volts, your good.  The extra volts are not wasted or lost, they are just not used.

 

 A 1 volt rms input signal, of course,  will be amplified to +- 28.29 volts, and clip.

 

Likewise, just because an amp can apply a certain voltage doesn't mean it supply the current. That's why the D75's 4 ohm rating isn't twice the 8 ohm rating.

 

 

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Thanks to all. Anyone got any links to articles that explain this any better? I found Rod Elliots pages, any others?

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13 hours ago, Deang said:

You remind of someone I miss very much around here.  

Who?

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1 hour ago, babadono said:

Thanks to all. Anyone got any links to articles that explain this any better? I found Rod Elliots pages, any others?

 

This is pretty good.

 

https://en.wikipedia.org/wiki/Bridged_and_paralleled_amplifiers

 

It was hard for me to understand the concept that each of the bridged amps "sees" one half of the speaker's impedance, until I worked backwards from a mono amp.

 

1)  A single-ended mono amp with a +-50 volt power supply delivers the same power to an 8 ohm load as two bridged amps with a +-25 volt power supply.  156 watts.

2)  If the two bridged amps combine to deliver 156 watts, then each amp must deliver half the power, or 78 watts.

3)  Each amp is applying 17.7 volts rms and 78 watts to it's side of the 8 ohm driver.

4)  Ohms = Volts squared / Watts.  17.7 squared / 78.  313.3 / 78.  4 ohms.

 

Not sure if that's the correct explanation, but it worked for me.

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That works for me.

 

Here's something to consider as well.  Draw a graph in your mind with the X axis being 0 volts and a waveform alternately going positive and negative; it's easy, you've seen it a million times.  'Normal' operation would consist of one lead hooked to the X axis and the other lead hooked such that it follows the curve.  When bridging, everything works the same, except that what would normally be a stationary X axis now moves in equal and opposite directions from the curve; that's to say it's no longer stationary.  Or you could consider it to remain stationary but the waveform is affected by a 2x multiplier at all times.

 

I've used bridged amps in the past, and wouldn't rule them out in the future, but there is one conceptual aspect that I've always considered.  It pertains to linearity between any two samples of a component; transistors in this case.  It may be that any two have exactly the same operational characteristics, but I rather doubt it (I mean exactly the same in all respects).  Compound that to include other circuitry as well.  The purist in me thinks that bridged mode must entail a certain amount of degradation.  It's likely below the threshold of notice for the most part, and in this particular case the Hypex stuff makes use of liberal negative feedback anyway, so my musing is likely unimportant.  But the thought remains in the back of my mind nonetheless.

 

I'm a fan of Hypex.  My NAD uses a UCD implementation.  I like how the following generation (forget what it's called) has a pinout for a clipping indicator, which I'd utilize in a heartbeat - just because I like to know such things.

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20 hours ago, Deang said:

You remind of someone I miss very much around here.  

 

I'll take that as a compliment.  So he was an opinionated *** as well?  ;)

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1 hour ago, glens said:

forget what it's called

Are you talking about NCore?

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1 hour ago, glens said:

I've used bridged amps in the past, and wouldn't rule them out in the future, but there is one conceptual aspect that I've always considered.  It pertains to linearity between any two samples of a component; transistors in this case.  It may be that any two have exactly the same operational characteristics, but I rather doubt it (I mean exactly the same in all respects).  Compound that to include other circuitry as well.  The purist in me thinks that bridged mode must entail a certain amount of degradation.  It's likely below the threshold of notice for the most part, and in this particular case the Hypex stuff makes use of liberal negative feedback anyway, so my musing is likely unimportant.  But the thought remains in the back of my mind nonetheless.

The topology In the Crown amp that I am testing on my bench just uses the second channels error amp as a unity gain inverter to make what i will call the negative side driver just an inverting slave of the high side drivers error amp. So it should behave as one composite amp. The Hypex set up will be different. Hypex says just hook up the low side drivers inputs out of phase from the high sides inputs. So there will be two error amps working on two feedback signals. Will this make a sonic difference? That you can hear?

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