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Old 7th May 2012, 08:23 PM   #11
Enzo is offline Enzo  United States
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Oops, posted while posts were posting...


Doesn;t matter what voltage rails you decide on, the principal is the same.

If you want a single 9v supply then yes, the output goes from 4.5v at rest to +9 or to 0. SO if you have two channels doing that and one is opposite in polarity from the other. Then connecting a speaker from the hot lead of one to the hot lead of the other results in the +9 on one end of hte speaker with 0 on the other, so you get the full 9v instead of just the 4.5.
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Old 7th May 2012, 08:28 PM   #12
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Because with a single supply rail and a single output the output is cap coupled. The output sits at some voltage potential and cannot be connected directly to a loudspeaker through a standard ground returned system. If you had a single supply amp powered by a 9 volt supply what you really see happening is the output swinging + and - about 9/2 volts. In other words you're power is defined by a waveform that swings +- 4.5 volts.

In the case of the BTL two outputs are tied together. Both of them sit at exactly 1/2 the rail voltage. This means that the + and - inputs to the loudspeaker are both connected to two individual amplifier channels and are held at the same potential (Vcc/2) and no current flows.

In this case as the +ve output swings high, up to +9V, the -ve output swings low down to 0 volts, giving you a positive going voltage peak of +9v. However this time around when the +ve output swings low, to zero volts, the -ve output swings high up to +9V and in this case gives you a negative going voltage peak of -9V. Here we end up having the potential to swing +9v and -9 volts.

Edit - It's like in the second case you're connecting a 9V battery up to the loudspeaker and the cone moves forwards. Then to get the cone to move backwards you reverse the polarity of the battery. In this case you're loudspeaker cone is moving as if it were energised one way by 9volts, then the other way by 9 volts, to a total peak to peak value of 18 volts.
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Last edited by 5th element; 7th May 2012 at 08:33 PM.
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Old 7th May 2012, 08:37 PM   #13
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Quote:
Originally Posted by HankMcSpank View Post
But how will amplifier 2 be at -4.5V?
Because the amplifier is biased to the mid-point of the rails by necessity. It is undesirable to apply DC to the load under a no-signal condition. Consider a single-ended supply amplifier which is AC coupled to the load. The steady state output of the amplifier (with no signal) is biased at +4.5V DC. AC coupling this to the load via a capacitor means that the load sees zero volts when no signal is present. A swing of between 0 and +9v at the amplifier output becomes transposed into a swing of -4.5V to + 4.5V in the load. When amplifier 1 is at +4.5V (instantaneous signal), amplifier 2 is at -4.5V. Amplifier 2 is fed from an inverted signal being fed to amplifier 1.
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Last edited by currentflow; 7th May 2012 at 08:43 PM.
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Old 7th May 2012, 10:50 PM   #14
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Thanks I get it now :-)
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Old 19th May 2012, 08:44 AM   #15
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For the purposes of calculating current through a speaker coil for single ended vs BTL

(assuming just DC resistance & ignoring impedance for simplicity)...

Coil = 8 ohm
peak to peak AC signal voltage swing = 8V (as measured with a scope on one leg)

To calculate the current through speaker coil, I need to convert the peak to peak swing to RMS, but since this is BTL, do I therefore double the swing ...therefore 16V peak to peak?

So for single ended driving amplifier, 8V peak to peak = 2.828 RMS...therefore divided by 8 Ohms = 354mA through the speaker coil.

For BTL I'm figuring it's 16V peak to peak therefore RMS = 5.657 RMS = 707mA through the speaker coil.

Are those calculations correct?
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Old 19th May 2012, 09:23 AM   #16
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Remember that voltage must be a potential between 2 points. Don't use peak-to-peak voltage for the single-ended output - the "single-end" can't be both positive and negative at the same time. So per your example, the voltage would swing 4V positive, then 4V negative. Use 4V to convert and calculate.
In bridged mode, one end of the speaker is sent 4V positive while the other end is sent 4V negative... at the same time. The voltage is doubled, so the current is doubled.
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Old 19th May 2012, 11:01 AM   #17
AndrewT is offline AndrewT  Scotland
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The rule for bridging and balanced amplifiers is simple.

A pair of amplifiers in bridged format will deliver twice the power into twice the load impedance.

eg.
two 100W into 4r0 amplifiers will deliver 200W into 8r0 when bridged.

That shows quite well that nothing is gained by bridging.
The total power from the two amplifiers is identical.

There is no magic 4times the power !!!!!

Having seen how easy it is, then it becomes obvious that to consider a single unbridged amplifier as the basis for all the investigations avoids all the confusion.
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Old 19th May 2012, 11:42 AM   #18
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Lies!! Keep the speaker impedance at 4 ohms and you will have four times the power, as each amp will see 2 ohms instead of 4 ohms.
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Old 19th May 2012, 12:19 PM   #19
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For the sake of clarity (since they're were two duplicate threads running...and unfortunately I reopened my latest line of questioning in the wrong one!), here's what I asked...

Quote:
For the purposes of calculating current through a speaker coil for single ended vs BTL

(assuming just DC resistance & ignoring impedance for simplicity)...

Coil = 8 ohm
peak to peak AC signal voltage swing = 8V (as measured with a scope on one leg)

To calculate the current through speaker coil, I need to convert the peak to peak swing to RMS, but since this is BTL, do I therefore double the swing ...therefore 16V peak to peak?

So for single ended driving amplifier, 8V peak to peak = 2.828 RMS...therefore divided by 8 Ohms = 354mA through the speaker coil.

For BTL I'm figuring it's 16V peak to peak therefore RMS = 5.657 RMS = 707mA through the speaker coil.

Are those calculations correct?

So in my above example, for a BTL amplifier for a signal measuring 8V peak to peak on one of the amplifier outputs (and ignoring impedance just for simplicity of discussion(!), for a coil with a resistance of 8 Ohms what would be the RMS AC current through the coil?


Since I'm scoping 8V peak to peak on one amplifier output, is the actual voltage scoped doubled? .....therefore 16V peak to peak, which is 8V peak, therefore 5.657V RMS, there 5.657V/8Ω = 0.707125A?

Last edited by HankMcSpank; 19th May 2012 at 12:29 PM.
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Old 19th May 2012, 12:30 PM   #20
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Did you see my post in the duplicate thread? There is no AC "peak to peak on one of the amplifier outputs" at the same instant in time.

edit: I'm taking that as a yes.
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Last edited by sofaspud; 19th May 2012 at 12:44 PM.
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