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How does BTL double the voltage swing?
How does BTL double the voltage swing?
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Old 19th May 2012, 01:44 PM   #21
Pano is offline Pano  United States
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How does BTL double the voltage swing?
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Old 19th May 2012, 01:57 PM   #22
sofaspud is offline sofaspud  United States
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Thank you, kind sir.

Hank, I didn't check your math. It looks right.
One amp's output simply alternates above and below a reference. In bridge mode, the other amp's output does the same, but with inverted polarity. So there is twice the voltage at the load, and the +/- reference becomes somewhat "virtual", for lack of a better term. It's no longer ground or the midpoint of either amp individually.
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Old 19th May 2012, 02:09 PM   #23
HankMcSpank is offline HankMcSpank  United Kingdom
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Originally Posted by sofaspud View Post
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.
I saw it, but I'm not understanding why not, as one leg traverses up, the other leg traverses down....I'll do some more scoping...in the words of Arnie, I'll be back....
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Old 19th May 2012, 02:25 PM   #24
HankMcSpank is offline HankMcSpank  United Kingdom
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Ok, just revisited this...

yellow trace is the amp's +ve leg

red trace is the amp's -ve leg

purple trace is the 'difference' between the two (USB scope setting A-B), so it looks to me now as if the voltage peak to peak is doubled...

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(yes, I know there's ringng/noise at the peaks, but it was just a quick lash up!)

So for the above, 7.22V(-ish) is peak, therefore 5.105V RMS ...coil resistance is 8 ohms (again for simplicity, not taking into account its reactance!) - therefore 5.1505/8 = 0.644A through the speaker coil??

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Old 19th May 2012, 03:08 PM   #25
AndrewT is offline AndrewT  Scotland
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Originally Posted by Tekko View Post
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.
completely untrue.

A 4ohm capable amplifier will not be capable of driving a 2ohm speaker.
In a similar vein, when you bridge the two 4ohm capable amplifiers and connect them to a 4r0 load you will not get four times the output power.

The rule is as I stated:
Quote:
A pair of amplifiers in bridged format will deliver twice the power into twice the load impedance
No matter how many times you see "4times the power" it is still wrong.
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Old 19th May 2012, 03:09 PM   #26
sregor is offline sregor  United States
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Calculations look correct. Voltage across load doubles. Current doubles. Max power goes up by 4. Note it is voltage across load that doubles, not voltage relative ground or center voltage. My 2cents.
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Old 19th May 2012, 03:10 PM   #27
HankMcSpank is offline HankMcSpank  United Kingdom
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Originally Posted by HankMcSpank View Post
Ok, just revisited this...

So for the above, 7.22V(-ish) is peak, therefore 5.105V RMS ...coil resistance is 8 ohms (again for simplicity, not taking into account its reactance!) - therefore 5.1505/8 = 0.644A through the speaker coil??
Just confirmed the above caclulation with a real world reading by putting my DVM in series with the power supply & reading the DC current drawn by the overall circuit ...it all tallies (give or take a couple of mA). So for the purposes of calculating current through the coil, whatever peak to peak voltage is scoped on one leg of the BTL amp is Vpeak.

Last edited by HankMcSpank; 19th May 2012 at 03:13 PM.
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Old 19th May 2012, 03:12 PM   #28
AndrewT is offline AndrewT  Scotland
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Hank,
consider one amp alone.
Find out how it behaves. Find out what the recommended load impedance is.

When this is all complete then extend all the data you have for the single amplifier and apply the "rule".

There is no complicated mental dexterity required.
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Old 19th May 2012, 03:15 PM   #29
HankMcSpank is offline HankMcSpank  United Kingdom
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Quote:
Originally Posted by AndrewT View Post
Hank,
consider one amp alone.
Find out how it behaves. Find out what the recommended load impedance is.

When this is all complete then extend all the data you have for the single amplifier and apply the "rule".

There is no complicated mental dexterity required.
It may well be that there's no complicated mental dexterity needed ...but the differing/conflicting info in some of the replies suggests it's not done & dusted in everyone's else's heads either!

It actually complicates it (to my eyes at least) as considering this a BTL as one amp first ....because it's not one amp, but two amps operating 180 degrees out of phase...which is why I wanted to square this away in BTL terms (vs SE)...which I've just done by confirming the 'on paper' calculations with actual real world outcome - I'm happy with this now :-)

Last edited by HankMcSpank; 19th May 2012 at 03:17 PM.
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Old 19th May 2012, 03:26 PM   #30
jcx is offline jcx  United States
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each amp does "see" 1/2 of the attached load, with the opposite end of the load going the other way the midpoint of the load is a virtual gnd

if each amp in the bridge can drive the 2x current into the 1/2 load Z it "feels" on its output then you do get 4x power into the same load Z

having current reserve to double current for load dips to 1/2 the nominal load is the bare minimum reserve for a decent audio power amp given the demands of some loudspeakers (worse with some crossovers)

so if you want the same current drive reserve, say reasoning form a audio power chip amp's datasheet bullet item "8 Ohm" power rating then you probably want to parallel 2x of each chip in the bridge for a total of 4 amp chips to give 4x the power into the same nominal impedance load
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