power/heat from chip amp

[xplod1236]

How much power can I expect from a bridge/parallel LM3886 chip amp? The supplies would be +/- 35, and the load would be 1.35 ohms (4 x 5.4 ohm coils in parallel). I should get 64v across the load, so mathematically, the max power output comes out to 3,034 watts. Would I get 3 kw to the speakers plus heat, or will the power draw be 3 kw, and that will be split up for output/heat?

[janneman]

Lets do some educated guessing. No signal, the LM takes, what, 50mA at 70V total, that's 3.5watts to begin with.

Now, suppose your average power output is 10 watts RMS (I'm being generous here) in 1.35 ohms, so I=sqr(10/1.35) which is about 2.7A RMS. Over the load you will find 2.7 * 1.35 = 3.6V RMS. The remainder (35V supply is about 25V rms for comparisons), over the chip's output transistors (thats where the heat will be) is about 21.5V rms. This will give a dissipation of P = 21.5 * 2.7 = 58 Watts, bringing the total just over 60 watts. Which is too much for continues 10W output.

Did you spot the problem? Your supply voltage is way too high for a 1.35 ohms load. You will have the chip limiting in current and/or dissipation long before you reach the max supply. In fact, if you lower the supply to 20V or so you can get more power out of this chip, at least in 1.35 ohms. But still, 4 ohms would be much better as far as using the max chips capabilities is concerned.

Jan Didden

[xplod1236]

Originally posted by janneman
You will have the chip limiting in current and/or dissipation long before you reach the max supply.

That's why I'm doing bridge AND parallel. The current and dissipation will be shared between the chips.

Edit: I will be paralleling at least 5 chips on each side of the bridge, so at least 10 chips for the whole amp.

[tlmadsen]

janneman is right.

However I would like to look at it from two different angels:

A. You have a load of 1.35 ohm for a bridge/parallel amplifier. This means that each "arm" sees a load of 0.675 ohm !!!. With 5 LM3886 in parallel in each arm they each see a load of 5 x 0.675 = 3.375 ohm !!!, Not imposible to drive for a LM3886, but with a supply of +/- 35 V it is very much on the limit.

B. Let say for a moment that you have a "perfect" PSU and a "perfect" cooling of all your chips. How much power can you get in a 1.35 ohm load ??. A specific power level in a specific load requires a specific level of current and voltage. You CAN'T "supstitute" one for the other. For a 1.35 ohm load you need:
A Vrms Vpp
3000W: 47.1 63.6 180.0
1000 W: 27.2 36.7 103,9
500W: 19.2 26.0 73.5
200W: 12.2 16.4 46.5

With +/-35 V, I don't think you will get much over 100 Vpp, so if everthing else is "perfect" you MIGHT get around 1000 W in your 1.35 ohm. 3000 W is simply not possible no matter how "perfect" your PSU and cooling is.

Have fun

Thomas

[xplod1236]

Originally posted by tlmadsen
With 5 LM3886 in parallel in each arm they each see a load of 5 x 0.675 = 3.375 ohm !!!, Not imposible to drive for a LM3886, but with a supply of +/- 35 V it is very much on the limit.

I said with AT LEAST 5 chips in parallel. With only 5 per side, each one will be handling a 10 amp current, which is on the limit.

How much power can you get in a 1.35 ohm load ??. A specific power level in a specific load requires a specific level of current and voltage.

Yes, I know that. Voltage = current x resistance, so 64v p-p = 1.35c, solve for current and you get 47.4 amps. 47.4 times the voltage (64) yields 3034 watts max into the 1.35 ohm load.

For a 1.35 ohm load you need:
A Vrms Vpp
3000W: 47.1 63.6 180.0
1000 W: 27.2 36.7 103,9
500W: 19.2 26.0 73.5
200W: 12.2 16.4 46.5

I don't understand your table. The column labeled A seems to be the Vrms, and one labled Vrms seems to be voltage p-p, and the last column--I have no idea what those numbers are.

With +/-35 V, I don't think you will get much over 100 Vpp

How is that possible? With supplies of only 35v, the max voltage you can get is 70v p-p. The chips are not rail to rail, so I used 64v p-p for my calculations. How did you get 100v p-p?

[tlmadsen]

The tabel is

Power Current(A) Vrms Vpp

So for 1000 W you ned

27.2 A

36.7 Vrms

103.9 Vpp

Remember it is a bridge design.

Regards

Thoms

[moamps]

Originally posted by xplod1236
How much power can I expect from a bridge/parallel LM3886 chip amp? The supplies would be +/- 35, and the load would be 1.35 ohms (4 x 5.4 ohm coils in parallel). I should get 64v across the load, so mathematically, the max power output comes out to 3,034 watts. Would I get 3 kw to the speakers plus heat, or will the power draw be 3 kw, and that will be split up for output/heat?

Hi,
the best way IMO is using standard BPA200 application (four IC's for one amp) for each speaker. (why paralleling them?)
You can get ca 250W from each amp, and overall ca 1kW.
Paralleling amps technique isn't ideal, and allways you have small disballance what can be huge problem.

Regards
Milan

[carlosfm]

Originally posted by moamps
Paralleling amps technique isn't ideal, and allways you have small disballance what can be huge problem.

Regards
Milan

Not to talk about layout, ground arrangement... too many chips will probably give you hummmmmmmmmm...

[xplod1236]

Originally posted by tlmadsen

A Vrms Vpp
3000W: 47.1 63.6 180.0
1000 W: 27.2 36.7 103,9
500W: 19.2 26.0 73.5
200W: 12.2 16.4 46.5

here, you are talking about rms wattage. I was talking about peak wattage.

[theChris]

"4 x 5.4 ohm coils in parallel"
-5.4 ohm speakers?

my guess is for 6-8ohm nominal impedance. that brings the load impedance up to 1.5 to 2, unless the woofers really are 5.4 ohm nominally.

i'm not sure a 10 IC (minimum) GC would beat out a simple class B amp. i assume this is for a set of woofers?

in anycase, i'd go with seperate channels with less paralleling if possible. this means running extra wire though. also if one chip dies, well, it won't be hard to fix -- the woofer that doesn't work... that's the channel to fix!