Resistors required on parallel amp

[Arx]

I'm building an amp with a paralleled 3875, and I was wondering if 2 of the ,1 ohm resisors are required.

It seems to me like one of the chips could be connected directly for marginally better efficiency and damping, and the other run through a 5W 0r1 resistor. Any thoughts?

[jcx]

due to high negative feedback each chip will have miliOhm output Z at low audio frequencies - leave out one of the 100mOhm R and the direct connected chip amp will supply 99% of the output current

[star882]

They're required for bipolar and certain IGBT output stages, but not MOSFETs. I think the chip in question uses bipolars so yes, you'll need them. And you'll need to synchronize the drive signals.

[sasmit]

when you're paralelling the idea is that each amp shares exactly..nearly half of the load current. In practice due to unavoidable differences between output devices, one of the devices is going to provide all the current. To avoid this and force load sharing these resistors are required, without these you'll end up using only one amp.

[lgreen]

yeah, what they said!

And make sure to use at least 1% resistors, they need to match closely or one device will still hog all the power. Your input and feedback resistors should be matched to about 0.1% or better or the chips will have different gain, different outputs and all he** will break lose.

[abraxalito]

Originally Posted by lgreen

And make sure to use at least 1% resistors, they need to match closely or one device will still hog all the power.

1% 0.1R resistors will be expensive, if you can find them. Fortunately they're not necessary, 5% is fine.

Your input and feedback resistors should be matched to about 0.1% or better or the chips will have different gain, different outputs and all he** will break lose.

Even with 0.1% gain setting resistors, any imbalance in current sharing is going to be mainly due to these resistors and not the output sharing resistors. (With 0.1R sharing resistors, I recommend 0.05% gain matching.) 5% is plenty good enough for the output resistors - I've simmed it

[janneman]

Originally Posted by lgreen

yeah, what they said!

And make sure to use at least 1% resistors, they need to match closely or one device will still hog all the power. Your input and feedback resistors should be matched to about 0.1% or better or the chips will have different gain, different outputs and all he** will break lose.

Of course not. Let's put in some numbers: If the gain of one amp is 10% higher than the other, and the sharing resistors are exactly the same, one amp will provide 10% more current than the other. No ideal sharing, but more than good enough (and of course 10% gain difference is huge and will never happen with 1% feedback resistors).

Now let's take the sharing resistors. Assume perfect gain matching but 10% share resistors accuracy. That means, assuming 4ohms nominal load, one amp sees 4.1 ohms and the other, say, 4.11 ohms load. So the sharing inaccuracy is tiny and you would have a hard time to even measure it.

So it's clear that these are non-issues.

jd

[abraxalito]

Originally Posted by janneman

Let's put in some numbers: If the gain of one amp is 10% higher than the other, and the sharing resistors are exactly the same, one amp will provide 10% more current than the other.

Looks like you put in the wrong numbers. The gain matching is substantially more sensitive than this - to see though, we need to convert the gain mismatch into a voltage mismatch at the outputs. 10% gain mismatch (not the worst case with 10% resistors, which is 20% mismatch) becomes 10% voltage mismatch, take the peak output voltage to be 28V (a 50W/8R amp). 10% of 28V is 2.8V, this appears across two 0.1R resistors in series giving a current of 14A. Its clear that both amps are well into mutual destruction - the paralleled amp looks just like a load resistor of 2R to the other amp. Even with 1% resistors the current can be 2.8A (one gain can be +1%, the other -1%).

<edit> seems like a quick rule of thumb is multiply the sharing resistor's value by 100 and divide by the tolerance of the gain resistors to see the worst case effect of mismatch. e.g. with 1% we get an equivalent 10R load seen by the other amp.

[janneman]

Originally Posted by abraxalito

Looks like you put in the wrong numbers. The gain matching is substantially more sensitive than this - to see though, we need to convert the gain mismatch into a voltage mismatch at the outputs. 10% gain mismatch (not the worst case with 10% resistors, which is 20% mismatch) becomes 10% voltage mismatch, take the peak output voltage to be 28V (a 50W/8R amp). 10% of 28V is 2.8V, this appears across two 0.1R resistors in series giving a current of 14A. Its clear that both amps are well into mutual destruction - the paralleled amp looks just like a load resistor of 2R to the other amp. Even with 1% resistors the current can be 2.8A (one gain can be +1%, the other -1%).

Yes, you are right - didn't realise that the outputs are connected through the 0.1R.
In that light, maybe 0.1R is a bit too low for safe sharing with realistic gain matching?

jd

[abraxalito]

Originally Posted by janneman

In that light, maybe 0.1R is a bit too low for safe sharing with realistic gain matching?

That's my view yeah, I suggest 0.2R as a minimum when using 0.1% types. But that's the ultra-conservative engineer in me, in practice few people will encounter the worst case scenario.

<edit> the other thing people often forget (even the writers of NS app notes on paralleled amps) is that the gain often has frequency dependence at LF introduced by electrolytics, these have far worse tolerance than even lousy resistors