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Old 27th May 2005, 01:23 AM   #1
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Default paralleling amplifiers.

I was wondering if anyone here knows How to calculate what resistor you need when paralleling amplifiers? it seems like one of those things that could go drastically wrong if the wrong choice is made.

I had an idea to make a low watt amp of a bunch of TPA6120s

they are current amplifiers so would that make any difference when paralleling them? (or make it impossible)
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Old 27th May 2005, 03:48 AM   #2
Leolabs is offline Leolabs  Malaysia
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I dont know about any calculation regarding this,but i have seen many using the value of 0.1ohms
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Old 27th May 2005, 05:43 AM   #3
methar is offline methar  Thailand
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BTW, if individual amp has higher DC offset the higher value may be used. I saw some thread long time ago, that the later version of Linn use 0.5 Ohm ballard res.
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Old 27th May 2005, 06:28 AM   #4
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Minimal load impedance by TPA 6120 is 8 Ohm. So if you use two pieces of this circuit ( four amps ), which will have connected outputs together through this resistor ( each ), you can get output power into 8 Ohm load cca 6 W. Distortion will be in this case very good ( bellow .001 % ), but DF will be low, only four ( 8/2 ), so sound will be similar to single ended tube amp. If you like similar sound, go on.
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Old 27th May 2005, 05:08 PM   #5
macboy is offline macboy  Canada
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Quote:
Originally posted by Upupa Epops
Minimal load impedance by TPA 6120 is 8 Ohm. So if you use two pieces of this circuit ( four amps ), which will have connected outputs together through this resistor ( each ), you can get output power into 8 Ohm load cca 6 W. Distortion will be in this case very good ( bellow .001 % ), but DF will be low, only four ( 8/2 ), so sound will be similar to single ended tube amp. If you like similar sound, go on.
The load-sharing resistor is not limited by the minimum impedance of the amp. So you could use much lower load sharing resistors than 8 ohms, like the typical 0.1 to 0.5 ohms. Just make sure that the gain of all of the paralleled amps is very closely matched so that current does not flow between the amps via these resistors.

And if you want to keep the Damping Factor high (low output impedance), then you can put the paralleled amplifiers inside the feedback loop of an opamp. This will effectively remove the load sharing resistors from the load - the opamp will adjust the output voltage to remove the effect of the voltage drop across these resistors.
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Old 27th May 2005, 05:25 PM   #6
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This circuit is very fast, so serial resistor protect output against capacitive load and at datasheet is 8 Ohm recomended as minimal value. Sure, if you all connect into loop, you get lower DF, but circuit will be relatively complicated and PCB will must be designed closely.
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Old 28th May 2005, 03:31 AM   #7
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Thanks for all the replies!

I was actually thinking of using more chips per channel (4 or 5) but 2 is a good place to start.

So if i put the load sharing resistors inside the feedback loop, then all of the chips would be getting their feedback from the same node (the final output) ? If they are sharing that node, could they also share the actual feedback resistor, so it is exactly the same value for all the amps?
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Old 30th May 2005, 01:31 PM   #8
macboy is offline macboy  Canada
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So if i put the load sharing resistors inside the feedback loop, then all of the chips would be getting their feedback from the same node (the final output) ? If they are sharing that node, could they also share the actual feedback resistor, so it is exactly the same value for all the amps?
Ok, I don't quite follow your line of thinking, but basically what I suggested above entails enclosing the parallel amp inside the feedback loop of an opamp. (see attachment). Each individual amp in the parallel amp still has its own output (load-sharing) resistor, and these absolutely must be outside of each chip's feedback loop.

In the attached schematic I have drawn a simplified four-way parallel amp inside the feedback loop of an opamp so that you can get the idea.

Discalimers: This is intended to be representative only and is not a tested design. There are no component values; you will have to choose them appropriately. Most of the National chips are stable at gains of 10 or higher, so set the gains appropriately, and match them very closely. Also note that the overal gain for the input opamp loop must be greater than the gain for the chipamps, even if the opamp is unity-gain stable. As drawn, the circuit probably has some nasty DC offset problems (especially with all the + inputs of the chipamps all tied directly together), and I would personally take extra steps to reduce DC offset.
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File Type: gif parallel-amp.gif (4.6 KB, 711 views)
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Old 30th May 2005, 02:20 PM   #9
methar is offline methar  Thailand
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Quote:
Originally posted by macboy

Ok, I don't quite follow your line of thinking, but basically what I suggested above entails enclosing the parallel amp inside the feedback loop of an opamp. (see attachment). Each individual amp in the parallel amp still has its own output (load-sharing) resistor, and these absolutely must be outside of each chip's feedback loop.

In the attached schematic I have drawn a simplified four-way parallel amp inside the feedback loop of an opamp so that you can get the idea.
[/i]

I agreed with macboy. In addition, I would add Cap in the feedback to ensure unity gain of each chip. Also, individual zobel of each chip will be helpful.

Good luck,
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Old 30th May 2005, 02:33 PM   #10
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Quote:
Originally posted by methar
In addition, I would add Cap in the feedback to ensure unity gain of each chip.
You can't use a cap on the feedback loop of a current-feedback op-amp.
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