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Old 11th December 2005, 06:18 AM   #11
Nordic is offline Nordic  South Africa
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Posted this exact question before... got no answer.... good work
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Old 11th December 2005, 02:16 PM   #12
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so let me get this straight you can do it, as long as you have a 47uF in the fb? would that be parallel to the fb resistor or series with it? im a bit confused.

the output .1 ohm resistors i kinda figured they would be used in the first place for current sharing but i do not understand the 47uf in the feedback loop.
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Old 12th December 2005, 05:11 AM   #13
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well could you at least use a electrolytic instead of the tantalum one? and put a nice non polarized cap in parralel with it.
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Old 14th December 2005, 12:05 AM   #14
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Default Re: paralleling chips

Quote:
Originally posted by tiltedhalo
Holy chip, it works! I tried it with two LM3875,S I had laying around. Split the input with 2 1K resistors and 1 22K to ground used a common feed back loop with a 22.1K .1%, 47uf tantalum cap, and a 1k .1% to ground fed the outputs to two .15 ohm 1% soldered directly two the pins and took the feedback of the combination of outputs. Might have to take a closer look at this! I got the least amount of DC offset of any chip amp Ive made!!
How hot does the chips get? I have a hard time believing it is that simple . I mean why would manufacturer who uses chips in parallel use so many extra parts? I hope you are right and your amp can be duplicated by other.
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Old 14th December 2005, 02:19 PM   #15
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Default parallelling chips

In all honesty your only saveing 2 resistors, everything else remains the same. You still need current shareing resistors ect., the only differance is your useing a global instead of local feedback network, this has been used for years with other op amps.
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Old 14th December 2005, 07:57 PM   #16
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Ok the tantalum cap can correct the internal dc offset of each chip? Have you measured each chip's offset separatly? It seem to me that if you dont get each chip's offset down before paralleling them,you will still get quite a bit of curent share even if you have resistors at the output.
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