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Amplifier power supply capacitor removal advice needed!

Chaggy78

Dec 30, 2012
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Ok thanks i'll try the resistor and get back to you. Does it matter which end of the resistor goes where? Thanks : )
 

tedstruk

Jan 7, 2012
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By the way. The power caps are the easiest part to change on the board...
 

Chaggy78

Dec 30, 2012
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Hi Kris here is the photo of the resistors I was telling you about:
 

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KrisBlueNZ

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Those are two other resistors that Chaggy78 wants to replace. They were mounted raised off the board, and were visibly flaking due to long term overheating. They are marked 620 ohms 5%.

My guess is that they're wirewound, 2W or 3W rated. I'm suggesting something like the Vishay BC AC04 series, 3W wirewound, see http://www.digikey.com/product-detail/en/AC03000006200JAC00/PPC3W620CT-ND/596668

--- ANYONE ELSE HAVE A DIFFERENT OPINION? ---
 

KrisBlueNZ

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Those are 620k (kilohms), i.e. 620,000 ohms, not 620 ohms. Apart from that, they should be suitable.
 

Chaggy78

Dec 30, 2012
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Ok thanks Kris i found some the same but 620ohm versions i might buy them.
 

Chaggy78

Dec 30, 2012
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Now i can't order the ones i just found from farnell because the minimum order must be £20!
Having trouble finding any in UK with low postage! dohhh!
 

KrisBlueNZ

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Yes, I would use 3W for safety. The originals might only be 2W but they have obviously been pretty hot. Yes, you need a tolerance of 5% or better.

You could make them up from several smaller resistors in series or parallel.

For example, you could replace each one with three 1.8k 1% 1W metal film resistors in parallel. This works out to 600 ohms +/- 1% which falls within the range of 620 ohms +/- 5%.

Likewise you could use four 2.49k 1% 1W metal film resistors in parallel.

More resistors will also give you a larger surface area for heat dissipation so they won't run as hot.
 

Chaggy78

Dec 30, 2012
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Ok thanks Kris, much appreciated info!
Sorry for the newbie questions! lol but what do you mean by putting the resistors in parallel?
Please can you explain so i don't blow anything up! haha cheers
 

Chaggy78

Dec 30, 2012
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I have managed to find 2.7K 1% 1W resistors (metal film) could i use four of these in parallel?
 

KrisBlueNZ

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In parallel means they're all connected "across each other". For example, put them all side-by-side and twist all the wires on one side together, then twist all the wires on the other side together.

When you put several resistors of the same value in parallel, the resulting resistance is the resistance of one resistor divided by the number of resistors.

So four 2k7 resistors in parallel would be 2700 / 4 ohms, which is 675 ohms, which is outside the range for 620 ohms +/- 5%. So no, you can't.

With that value you could make up series-parallel arrangements to get closer to the value you need. To do this, you need the extended version of the parallel resistor formula, which is:

1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ...

Start with four 2k7s in parallel, that's 675 ohms. To reduce that to 620 ohms you use that formula:

1/620 = 1/675 + 1/needed_resistor.

This rearranges to needed_resistor = 1/ (1/620 - 1/675)
which is 7609 ohms. Three 2k7s in series will get you 8k1 (resistors in series just add). Substituting 8100 into that formula gives:

1/Rtotal = 1/675 + 1/8100
Rtotal = 623 ohms - good.

This arrangement needs seven 2700 ohm resistors. Four of them are directly in parallel with each other, and the other three are connected in series, and the series string is connected in parallel with the other four resistors as well.

There is a shorthand for writing resistors in parallel, using double vertical line symbols. The formula for that resistor combination is:

2700 || 2700 || 2700 || 2700 || (2700 + 2700 + 2700)

Now you can double-check it using the full formula for parallel resistors:

1/Rtotal = 1/2700 + 1/2700 + 1/2700 + 1/2700 + 1/(2700+2700+2700)
Rtotal = 623 ohms.

I hope this makes some sense. It is an option, although it will look pretty messy.
 
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