Resistor wattage for stepped attenuator
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 23rd February 2011, 10:04 AM #1 Dagwood   diyAudio Member     Join Date: Feb 2010 Location: North of Auckland Resistor wattage for stepped attenuator Hi all I'm working on a simple passive pre-amp with just a 25K stepped attenuator and a 4 pole, 5 way rotary switch. Questions??? 2 mains sources will be Arcam CD player and an Ipod, is 25K ok feeding into Pete Millets DCPP amp? Also,I am having trouble matching the values required using Dale vishay RD60N 1/2 watt resistors, will there smaller brother at 1/8 watt be ok for this setup Part attached (RN55D5101FRE6 Vishay/Dale Metal Film Resistors - Through Hole) Cheers __________________ If you keep digging in the garden of failure you will eventually find success
 23rd February 2011, 10:46 AM #2 richie00boy   Did it Himself diyAudio Member     Join Date: Nov 2003 Location: Gloucestershire, England, UK 1/8 watt is many times more powerful than needed, so ok. __________________ www.readresearch.co.uk my website for UK diy audio people - designs, PCBs, modules and more.
 23rd February 2011, 12:21 PM #3 AndrewT   diyAudio Member   Join Date: Jul 2004 Location: Scottish Borders I hope you were asking in jest. P = V*I = V^2 / R = I^2 * R For an input attenuator you already know the maximum input voltage and the resistance of the attenuator. Choose the P = V^2 / R version and insert your numbers. example. For a CD player with a maximum output of 2.2Vac feeding a 50k attenuator the maximum power delivered to the attenuator is 2.2^2 / 50000 = 0.000097W ~ 100uW __________________ regards Andrew T. Last edited by AndrewT; 23rd February 2011 at 12:24 PM.
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Quote:
 Originally Posted by AndrewT I hope you were asking in jest. P = V*I = V^2 / R = I^2 * R For an input attenuator you already know the maximum input voltage and the resistance of the attenuator. Choose the P = V^2 / R version and insert your numbers. example. For a CD player with a maximum output of 2.2Vac feeding a 50k attenuator the maximum power delivered to the attenuator is 2.2^2 / 50000 = 0.000097W ~ 100uW
Would that not be the *minimum* (typical) power in W, ie at 50 Kohm resistance? What about at near 0K resistance (full clockwise, with the lowest value resistor, or nearest "10")?

In any case, 1/8 w should be fine over most, if not all, values; there will be an input impedance at your amplifier to load down the source. Use the value for your amp's input impedance in the above formula to determine if your lowest value resistor is robust enough.
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Last edited by Johnny2Bad; 23rd February 2011 at 06:26 PM.

 23rd February 2011, 06:54 PM #5 Dagwood   diyAudio Member     Join Date: Feb 2010 Location: North of Auckland Thanks guys, I'm not an EE so appreciate any help with formulas, will know for future reference now. So, lowest point on the step will be a 7ohm resistor, 2.2^2 / 7 = 0.6914285W of dissipation? Cheers __________________ If you keep digging in the garden of failure you will eventually find success
 23rd February 2011, 08:19 PM #6 Dagwood   diyAudio Member     Join Date: Feb 2010 Location: North of Auckland Thanks Johnny2Bad I wanted to complete the passive pre-amp before I completed the power amp for testing. But I guess it's chicken and egg really, I could always just rig the power amp up on a standard dual pot to check the levels? I'm not really sure how to measure the amp's input impedance sorry?? can it be worked out from the schematic?? DCPP Amp Cheers for your help __________________ If you keep digging in the garden of failure you will eventually find success
 23rd February 2011, 08:53 PM #7 richie00boy   Did it Himself diyAudio Member     Join Date: Nov 2003 Location: Gloucestershire, England, UK No the lowest point still has all the preceeding resistors before it, so the calaculation Andrew did stands and is the dissipation of the entire bunch of resistors, so as you can see 1/8 watt really is many times more than is needed as 5 ways will give over half a watt of capacity. __________________ www.readresearch.co.uk my website for UK diy audio people - designs, PCBs, modules and more.
 23rd February 2011, 09:51 PM #8 bear expert in tautology diyAudio Member     Join Date: Apr 2002 Location: New York State USA Fuel to the fire, and meat to the dogs... I'll disagree. The 1/2 watt resistors will simply sound better in most instances. Why? All sorts of reasons, and also the construction of the resistors counts as well. There was a discussion of resistors in the John Curl Blowtorch part II thread - might be worthy of a review... Since people generally report a difference in sound between vintage Holco metal films and other metal films, and again a difference with things like bulk film (Vishay for example), I'd not consider "resistors to be resistors". Feel free to chomp and thrash... _-_-bear __________________ _-_-bear http://www.bearlabs.com -- Btw, I don't actually know anything, FYI -- every once in a while I say something that makes sense... ]
 24th February 2011, 04:43 AM #9 Dagwood   diyAudio Member     Join Date: Feb 2010 Location: North of Auckland Thanks Bear I had a scan through JohnCurlBlowtorchpartIIthread, man my head is sore! Ok I get Simons results (sort of) and have read through a few of the data sheets I still think I will use the Vishey Dale RN55D's as I have to consider product cost, availability at required values, cost of shipping to NZ, size and overall performance (datasheet attached) http://www.vishay.com/docs/31027/cmfmil.pdf Cheers Ben __________________ If you keep digging in the garden of failure you will eventually find success
 24th February 2011, 05:12 AM #10 bear expert in tautology diyAudio Member     Join Date: Apr 2002 Location: New York State USA Ya ok, everyone makes choices in this thing... as I said I prefer the 0.5W part... I think it was also said that paralleled resistors had benefits as well... But it may be that for most folks this is minutia that matters not. I've had good results with lowly asian made (good quality) 0.5w MFs... I tend to attribute that to the fact that they are 0.5w devices, but I also use a super high quality switch and tend to build "discrete L" attenuators, not the usual "pot" type units. The discrete L puts only two resistors at a time in the signal path, not a ladder. That may make a difference too. Btw some people say it was the copper end caps and the way they were made that made the original vintage Holco resistors sound the way they did... dunno. _-_-bear __________________ _-_-bear http://www.bearlabs.com -- Btw, I don't actually know anything, FYI -- every once in a while I say something that makes sense... ]

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