Resistor wattage for stepped attenuator

[Dagwood]

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

 

[richie00boy]

1/8 watt is many times more powerful than needed, so ok.

 

[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

 

[Johnny2Bad]

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.

 

[Dagwood]

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

 

[Dagwood]

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

 

[richie00boy]

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.

 

[bear]

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

 

[Dagwood]

Thanks Bear

I had a scan through John Curl Blowtorch part II thread, 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

 

[bear]

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

 

[Dagwood]

Bear, they don,t make em like they used to!

Yeah sorry, Just lost my job and got 3 kids to feed so wife would go off like a firework if I spent the kind $$$ on PRP or sourced the .5 watt RN60D's from loads of different suppliers( meaning loads of postage ).

I guess you spend as much as you can afford at the time? It will still sound better than any of the carbon pots I have used in the past!

Cheers

 

[AndrewT]

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.

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

NO !!!! NO !!!! NO !!!!

The attenuator impedance/resistance seen by the Source is fixed at 10k or 50k or 100k.
The impedance/resistance seen by the attenuator as presented by the receiver is fixed. It could be 22k//100pF, 47k//50pF, 100k//220pF

As the wiper (or switch) moves, the impedance presented to the receiver changes. It is different for every different volume setting.

When the attenuator is chosen to suit the receiver it is usual to select Attenuator resistance ~ < Rin/10.
If Rin=100k//220pF then a 10k attenuator would be chosen.

Now let's look at what the Source will see.
At minimum volume the wiper is at the bottom and the full 10k alone is presented to the source. You can calculate the total power dissipated in that attenuator total resistance.
Turn the attenuator to max volume.
Now you have the 10k attenuator parallel to the 100k//220pF seen by the source.
Since these are in parallel they both see the same voltage. You can calculate the dissipation in the attenuator resistance and the dissipation in the Rin resistance. (have you ever felt the need to calculate the Rin power dissipation?)

The arithmetic is slightly more complicated for a volume setting between Max and Min.

Let's take a -6dB volume setting.
The attenuator is split 5k to upper half and 5k to lower half.

The load on the wiper is still 100k//220pF.
The Source first passes through the upper 5k. Then the current has a choice to pass through the lower 5k or to divert to the 100k//220pF. These routes are effectively in parallel. 100k//5k = 4k762.
The total resistance seen by the source is 5k + 4k762 = 9k762.
Calculate the power dissipated in this new low impedance. Now proportion that power loss over the 5k upper and the 5k lower and the 100k Rin.

This is very basic arithmetic. There is absolutely no reason not to pick this up as you go along with your hobby. Alternatively Members can actively research the simple arithmetic that is NEEDED to do any electronics component selection.

Johnny,
you are a bad example of what happens on the internet. You have not bothered to research the subject and you come along and post complete inaccuracies. Then other Members who are starting out see your misleading lies and don't know who to believe.
Do your homework or ASK !!!!

Sorry to all those readers to have to witness my rant.