Dummy load with output to power amp
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 6th February 2010, 03:07 AM #1 acio   diyAudio Member   Join Date: Jun 2007 Dummy load with output to power amp I apologize if i missed it somewhere but I have been searching high and low through Google, endless forums, and so forth trying to figure this one out. Either it is so darn simple that it is right in front of my face and I missed it or the information is next to impossible to find. I am trying to custom design my tube amp to go into a dummy load and from there into a SS power amp or other line level device. All of the schematics I find that have dummy loads go nowhere but end there. They seem to have a switch to go to line out for sound or to dummy load for testing or what have you but no output. What does it take to go from a resistive dummy load to a line level in order to plug straight into an amp, fx unit, eq, etc.? Thanks in advance
 6th February 2010, 03:57 AM #2 raudio1969   diyAudio Member   Join Date: Sep 2007 Location: N38 You could try a resistor voltage divider after the dummy load. - Richard
 6th February 2010, 06:38 AM #3 acio   diyAudio Member   Join Date: Jun 2007 Thanks Rich. I'll have to look into that. I was hoping to find a schematic or layout but I suppose I will have do some research , Googling, and Lord forbid........possibly some math on that divider idea lol!
Mooly
diyAudio Moderator

Join Date: Sep 2007
Take the output of the amps in watts RMS... say 10 and decide what dummy load you want, say 16 ohms.
So you can now find the voltage the amp puts out.
That equates to W=Vsqrd/R so rearranging gives V = sqr root W*R which is 12 volts approx (RMS)

So if you need 1 volt RMS to feed into another device, you need a divider that gives a drops 9 volts across the "top" half and 1 volt across the "bottom" half, so a 90 ohm and 10 ohm or a 900 ohm and 100 ohm etc would be fine.
But it's all so non critical I would use something like a 10K and a 1K in that example.
Make them too low and they dissipate power (so you need bigger than 0.5watt etc) make them too high and the "output impedance" of the divider is high and may be affected by capacitive loading on the device you are connecting too.
You could also fit a 2k2 pot in place of the 1K and find the ideal value by that way, then just measure and replace with a fixed resistor of the nearest value.
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 13th March 2010, 08:32 AM #5 acio   diyAudio Member   Join Date: Jun 2007 Great info Mooly! Sorry I just now caught your post. I have been away from the forum for quite a while. Actually ever since I got the thing working lol! I am currently using this method : http://www.singlecoil.com/docs/out.pdf But I have pots where the resistors are so I can find the values I prefer. I still cannot figure out what exactly is doing what. For instance, what is the difference between changing the value on the resistance going out from one jack to the other, vs changing the resistance from the other jack to ground? Anyone know this? It seems to have more sustain and guts when I turn the resistance all the way down between the two jacks and from there work with the resistance to ground vs doing it the other way around.
 13th March 2010, 07:37 PM #6 Mooly   diyAudio Moderator     Join Date: Sep 2007 The two resistors just form a divider. If you have presets and turn the "560" ohm to zero, then turn the "100 ohm" down you will probably damage/burn out the pots. They will probably only be rated for perhaps 50 to 100 milliwatts max. Keeping the values low as in your link minimises the output impedance... which if feeding the input to any "normal" amp isn't an issue within reason.

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