# Traynor PM-100



## MarkM (May 23, 2019)

I picked up a Traynor PM-100 cheap, I tried it to see if it works with a guitar into an 8 ohm speaker and it isn't very loud. No hum or hissing and sounds clear. Is my test equipment not enough to get the full potential of this amp, if not what should I look at?


----------



## Paul Running (Apr 12, 2020)

If you are directly connecting a guitar to the PM-100, chances are the guitar PU is being loaded-down...32KΩ input impedance for the PM-100. Here's the schematic...make sure that you can measure + and - 44VDC from the supplies.


----------



## dtsaudio (Apr 15, 2009)

The PM100 is a basic power amp. It is designed to be driven by a mixer or preamplifier.
The gain is too low for a guitar.


----------



## jb welder (Sep 14, 2010)

You can test it with signal from the pre-out or FX send of a guitar amp. Like Dan said above, it's a power amp designed for line level input signal.


----------



## MarkM (May 23, 2019)

Right on that's why I thought I would ask you, it's also a 4 ohm output into an 8 ohm speaker so I am sure that will contribute too.

Thanks guys!


----------



## Paul Running (Apr 12, 2020)

MarkM said:


> t's also a 4 ohm output into an 8 ohm speaker


Something seems off with the numbers on that schematic and the spec sheet for that model.
Theoretically with an ideal PS, that amp has a rail to rail headroom of about 80Vp-p. With an 8Ω load, you should expect 80V×10A =800W peak to peak power or about 280WRMS.
With a 4Ω load, you should expect 80V×20A =1K6KW peak to peak power or about 565WRMS.
With adequate heatsinking and forced air cooling, I believe that the output devices could handle those power capacities.
The power supply is far from ideal, with those power losses.


----------



## MarkM (May 23, 2019)

Paul I appreciate you sharing that with me, please accept my ignorance in such manners. I build roads, paved, installed water sewer and was a Civil Technologist. I am going to help my Electronics guru build his driveway, I will ask him to sifur that for me. The other option was to get @Latole to use goggle translate to help me out!

I really appreciate your technical help with my basic knowledge as well as all the other amp gurus that help me out.

Cheers guys


----------



## dtsaudio (Apr 15, 2009)

Paul, hate to tell you this but your math is incorrect.
You need the peak voltage not P-P to find the output power.
So you need one rail (44vdc), divide that by sq root 2 to get peak voltage.
RMS is then peak x .707.
So for this amp 44/1.414=31.1
31.1x.707=21.99.
RMS power is 21.99squared/4=120.9 watts.
Pretty close to rated power.


----------



## jb welder (Sep 14, 2010)

Yep, you have to convert the voltage (from p-p to rms) before you calculate the power.
The squaring is where things get messed up if you try to convert later.
Also, the supply rails usually sag somewhat at full power @ minimum rated load.


----------



## jb welder (Sep 14, 2010)

Still something funny with the math though. @dtsaudio , didn't you convert that 44peak to RMS twice there? Peak to peak should be same as rail to rail (88V) so 44V peak.
Using a theoretical (no sag) 44Vpeak, I get approx. 240W into 4, 120W into 8 ohm load.
(edit: listed spec is [email protected], [email protected])


----------



## Paul Running (Apr 12, 2020)

Sorry about the errors, many different views on power or energy. I was allowing 8Vp-p for losses in the output devices; I did set the power supply as ideal.


----------



## dtsaudio (Apr 15, 2009)

jb welder said:


> didn't you convert that 44peak to RMS twice there? Peak to peak should be same as rail to rail (88V) so 44V peak.


No, I took the 44VDC rail as an approximation to get Vpeak, and used that number to get Vrms.
Something I once read, but can no longer find the reference.


----------



## jb welder (Sep 14, 2010)

dtsaudio said:


> Something I once read, but can no longer find the reference.


Maybe they were trying to account for supply sag or something? Anyway, I do know that 'divide by 1.414' is exactly the same as 'times .707'  , so not sure why both operations would be performed.


----------



## dtsaudio (Apr 15, 2009)

jb welder said:


> I do know that 'divide by 1.414' is exactly the same as 'times .707'  , so not sure why both operations would be performed.


The two are the same thing. That's what happens when two formulas are written down in different folders on the computer.
44VDC is not the same as Vpeak.
I was taught many moons ago that an approximation for Vpeak of an amps output is the DC rail x .707.
You then need to convert to Vrms which is as you noted the same formula.
This isn't any law, it's just an approximation that seems to come reasonably close.
Another way I was taught, and it only works for 8 ohm load is the 60% approximation. 
Take one rail, multiply by 60%, and then the same power calculation.
For this amp you get about 87 watts into 8 ohms.


----------



## Paul Running (Apr 12, 2020)




----------

