Lowering B+ is this method safe ??

[dansamp]

So I have and old Masco P.A. amp that has been kicking around for a while
I turned it into a PTP Express inspired amp but it was not laid out well
and it buzzed and hummed
Yesterday I turned it into a PTP inspired JTM45 and it turned out surprisingly well
but I have about 490-500 volts on the plates of the power tubes
and tried several ways of lowering the B+

I didn't want to wait for any parts and I know the best way to lower it is with the VVR
but today I wired one 5W 100 ohm resistor on each of the leads from the PT going to my rectifier and have succeeded in dropping about 45-50 volts so now with EL 34 tubes biased at around 41ma I have 465 v on the plates

any idea if this will hold up
the PT is putting out more B+ than I was expecting


Dan.....

[tkdrew]

Is the 45-50 volts you dropped DC after the recto? Need to measure the voltage drop across the resistor themselves. As long as you don't exceed the 5 watt rating should be "safe", although I would think it would make for a lot of sag from the power supply as full throttle.

[CB Amp]

I agree with tkdrew...
It would be better to shift the high voltage points : the output trans after the choke (better filtering, but different from a JTM of course), etc... And you will certainly have to change the values of the resistors, perhaps add a RC filter...
The most important is the voltage value on the tubes, not the value of the components.

[dansamp]

thanks for the input
the resistors I have installed on the A/C side going to the rectifier
they are on each of the red wires
I tried putting a 1k resistor from the DC side of the rectifier going to the B+1 put it would get too hot and started cooking

I also thought of putting a cap right on pin 8 then running a resistor from there but not sure that would work

the two resistors idea I got off of the 18watt website where I remember seeing a post about subbing two resistors and a diode instead of the rectifier tube

I don't think I will be getting any sag from the resistors since the resistors are before the rectifier

am I incorrect ??

CB Amp,
where are you suggesting adding the RC filter ??

again thanks for the input

Dan.......

[drz400]

So I have and old Masco P.A. amp that has been kicking around for a while
I turned it into a PTP Express inspired amp but it was not laid out well
and it buzzed and hummed
Yesterday I turned it into a PTP inspired JTM45 and it turned out surprisingly well
but I have about 490-500 volts on the plates of the power tubes
and tried several ways of lowering the B+

I didn't want to wait for any parts and I know the best way to lower it is with the VVR
but today I wired one 5W 100 ohm resistor on each of the leads from the PT going to my rectifier and have succeeded in dropping about 45-50 volts so now with EL 34 tubes biased at around 41ma I have 465 v on the plates

any idea if this will hold up
the PT is putting out more B+ than I was expecting


Dan.....

Resistor drop will be more dependent on current draw so it might fluctuate more than you would like. It will also get blazing hot !
I would try and wire a small transformer out of phase on the secondary. Or just lower the screen voltage

[Structo]

Typically you would put your dropping resistors after the screen supply caps.
In series in between filter caps.

Such as this:

Don't pay any attention to the values on this schematic, it is simply to show a dropping string.

[dansamp]

OK.
so I don't quite understand why putting a dropping resistor directly from the rectifier to B+1 gets so hot
is it because I don't have a cap connected to it ??

if so then my thinking of adding a cap directly from the rectifier and a dropping resistor then that becomes B+1 and then every thing else gets connected to B+2 , B+3, etc, etc

the way I have it now with the two resistors coming off of the AC side
do not seem to get very hot at all

but when I had the one larger value resistor from the rectifier to B+1
I could smell it cooking, I removed it and you can see a dark spot on the
underside of it

anyway hope this makes sense

thanks

Dan......

[benoit]

You've gotta do the math. You've got a 5W 100 Ohm resistor.

Power = Voltage^2 / Resistance

Plug in your resistor's specs and you find that 500V ought to get you pretty close to that limit. You said without the resistor you were seeing 500V on the plates, which means a 100R resistor is going to dissipate about 5W, if I'm not mistaken. Unless I'm very mistaken, this is what's happening with your resistor.