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Old 25th June 2013, 08:53 PM   #11
fabioab is offline fabioab  Brazil
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Quote:
Originally Posted by GoatGuy View Post
I think funk1980 got it. No heat sink on the resistor. By itself, it probably only has a rating of 10 to 15 watts with no-heat-sink air cooling.

The 10 ea / 1.8K-in-parallel resistors will work fine. Personally, i wouldn't worry about the over-voltage, as long as your tubes aren't arcing or exhibiting red-glowing anodes. They're tough, power tubes are. I think you'd get more traction by simply adjusting the BIAS of the tubes if you're concerned for the total dissipation.

GoatGuy
Thanks, but I feel the ampli more stable and more musical working at 460V on plates, and give-me more range to vary the bias voltage and the current into the tube.
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Old 25th June 2013, 09:34 PM   #12
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100watt resistors usually need a heatsink.
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Old 25th June 2013, 09:39 PM   #13
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Originally Posted by fabioab View Post
Thanks, but I feel the ampli more stable and more musical working at 460V on plates, and give-me more range to vary the bias voltage and the current into the tube.
Could it be that you like the effect of the HT sagging when playing loud? Just like using valve rectifiers instead of silicon? The current goes from maybe 120mA idle to 500mA at full power.
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Old 25th June 2013, 09:58 PM   #14
llwhtt is offline llwhtt  United States
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The '67 Plexis had a B+ of 460VDC, those are the magical ones. EVH comes to mind if that's your thing.

Craig
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Old 26th June 2013, 12:53 AM   #15
fabioab is offline fabioab  Brazil
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Originally Posted by llwhtt View Post
The '67 Plexis had a B+ of 460VDC, those are the magical ones. EVH comes to mind if that's your thing.

Craig
That is what my ears can hear !
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Old 26th June 2013, 08:46 AM   #16
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Just use the 100W resistor, mounted on a suitable heatsink to get the ful 100W capability. You should be able to even drop 134V across it!

Using the resistor is this way will cause sag, as many have stated above already. Sag depends of the amount of current drawn from the PSU, so B+ will vary accordingly and the amp sounds different depending on how hard it is driven.
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Old 26th June 2013, 02:37 PM   #17
kevinkr is offline kevinkr  United States
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All musical instrument amplifier related threads belong in Instruments & Amplifiers. Please read the forum subheadings for guidance on where to post your thread. Thank You.
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Old 26th June 2013, 04:53 PM   #18
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A sag resistor will not accomplish what you want. If the B+ drops 20V at idle, it will drop way too much at full power. Power Scaling or Zener diodes are the way you need to go.

Because transformer current flows in pulses where the RMS value is much higher than average value measured by your meter, the power dissipated by the sag resisitor can not be calculated by simple ohms law.

Consider how this works. If 1 Amp flows for 1 second in a 1 ohm resistor, 1 Watt is dissipated. If 2 Amps flows for 1/2 a second and falls to zero for 1/2 a second, is that the same? NO!!! 2 Amps causes the resistor to dissipate 4 Watts for 1/2 a second and zero for half a second so averaged over the whole second, the dissipation is 2 Watts. If 4 Amps flows for 1/4 second and falls to zero for 3/4 second, the dissipation is 16W for 1/4 the time and averages out to 4 Watts. In this case the RMS current is 2 Amps. The RMS current squared times the resistance equals the power dissipation.

Pop quiz: What is the RMS current when 2 Amps flows for half the time and falls to zero for half the time?

Off the top of my head 180 ohms is way too high for a sag resistor. The best thing to do is to model the circuit with a SPICE like progran. Duncan's PSU Designer II works well and is simple to use. You will need to know:

1) What is the rectifier configuration? Full Wave (using a center tap) or the Full Wave Bridge
2) What is the open circuit (no load) voltage of the power transformer secondary
3) What is the resistance of the power transformer secondary winding(s)?
4) What is the power line frequency (50Hz?)
5) What is the current draw at idle?
6) What is the current draw at full power (rail to rail clipping)
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Old 27th June 2013, 11:17 AM   #19
fabioab is offline fabioab  Brazil
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Quote:
Originally Posted by Loudthud View Post
A sag resistor will not accomplish what you want. If the B+ drops 20V at idle, it will drop way too much at full power. Power Scaling or Zener diodes are the way you need to go.

Because transformer current flows in pulses where the RMS value is much higher than average value measured by your meter, the power dissipated by the sag resisitor can not be calculated by simple ohms law.

Consider how this works. If 1 Amp flows for 1 second in a 1 ohm resistor, 1 Watt is dissipated. If 2 Amps flows for 1/2 a second and falls to zero for 1/2 a second, is that the same? NO!!! 2 Amps causes the resistor to dissipate 4 Watts for 1/2 a second and zero for half a second so averaged over the whole second, the dissipation is 2 Watts. If 4 Amps flows for 1/4 second and falls to zero for 3/4 second, the dissipation is 16W for 1/4 the time and averages out to 4 Watts. In this case the RMS current is 2 Amps. The RMS current squared times the resistance equals the power dissipation.

Pop quiz: What is the RMS current when 2 Amps flows for half the time and falls to zero for half the time?
Answer:
I_rms=SQRT{ 1/T [ (2^2)*0.5T + [0^2]*0.5T ] } = 1.41 amps
This is the value that the meter measure

Because of the peak dissipation I thank 100W could support the effect of peak dissipation. I'm using 118_Oms 80W (4x20W ceramic resistors 470_Ohms in paralell), and It goes well.


Quote:
Originally Posted by Loudthud View Post
Off the top of my head 180 ohms is way too high for a sag resistor. The best thing to do is to model the circuit with a SPICE like progran. Duncan's PSU Designer II works well and is simple to use. You will need to know:
It is true, using 118_Ohms the +B voltage drops to 350V in half volume! Without sag resistor the +B drops to 420V.

Quote:
Originally Posted by Loudthud View Post
1) What is the rectifier configuration? Full Wave (using a center tap) or the Full Wave Bridge
A: Full Wave (using a center tap)

Quote:
Originally Posted by Loudthud View Post
2) What is the open circuit (no load) voltage of the power transformer secondary ?
A: 516V

Quote:
Originally Posted by Loudthud View Post
3) What is the resistance of the power transformer secondary winding(s)?
A : 114_ohms

Quote:
Originally Posted by Loudthud View Post
4) What is the power line frequency (50Hz?)
A: 60Hz

Quote:
Originally Posted by Loudthud View Post
5) What is the current draw at idle?
A: 100mA

Quote:
Originally Posted by Loudthud View Post
6) What is the current draw at full power(rail to rail clipping)
A: I think 500mA

Thanks the hints Loudthud
FAB
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Old 27th June 2013, 01:56 PM   #20
GoatGuy is offline GoatGuy  United States
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Hey... There is a MUCH simpler way to do this than resistors!

1N5339 - a 5 watt, 5.6volt zener diode. (industry uses 5 watt zeners a lot in motor and panel work). 850 ma working current. Can take quite a bit more than that 'pulsed.'

So you want 20 volts of voltage drop? 4 of them in series. 50 volts? 9 of them in series. And ... they can "handle" more power collectively 'cuz of the series connection.

And here's the best part: $0.36 each from Mouser, "quantity 1 to 25". Buy 20 of 'em and save on shipping! The cool thing about zener voltage is that it is almost constant over a HUGE range of current flow. This is exactly what the "resistor idea" was supposed to accomplish, but has not.

ALL analysis above that points to the resistor getting hot 'cuz of the very large difference between idle current, and high-output driven current ... is correct. Go with the zeners, and get a triple-good solution. Constant voltage drop. Current independence. More than adequate power dissipation over full operating range. Cheap... oops, a 4th factor.

GoatGuy
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