Hello, I need some hint!
I wired a sag 180_OHMS resistor and te voltare dropped 20V between the poles of the resistor. This is the exact value I wanted. It's installed in my plexi 100W.
The resistor is 100W (yes 100Watt) and installed after the diodes. At the other side of the resistor is wired to the first capacitor , i.e.,
PT--->Diodes--->resistor--->rectification_capacitor (50uF 500V).
Ok, [(20V)^2] / 180_ohms = 2,3W.
After play 5 minutes the temperature of the resistor goes up extremely high.
I think 100W is sufficient to survive the variations of DDP caused by input frequency of the guitar.
Someone could explain waht are happening?
Best regards and thanks
FAB
I wired a sag 180_OHMS resistor and te voltare dropped 20V between the poles of the resistor. This is the exact value I wanted. It's installed in my plexi 100W.
The resistor is 100W (yes 100Watt) and installed after the diodes. At the other side of the resistor is wired to the first capacitor , i.e.,
PT--->Diodes--->resistor--->rectification_capacitor (50uF 500V).
Ok, [(20V)^2] / 180_ohms = 2,3W.
After play 5 minutes the temperature of the resistor goes up extremely high.
I think 100W is sufficient to survive the variations of DDP caused by input frequency of the guitar.
Someone could explain waht are happening?
Best regards and thanks
FAB
P = E * E / R
P = 20 * 20 / 180
P = 400 / 180;
P = 2 watts
A 100 watt rated resistor shouldn't be getting even barely warm to the touch, Fabio. You've got something else amiss... it could be
* the "20 volts" you're measuring is just an average delivered by your multimeter - but an oscilloscope trace across the same resistor might show substantially higher high frequency voltage.
-or-
* when you're playing your guitar, the voltage rises substantially - because the output to the speaker is taking substantial power to deliver. This wouldn't surprise me. With the above calculations:
P = E * E / R
P = 60 * 60 / 180
P = 3600 / 180
P = 20 watts
Again ... the resistor shouldn't be getting "terribly hot" - if it is 100 watt rated. Nice and warm, but not burning hot.
* You could have mistaken the resistance (or mis-stated it originally). Maybe take a measurement of the resistor's actual in-circuit value? If it is substantially LOWER than 180 ohms, of course the heat dissipation (for the respective runtime levels) would increase by the same proportion.
Y'all have a circuit diagram of where you actually inserted the resistor?
... and perhaps a rating of nominal output power of the amp?
GoatGuy
P = 20 * 20 / 180
P = 400 / 180;
P = 2 watts
A 100 watt rated resistor shouldn't be getting even barely warm to the touch, Fabio. You've got something else amiss... it could be
* the "20 volts" you're measuring is just an average delivered by your multimeter - but an oscilloscope trace across the same resistor might show substantially higher high frequency voltage.
-or-
* when you're playing your guitar, the voltage rises substantially - because the output to the speaker is taking substantial power to deliver. This wouldn't surprise me. With the above calculations:
P = E * E / R
P = 60 * 60 / 180
P = 3600 / 180
P = 20 watts
Again ... the resistor shouldn't be getting "terribly hot" - if it is 100 watt rated. Nice and warm, but not burning hot.
* You could have mistaken the resistance (or mis-stated it originally). Maybe take a measurement of the resistor's actual in-circuit value? If it is substantially LOWER than 180 ohms, of course the heat dissipation (for the respective runtime levels) would increase by the same proportion.
Y'all have a circuit diagram of where you actually inserted the resistor?
... and perhaps a rating of nominal output power of the amp?
GoatGuy
Thanks GoatGuy
I think the same you, and did some measures again
The resistor is 180_Ohms, 20V across the resistor and very hot! It happened with others e.g. 30Watt.
I really don't know.
The resistor comes direct after the rectification (after the diodes) in the power section.
Without the resitor the ampli works well but , I think the plate voltage is so much to the El34 : 480V when biased with 30mA across the tube. I prefer around 455-460V. This is the reason to use sag.
Thanks again
I think the same you, and did some measures again
The resistor is 180_Ohms, 20V across the resistor and very hot! It happened with others e.g. 30Watt.
I really don't know.
The resistor comes direct after the rectification (after the diodes) in the power section.
Without the resitor the ampli works well but , I think the plate voltage is so much to the El34 : 480V when biased with 30mA across the tube. I prefer around 455-460V. This is the reason to use sag.
Thanks again
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
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 all,
As first approximation I tried 12 resistors (2w) 2.2k_ohms in parallel resulting in 0,2W dissipation per resistor. The max power of the core is around 21.8W. Playng, the resistor work very hot. If I try Using 12 of 5W it will support maximum 58W. It's almost fine to plaxi! It explain why my core of 12 x 2W is working very hot, and will not support long time!
As first approximation I tried 12 resistors (2w) 2.2k_ohms in parallel resulting in 0,2W dissipation per resistor. The max power of the core is around 21.8W. Playng, the resistor work very hot. If I try Using 12 of 5W it will support maximum 58W. It's almost fine to plaxi! It explain why my core of 12 x 2W is working very hot, and will not support long time!
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.
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.
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.
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.
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)
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)
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.
It is true, using 118_Ohms the +B voltage drops to 350V in half volume! Without sag resistor the +B drops to 420V.
A: Full Wave (using a center tap)
A: 516V
A : 114_ohms
A: 60Hz
A: 100mA
A: I think 500mA
Thanks the hints Loudthud
FAB
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
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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