I've just upgraded my EL34 P-P amp's power supply and I am concerned about the heat being generated by the filter resistor.
Before the upgrade, this was a straight C filter - rectifier straight into a 470uF 450V cap.
What I did was put a 200uF 450V cap and 100ohm 25W resistor in front of the existing cap. So a CRC filter.
It works. My B+ is down where I want it and I am sure there is far less ripple. However, the 25W resistor is running at 100 Celsius + after only a few minutes (amp running upside down with the bottom open).
I used this site Ohms Law Calculator to calculate the power resistor dissipation. My voltage drop across the resistor is 32 volts (based on the fact that my B+ is dropping from 426V before this change, to 394V after). So I plugged in 32V and 100 ohms on that site and it tells me 10.24 watts.
Did I screw this up, or is it normal for the 25W ceramic resistor to run this hot? I can imagine it is going to get considerably hotter when the bottom plate is installed and the amp is running upright for a few hours.
Before the upgrade, this was a straight C filter - rectifier straight into a 470uF 450V cap.
What I did was put a 200uF 450V cap and 100ohm 25W resistor in front of the existing cap. So a CRC filter.
It works. My B+ is down where I want it and I am sure there is far less ripple. However, the 25W resistor is running at 100 Celsius + after only a few minutes (amp running upside down with the bottom open).
I used this site Ohms Law Calculator to calculate the power resistor dissipation. My voltage drop across the resistor is 32 volts (based on the fact that my B+ is dropping from 426V before this change, to 394V after). So I plugged in 32V and 100 ohms on that site and it tells me 10.24 watts.
Did I screw this up, or is it normal for the 25W ceramic resistor to run this hot? I can imagine it is going to get considerably hotter when the bottom plate is installed and the amp is running upright for a few hours.
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Indeed 32 x 32 / 100 is 10.24
To double check, measure the current through the resistor, it should be 32V / 100 Ohm, so 320 mA.
Ceramic can stand high temperatures, look at the datasheets. Could be it needs a heatsink. Could be so hot, solder joint melt.
To double check, measure the current through the resistor, it should be 32V / 100 Ohm, so 320 mA.
Ceramic can stand high temperatures, look at the datasheets. Could be it needs a heatsink. Could be so hot, solder joint melt.
Resistor economics is about the most heat in the least material. It's gonna get HOT.
Why do you think the amp wants less voltage? 30V drop is hardly anything. The EL34 is good for twice the voltage you have now.
Why do you think the amp wants less voltage? 30V drop is hardly anything. The EL34 is good for twice the voltage you have now.
My goals were to reduce ripple, reduce power tube dissipation, and lighten the load on the power transformer (smaller first cap reduces current peaks a bit).
I realize now that all I did was move the heat from the tubes to the resistor. 🙂
I realize now that all I did was move the heat from the tubes to the resistor. 🙂
Reminds me of the cheap method of voltage reduction used in small transformer-less Bakelite radios --big ceramic resistor --heat went to top of casing --cracked Bakelite - radio worthless .
Carried on in early UK CRT TV,s .
If you intend to leave it do as post #2 says - fit a heat-sink.
Carried on in early UK CRT TV,s .
If you intend to leave it do as post #2 says - fit a heat-sink.
10 watts of power dissipated will heat up any resistor, no matter what its power rating is. It's just a high power resistor won't fail. One thing to check is what the resistor's power rating is at high temperatures. Its power rating is only for a specified design temperature.
Thanks everyone.
I've changed the resistor mounting location to a top corner inside the chassis, set a piece of heat sink on top of it, and secured it all with a bracket of sturdy spring steel that I bent into a shape that provides constant pressure to the whole works - pushes it into the corner of the (all metal) chassis and forces the heat sink down onto on surface of the ceramic.
I popped the bottom cover on it and have been running the amp for several hours. The exterior corner of the chassis (opposite where the resistor is located) is obviously radiating some heat, but it isn't painfully hot or anything.
I am going to continue to run it all day and evening, and then shut it down for the night and open it up tomorrow morning to inspect it. If all is well, I'll consider it good enough.
The amp sounds better than it ever has. I also modified it from a shared cathode resistor / bypass cap topology to dedicated cathode resistors and bypass caps. At this point the EL34s are operating very close to the Mullard spec for this topology, except plate voltage is running at 390v instead of 400v and I installed 470ohm cathode resistors vs. 440ohm.
One more thing: The bloody power transformer still runs as hot as it ever did. ~ 77 Celsius.
I've changed the resistor mounting location to a top corner inside the chassis, set a piece of heat sink on top of it, and secured it all with a bracket of sturdy spring steel that I bent into a shape that provides constant pressure to the whole works - pushes it into the corner of the (all metal) chassis and forces the heat sink down onto on surface of the ceramic.
I popped the bottom cover on it and have been running the amp for several hours. The exterior corner of the chassis (opposite where the resistor is located) is obviously radiating some heat, but it isn't painfully hot or anything.
I am going to continue to run it all day and evening, and then shut it down for the night and open it up tomorrow morning to inspect it. If all is well, I'll consider it good enough.
The amp sounds better than it ever has. I also modified it from a shared cathode resistor / bypass cap topology to dedicated cathode resistors and bypass caps. At this point the EL34s are operating very close to the Mullard spec for this topology, except plate voltage is running at 390v instead of 400v and I installed 470ohm cathode resistors vs. 440ohm.
One more thing: The bloody power transformer still runs as hot as it ever did. ~ 77 Celsius.
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All that ripple voltage is also dropped across the resistor, besides just the DC drop.
Often that will be more power than the IR loss.
Often that will be more power than the IR loss.
Interesting point, rayma! That never occurred to me.
According to PSUD2, the original configuration had 3.9V (!!!) of ripple after the 470uF cap, and the new configuration has 171mV of ripple after the 470uF cap. So there's another 3.88-ish volts that the filter resistor is converting to heat. Added to the DC that all works out closer to 13 watts being dissipated.
According to PSUD2, the original configuration had 3.9V (!!!) of ripple after the 470uF cap, and the new configuration has 171mV of ripple after the 470uF cap. So there's another 3.88-ish volts that the filter resistor is converting to heat. Added to the DC that all works out closer to 13 watts being dissipated.
Use a 47 ohm resistor but still 10 watts to bring power down to 5 watts.
The danger with very hot resistors is they can melt the solder and drop out.
The danger with very hot resistors is they can melt the solder and drop out.
Your measured DC voltage drop probably already includes some component of the AC ripple voltage. It depends on how your DC voltmeter handles AC ripple on top of DC, so it is likely that it is not as bad as you calculate.
You can use paralleled resistors to share the power burned, and to lower the temperature of the resistors.
The input ripple is roughly a sawtooth waveform, with an average value of about 2V (if peak is about 4V),
so add that AC loss to the DC drop loss.
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This thought occurred to me as well, but I don't have 2 200ohm resistors.
However, I do have a couple of 20W 250ohm resistors that I just removed from the amp (they were shared cathode resistors before).
If I parallel the 100ohm with one of the 250ohm, I'll have about 71 ohms. PSUD2 shows my ripple increase by about 100mV but my B+ and plate voltage will be pretty-much bang on Mullard spec (as shown earlier). The two resistors will share the job, and there will be less overall dissipation to share.
I'll do it.
However, I do have a couple of 20W 250ohm resistors that I just removed from the amp (they were shared cathode resistors before).
If I parallel the 100ohm with one of the 250ohm, I'll have about 71 ohms. PSUD2 shows my ripple increase by about 100mV but my B+ and plate voltage will be pretty-much bang on Mullard spec (as shown earlier). The two resistors will share the job, and there will be less overall dissipation to share.
I'll do it.
Done. Plate voltage is right at 404V now and the two resistors are warm after a few minutes. I am sure they are still going to get hot, but nothing like before. This is all around a better solution. I very much doubt I will be able to hear any difference.
Thanks again everyone!
Thanks again everyone!