John,
Changing the regulator to a 7A version will not solve the problem. The issue seems to be that the regulator goes into thermal shut-down. I.e. for the power currently dissipated in the regulator, the die temperature reaches the trip point for thermal shut-down function (probably set at 140-150 deg C). Changing the regulator to a different type will not change that.
Personally, I would shy away from CPU/GPU heat sinks as they tend to be made from rather skinny aluminum. They are made for forced air cooling. You need something with a little more thermal mass connecting the regulator to the fins of the heat sink. It's a bit tough to guess how much heat sink is big enough without actually doing the math.
If I recall correctly, the 300B draws around 1.2~1.5 A on the heater and you're running two tubes off of one regulator. Figure an input voltage to the regulator of 1.1*6.3*sqrt(2) = 9.8 V. This leaves a drop of 9.8-5 = 4.8 V across the regulator IC. So the dissipated power will be P = I * E --> P = 2*1.5 * 4.8 = 14.4 W.
Let's say the air around the heat sink is 30 deg C (86 F) and air is allowed to flow freely across the heat sink. Let's say you want the max heat sink temperature to be 60 deg C. That's pretty hot to the touch but not so hot that you burn yourself instantly if you touch it. That's a 30 degree temperature difference between ambient and heat sink. 14.4 W is dissipated in the heat sink. Thus, you'll need a heat sink with a thermal resistance of NO HIGHER than 30/14.4 = 2.08 deg/W. That's actually a fairly substantial heat sink. This one has a slightly higher thermal resistance (2.4 deg/W), thus, will be slightly warmer than the 60 degrees designed for (actual temperature = 30 + 2.4*14.4 = 64.6 deg C). But it'll probably work.
Note that the regulator IC will be even warmer than the heat sink. Assuming a thermal resistance of 1 deg/W from junction to heat sink (probably not too far off), it'll be 1*14.4 = 14.4 degrees warmer. So 79 degrees with the heat sink linked to above. Or slightly below spec. Most semiconductors these days are spec'ed at +85 deg C. Some only 0-70 deg C. I wouldn't run them hotter than 125 deg C under ANY conditions.
Now you know how to calculate the needed heat sink. I have used the figures I normally feel comfortable with. I'll leave it to you to substitute yours if need be.
Good luck.
~Tom
Changing the regulator to a 7A version will not solve the problem. The issue seems to be that the regulator goes into thermal shut-down. I.e. for the power currently dissipated in the regulator, the die temperature reaches the trip point for thermal shut-down function (probably set at 140-150 deg C). Changing the regulator to a different type will not change that.
Personally, I would shy away from CPU/GPU heat sinks as they tend to be made from rather skinny aluminum. They are made for forced air cooling. You need something with a little more thermal mass connecting the regulator to the fins of the heat sink. It's a bit tough to guess how much heat sink is big enough without actually doing the math.
If I recall correctly, the 300B draws around 1.2~1.5 A on the heater and you're running two tubes off of one regulator. Figure an input voltage to the regulator of 1.1*6.3*sqrt(2) = 9.8 V. This leaves a drop of 9.8-5 = 4.8 V across the regulator IC. So the dissipated power will be P = I * E --> P = 2*1.5 * 4.8 = 14.4 W.
Let's say the air around the heat sink is 30 deg C (86 F) and air is allowed to flow freely across the heat sink. Let's say you want the max heat sink temperature to be 60 deg C. That's pretty hot to the touch but not so hot that you burn yourself instantly if you touch it. That's a 30 degree temperature difference between ambient and heat sink. 14.4 W is dissipated in the heat sink. Thus, you'll need a heat sink with a thermal resistance of NO HIGHER than 30/14.4 = 2.08 deg/W. That's actually a fairly substantial heat sink. This one has a slightly higher thermal resistance (2.4 deg/W), thus, will be slightly warmer than the 60 degrees designed for (actual temperature = 30 + 2.4*14.4 = 64.6 deg C). But it'll probably work.
Note that the regulator IC will be even warmer than the heat sink. Assuming a thermal resistance of 1 deg/W from junction to heat sink (probably not too far off), it'll be 1*14.4 = 14.4 degrees warmer. So 79 degrees with the heat sink linked to above. Or slightly below spec. Most semiconductors these days are spec'ed at +85 deg C. Some only 0-70 deg C. I wouldn't run them hotter than 125 deg C under ANY conditions.
Now you know how to calculate the needed heat sink. I have used the figures I normally feel comfortable with. I'll leave it to you to substitute yours if need be.
Good luck.
~Tom
Oh... I assumed that you're using a 6.3 V winding on the transformer. If you're using a 5 V winding, you need to modify your equations a bit as the worst case voltage will be 1.1*5*sqrt(2) = 7.8 V.
~Tom
~Tom
This I agree with.
This I don't understand. True the peak voltage across the filter cap can go to sqrt(2) * 6.3 volts under unloaded conditions.
When loaded the filter cap does not charge to the full peak voltage. The conduction angle would be zero if it did. The accepted value is from 1.2 to 1.3 times the transformer voltage. I have found that the distortion on todays power line causes wide variatiations in the rectified voltage. 10% distortion is common, but the waveform of the sine wave shows some flat topping (lowers the voltage), or worse, signs of power pole distribution transformer saturation (riases the voltage). This changes the conduction angle of the rectifier diodes which can result in a higher or lower rectified voltage than expected. The type of power line distortion varies during the day.
The voltage will be also be lower under load due to the drop in the rectifier diodes. The Tubelab SE uses a full wave bridge when configured for 300B's. One pair of diodes are conventional silicon diodes, while the other pair are Schotkeys. This will add about 1 volt of drop under normal operating conditions.
I should also mention that many power transformers routinely supply a higher AC voltage output than what is specified on the label. It is not uncommon for a Hammond transformer to provide nearly 7 volts on the 6.3 volt winding when loaded with a resistive load of the rated current.
So the actual rectified voltage on the Tubelab SE is a bit more difficult to predict, but is relatively easy to measure. I have two Tubelab SE's here. The little guy with the Allied 6K56VG transformer and a pair of NX-483 tubes (small but unknown filament current) measures in at 6.6 volts and the big guy with a Hammond 272JX and Chinese 300B's measures 7.4 volts (right now at 10PM). These numbers vary by several tenths of a volt during the day.
The newer stuff is pretty thin, but I use hamfest quality Pentium 1 or 486 vintage stuff that didn't come with a fan. The big green heatsink in my Lexan amp came off of a 66MHz (blazing speed) 486 chip. I busted off a few fins and drilled a hole through the center of it. Leave the little black heat sink on the regulator and Schotkey rectifier. It helps keep the rectifier cool. I removed the screw from the regulator and bolted the heat sink up to the little black one using a longer #4 screw. I used a thin piece of aluminum as a spacer so the big heat sink would clear the edge of the PC board. There is nearly zero airflow in that Lexan box, and I can run 300B tubes in that amp although the little power transformer gets pretty hot when I do. I am pulling nearly 200 mA from a 125 mA transformer. Its 5 years old, and still smokin......
Thanks Tomchr and Tubelab. Although I'm not versed with the math you guys discussed, my background is medicine, I can picture what you guys are referring at. I'm trying a 3 inch by 3 inch square heatsink from a busted computer, 2 inches thick and I played 2 albums and 2 CDs already and the filaments are still lit. The heatsink is 3 times larger than what used to be in there. I think I got the problem under control. I hoped. If I'll encounter the same problem in the future I least I know the culprit. Godspeed.
John Revilla
John Revilla
I wasn't aware of the 1.2~1.3x factor. I've always used sqrt(2), although, for the reason you mention it is a pessimistic estimate. I included the 1.1x to account for +10 % variation in mains voltage.
Correct. And ripple voltage will result in a varying drop-out across the regulator, thus, lower the dissipated power.
One can get fancy with the math or just run a quick back-of-the-envelope worst-case estimate like I did. If I wanted to be fancy, I'd probably throw the circuit into SPICE and run a quick sim. Or measure in the lab like you did.
Why one pair of each type?
Yep! Actually, one of the things I like about the Antek transformers is that the no load and full load voltages are specified. And at least for the two samples I've seen this far, I've been able to reproduce the data sheet results.
~Tom
The amp was originally used for 45s and in that case the diode pair used for the lower half of the 6.3V was only used for the input tube heaters. I think that's why he didn't use the more expensive Schottky package for those.
In fact, when configured for 300Bs, those two diodes are pushed hard. The PCB under mine is becoming quite brown...I probably should have elevated them more than I did. I may swap them for a pair of Schottky's at some point.
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