Question for those with more building experience. Working on my first from-scratch SET amp, building is about to begin. I need to drop the filament voltage for my power tubes, it is a little too high for comfort (7.1V on 6.3V rated tubes). Values and whatnot are already determined, going to use a pair of 0.25ohm Ohmite wirewound resistors to knock 0.5V off the 1A filaments.
I am using a Lundahl mains transformer which does not come pre-wired. My question is, rather than connecting the resistors at the tube socket pins, is there any reason I can't solder them directly to the transformer winding pins? That way I'd be able to keep a nice tight twist all the way to the heaters rather than it ending in a pair of resistors at the tube socket.
I konw this is a novice question, I don't see why not but I'd rather ask and than get a surprise 😀 thanks.
I am using a Lundahl mains transformer which does not come pre-wired. My question is, rather than connecting the resistors at the tube socket pins, is there any reason I can't solder them directly to the transformer winding pins? That way I'd be able to keep a nice tight twist all the way to the heaters rather than it ending in a pair of resistors at the tube socket.
I konw this is a novice question, I don't see why not but I'd rather ask and than get a surprise 😀 thanks.
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Be careful measuring transformer on and off load.
It could be designed to 6v3ac when taking current.
If its really 7v1ac on load then obviously high power resistor required.
Try to keep value as low as possible to keep heat down.
It could be designed to 6v3ac when taking current.
If its really 7v1ac on load then obviously high power resistor required.
Try to keep value as low as possible to keep heat down.
Good advice, no numbers.
Ohm's Law
Use this 75¢ resistor:
There you go.
⋅-=≡ GoatGuy ✓ ≡=-⋅
Ohm's Law
E = I R … volts = amps times Ω.
(7.1 - 6.3) = (1.6 × 2) R … 1600 mA for 6550 triode-wired outputs, 1 per channel.
0.8 ÷ 3.2 = R
0.25 = R.
Power Law(7.1 - 6.3) = (1.6 × 2) R … 1600 mA for 6550 triode-wired outputs, 1 per channel.
0.8 ÷ 3.2 = R
0.25 = R.
P = IE … watts equals amps times volts.
P = 0.8 V × 3.2 A
P = 2.56 W
Now we know the resistor sizing. Rule of thumb … 2× the expected power dissipation. So, 5 watts. P = 0.8 V × 3.2 A
P = 2.56 W
Use this 75¢ resistor:
There you go.
⋅-=≡ GoatGuy ✓ ≡=-⋅
It depends on how much voltage they want to drop.
Personally I wouldnt have bothered with a dropper.
I built a valve pre amp with 12AX7 and used same 12 volts to power B+ and heaters.
Not thinking I decided to up the B+ a bit as it was a little low.
I wound it up to 30VDC before suddenly realising the heater was getting a bit bright ! Anyway, the heater survived 30VDC so I suspect 7v1ac wont cause too many problems.
You may end up adjusting the values a couple of times and don't want to damage the pin mountings with the soldering.
Thank you for your responses gentlemen, sounds like it is not an altogether horrible idea 😀
Thanks rayma, and thank you for your help with another question, I tried to respond via PM but your inbox was full 😉
Yes, this is at load. I knew it would be high since the windings of my transformer (Lundahl LL1648) are rated for 6.6V at 115VAC. My wall AC runs high at 123-124VAC. I hooked the transformer up to a variac and increased the voltage to 124VAC with a second DMM across the tube heaters, that's where I got the 7.1V.
Mr. GoadGuy sir, here are the numbers. My output tubes are 6A5G, which have a heater current of 1A, each tube will get its own heater winding, no daisy chaining. The heater of these tubes is center-tapped to ground through the cathode, so each leg will carry
7.1/2 = +/-3.55V
I had planned to drop the total voltage by 0.5V down to ~6.6V (this is because my mains is high, will be closer to 6.3V on more typical 120VAC outlets), so 0.25V dropped per leg.
R = 0.25V/1A = 0.25ohm (two 0.25ohm resistors, one per leg)
P = 1A * 0.25V = 0.25W
I ordered some 0.25ohm 1W Ohmite wirewound resistors, so four times the dissipated power. Hope that looks good 🙂
The transformer gives you a little more than a centimeter of pin to work with and they are quite sturdy, thanks for the advice, I will be careful and not go crazy with the solder, will fasten it close to the edge so it could be easily desolder with some wick if necessary.
Thanks rayma, and thank you for your help with another question, I tried to respond via PM but your inbox was full 😉
Yes, this is at load. I knew it would be high since the windings of my transformer (Lundahl LL1648) are rated for 6.6V at 115VAC. My wall AC runs high at 123-124VAC. I hooked the transformer up to a variac and increased the voltage to 124VAC with a second DMM across the tube heaters, that's where I got the 7.1V.
Mr. GoadGuy sir, here are the numbers. My output tubes are 6A5G, which have a heater current of 1A, each tube will get its own heater winding, no daisy chaining. The heater of these tubes is center-tapped to ground through the cathode, so each leg will carry
7.1/2 = +/-3.55V
I had planned to drop the total voltage by 0.5V down to ~6.6V (this is because my mains is high, will be closer to 6.3V on more typical 120VAC outlets), so 0.25V dropped per leg.
R = 0.25V/1A = 0.25ohm (two 0.25ohm resistors, one per leg)
P = 1A * 0.25V = 0.25W
I ordered some 0.25ohm 1W Ohmite wirewound resistors, so four times the dissipated power. Hope that looks good 🙂
The transformer gives you a little more than a centimeter of pin to work with and they are quite sturdy, thanks for the advice, I will be careful and not go crazy with the solder, will fasten it close to the edge so it could be easily desolder with some wick if necessary.
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