Drawmer 1960 12V Valve heater circuit - Page 1
Forum section: electronics / repair

author: keland_uk
author_rank: Contributor
timestamp: September 05, 2021, 04:16:58 pm
content: Can anyone please enlighten me as to the purpose of the 2W 56ohm resistor in this cct? (The LED illumitators also have a 2W56ohm resistor in series)Also IN4002's getting pretty warm, would a higher wattage diode help?Obviosly 7812 is maxxed out, especially when tubes are cold (heater resistance I believe are 15ohm cold, 80ohm hot)

author: Gyro
author_rank: Super Contributor
timestamp: September 05, 2021, 06:45:25 pm
content: The resistor is there to reduce the dissipation of the 7812 by acting as a bypass for some of the heater current (but not enough that it doesn't still regulate the voltage). It probably makes the difference between being able to get away with a 7812 or not at startup.The 1N4002 is rated for 1A at 75'C ambient. The 7812 is capable of a bit more than this before going into current limit, but unless the 4002 is running really hot, it's probably fine. At 1A it will be dissipating about 1.1W but into 80R load it should be way less than that. You could go to a 3A 1N5402 if you wanted, but it has larger body size and more importantly, thicker leads.

author: keland_uk
author_rank: Contributor
timestamp: September 05, 2021, 11:16:05 pm
content: many thanks.There are 4 valves tho... I think someone worked out that was 300mA each when warm, and 800mA when coldWould a 78S12 would be a bad idea,Ah it's150ma x 4 = 600mA total when warm, surprised 800mA fuse doesn't blow on start up if its 3.2A.... Good slow blow fuse technology.....“If you’re hip to Ohm’s Law (I = E/R, R = E/I) you can see that one 12-volt filament(or heater) consuming 150mA has an equivalent hot resistance of 80ohms. Imeasured about 15Ohms cold, so the start up current for ONE tube is about 800mA(1000mA = 1 Amp). That times four momentarily exceeds the LM7812’scapacity, considerably decreasing the Mean Time Before Failure or MTBF.(Translation: a long life is not in the cards.) It was not a surprise to find anormally green circuit board had turned brown from heat.”

author: james_s
author_rank: Super Contributor
timestamp: September 05, 2021, 11:37:05 pm
content: It might allow you to get away without needing the resistor. The total dissipation and heat output is not going to change though. A regulator capable of dissipating more current may need a larger heatsink depending on what is there already.

author: BrokenYugo
author_rank: Super Contributor
timestamp: September 06, 2021, 05:21:52 pm
content: While kind of hacky and hot it seems to work, leaving well enough alone is an option.I'm just confused as to why they're going to so much trouble to regulate a heater supply 10% low.

author: keland_uk
author_rank: Contributor
timestamp: September 06, 2021, 06:34:20 pm
content: it is a known problem with these drawmer 1960'sI found this article which proposes a "fix"https://www.mixonline.com/technology/techs-files-turning-down-heat-371428BTW actualy measuring amperage at start up when cold, it starts at 1.25A, and steady state is 640mA.Resistor is around 100C, I would have thought it would have dropped after start up and the regulator started up?

author: BrokenYugo
author_rank: Super Contributor
timestamp: September 06, 2021, 07:59:27 pm
content: We really need to know the input voltage (voltage across C2) to analyze this much. Voltage across the resistor, diode, and 7812 wouldn't hurt either.

author: Gyro
author_rank: Super Contributor
timestamp: September 06, 2021, 09:41:26 pm
content: Quote from: BrokenYugo on September 06, 2021, 05:21:52 pmYeah, I wondered that, especially as most 12V valve heaters are actually 12.6V.Edit: With 0.7 - 1.1V across the diode and 0.6V short on the heater spec, it's actually probably a bit more than 10% low.

author: BrokenYugo
author_rank: Super Contributor
timestamp: September 06, 2021, 11:18:14 pm
content: Quote from: Gyro on September 06, 2021, 09:41:26 pmThe partial schematic OP posted shows 11.3 going to the heaters, 10.3% low.

author: TimFox
author_rank: Super Contributor
timestamp: September 06, 2021, 11:28:45 pm
content: Most, but not all, nominal 12.6 V vacuum-tube heaters pull 150 mA, since they are the 12-V version of the 6.3 V heaters that typically pull 300 mA (both values warm).Power tubes and a few higher-current filament tubes are the exception, so look at the relevant data; seehttps://frank.pocnet.net/for international data.

author: keland_uk
author_rank: Contributor
timestamp: September 07, 2021, 04:09:16 pm
content: it starts at 1.25A, and steady state is 640mA.We really need to know the input voltage (voltage across C2) to analyze this much.17.2VVoltage across theresistor 6.07VDiode 0.82V7812 5.2VResistance when cold of heaters is 3.55Ohms (4 in Parallel across pins 4 & 5), Each ECC82 is around 14OhmsReally appreciate input on this . Pretty much Ohms law, would you expect the resistor and the 7812 to "share the current"? Would explain why it gets so hot (0.32Ax6V= 1.92W)A single 78S12 would probably work without the Resistor, and hence the Diode as well.... (in fact I don't see why the 7812 couldn't actually cope on its own...)

author: BrokenYugo
author_rank: Super Contributor
timestamp: September 07, 2021, 06:17:27 pm
content: The tricky thing here is how a linear regulator works, one way or another it turns the difference between Vin and Vout to heat. Pare the circuit down to just the 7812 and it doesn't matter what variant you use, it's still gonna dump all 3.3 watts (if I did the math right) of excess into the heatsink that may or may not be adequate. That is one trap with linear regulator ICs, you usually run into thermal problems before the current rating. Plus then you won't be sending the intended 11.3 volts, and they must be doing that for some reason, further research indicates this may be to induce more of the "tube sound" distortion.I suppose you could replace the resistor with something like a larger case 3W one or 2 100Ω@2W in parallel to help keep it from cooking the PCB, but the biggest liability I see is that little 47uF cap (which keeps the 7812 from oscillating) getting cooked, use a long life 105*C part. Another thing you can do with hot running axial diodes and resistors is space them off the board something like 3-4mm.

author: keland_uk
author_rank: Contributor
timestamp: September 07, 2021, 10:25:24 pm
content: Quote from: BrokenYugo on September 07, 2021, 06:17:27 pmIf I kept the diode in, the voltage drop would reduce it to 11.3V me thinks. The Heat sink is pretty big for 3.3W, especially if I use one of the newer high transfer thermal pads (Gelid Solutions GP-Ultimate 15W/mK) My thermal imaging picture didn't show up a particularly high temperature on the heatsink, and thats with only a small amount of very dried thermal paste only half way up the 7812.To be exact though1) the 7812 is taking half the current and the resistor the other half?2) If that's the case, how does the 7812 regulate the voltage from the resistor?Quote from: BrokenYugo on September 07, 2021, 06:17:27 pmgood advice Thank you, on the case for both. Strangly only the 2200µF capacitator was cracked/suffering...

author: james_s
author_rank: Super Contributor
timestamp: September 08, 2021, 03:34:30 am
content: At one time it was reasonably common to see 78xx regulators bypassed by a resistor to carry more current. In particular it was done in the Amplifone vector monitors on the HV board, and the Heathkit GC-1000 "Most Accurate Clock". At the time I don't think the higher current 3 terminal regulators existed, and also a ceramic power resistor can safely run at considerably higher temperature than a semiconductor. If you're going to dissipate the power, there's some reasonable logic in doing so in a part that can run very hot without problems.