Simple question. What would you use for a basic regulated DC heater supply? Right now I need about 2.5A at 4v but it would be good to have some general ideas.
AC isn't possible since I'm using dropper resistors from a 6v transformer and possible tubes to be used are 4v but have different currents, so the dropper resistors give different results.
I assume an LM338 is a simple solution. Do you need the input and output caps in the attached circuit, or could it be simplified if supply caps are inside 6" (says so in the data)? I'd be using a 6v transformer.
Using online calculators, with a 240R in R1 looks like you need 520R in R2 for 4v. Not sure what wattage though.
Suggestions for a heat sink size for 2.5A out?
AC isn't possible since I'm using dropper resistors from a 6v transformer and possible tubes to be used are 4v but have different currents, so the dropper resistors give different results.
I assume an LM338 is a simple solution. Do you need the input and output caps in the attached circuit, or could it be simplified if supply caps are inside 6" (says so in the data)? I'd be using a 6v transformer.
Using online calculators, with a 240R in R1 looks like you need 520R in R2 for 4v. Not sure what wattage though.
Suggestions for a heat sink size for 2.5A out?
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I would use the circuit as shown, but no pot. The wattage of the resistors is very small,
but use 2% or better. Heat sink needed depends on worst case input voltage and the
load current. However you are looking at around 11W, so significant sinking is required.
but use 2% or better. Heat sink needed depends on worst case input voltage and the
load current. However you are looking at around 11W, so significant sinking is required.
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I use regularly LT1038 (I have some).
Robust (up to 10A), adjustable, working even at 100C grade.
Cons it's have TO3 case.
Robust (up to 10A), adjustable, working even at 100C grade.
Cons it's have TO3 case.
I have some heat sinks 5cm x 5cm x 2cm. One should do?
I looked for a heat sink size calculator and found this...
Heat Sink Size Calculator | Celsia
Any other useful calculators?
I looked for a heat sink size calculator and found this...
Heat Sink Size Calculator | Celsia
Any other useful calculators?
The LM338 is ancient and you will waste a lot of heat. Go with the LT1038 or for something slightly more complicated the MIC5156 where you can use your own mosfet pass device.
I use an adjustable LM1084 fed from a CRC supply to heat the 6E5P in my 833C amps at 6.3V and 600mA, from a 10V transformer. LM1084 is adjustable and has low dropout voltage of about 1.3V. Good up to 5A, with overtemp and short circuit protection built in.
The whole circuit including an RC transformer snubber resides on the little perf board in the upper left of the attached picture. 2W heat sink on the LM1084, you'll need more.
The whole circuit including an RC transformer snubber resides on the little perf board in the upper left of the attached picture. 2W heat sink on the LM1084, you'll need more.
Attachments
Well, I rigged up a LM338 circuit on a large heatsink 5cm x 5cm and so far for 1A out it's barely warm. I'll leave it for a while. I used 220R and 470R for 3.9v.
So looks like LM317 (1.5A max), LM338, LM1084 (5A max) all work, same pinout.
Update - after about 20m the heatsink is quite warm but fine to hold the hand to. That was a single tube 1A. I have a pair to heat so will try 2A next. Certainly needs a decent heatsink.
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So looks like LM317 (1.5A max), LM338, LM1084 (5A max) all work, same pinout.
Update - after about 20m the heatsink is quite warm but fine to hold the hand to. That was a single tube 1A. I have a pair to heat so will try 2A next. Certainly needs a decent heatsink.
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Correction: supply is 10mF Cap-LM1084, not CRC. the other caps are on the output and to ground from the adjust pin. Memory lapse in the 8 years since I wired it up!
Are you using:
6.3VAC rectified, filtered with a cap, then passed to an LM338 or LM1084?
1. 6.3 x 11.4 = 8.8V peak, not counting the rectifier drop (Schottky Bridge?).
Suppose the bridge drops 2 x 0.35V, we have about 8V DC.
8V - 4V IC drop = 4V at the load.
2.5 A x 4V = 10 Watts dissipated in the regulator IC.
Make sure the IC burden voltage is 4V or less (will work with only 4V in to out).
Do not forget, that the 6.3V winding with a bridge rectifier and cap input filter to the IC, needs to be rated for about 1.6 to 1.8 times more AC current than the DC load current.
4A x 1.8 = 7.2A for the 6.3A secondary.
Use a 8A or 10A rated 6.3A secondary.
2. If your tubes can work with AC filaments, can be connected in parallel, and your power main voltages are constant over time of day, weather season, etc., then you can use a 6.3A filament winding.
6.3VAC - 2.3V = 4VAC
2.3V/2.5A = 0.92 Ohms dropping resistor (an automatic soft start).
2.5A Squared x 0.92 Ohms = 5.75 Watt (use a 20W or 25W rated resistor, to keep it cool).
3. It is ironic . . . The dissipation of the IC is 10Watts for regulated DC, the dissipation with a resistor is 5.75W, for AC.
How many of you expected that?
6.3VAC rectified, filtered with a cap, then passed to an LM338 or LM1084?
1. 6.3 x 11.4 = 8.8V peak, not counting the rectifier drop (Schottky Bridge?).
Suppose the bridge drops 2 x 0.35V, we have about 8V DC.
8V - 4V IC drop = 4V at the load.
2.5 A x 4V = 10 Watts dissipated in the regulator IC.
Make sure the IC burden voltage is 4V or less (will work with only 4V in to out).
Do not forget, that the 6.3V winding with a bridge rectifier and cap input filter to the IC, needs to be rated for about 1.6 to 1.8 times more AC current than the DC load current.
4A x 1.8 = 7.2A for the 6.3A secondary.
Use a 8A or 10A rated 6.3A secondary.
2. If your tubes can work with AC filaments, can be connected in parallel, and your power main voltages are constant over time of day, weather season, etc., then you can use a 6.3A filament winding.
6.3VAC - 2.3V = 4VAC
2.3V/2.5A = 0.92 Ohms dropping resistor (an automatic soft start).
2.5A Squared x 0.92 Ohms = 5.75 Watt (use a 20W or 25W rated resistor, to keep it cool).
3. It is ironic . . . The dissipation of the IC is 10Watts for regulated DC, the dissipation with a resistor is 5.75W, for AC.
How many of you expected that?
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You can use this, zero heat, soft start, no transformer or other components needed: Heating Module 4 V / 2 A Battery Post AL0001 : RADIOELEC : Electronics Components and Modules, specially for Audiophiles and Radio-Ham
I am waiting for the No Heat Power Supply.
Switchers are efficient, but . . .
I have always wanted to see perpetual motion.
Switchers are efficient, but . . .
I have always wanted to see perpetual motion.
Thanks for all the suggestions. I have 2 x tubes to heat at 4v/1A. I have 2 x transformers 10VA 6v. I could either:
1. Connect the transformers in parallel and use one LM338
2. Use 2 x LM338, one per transformer.
I'm inclined to go with 2.
1. Connect the transformers in parallel and use one LM338
2. Use 2 x LM338, one per transformer.
I'm inclined to go with 2.
10VA / 6V = 1.67 A, barely enough to run a bridge and cap input filter, to run the IC.
(1A DC requires 1.61A secondary rating)
10VA / 6.3V = 1.59A, not quite enough to run a bridge and cap input filter, to run the IC.
(1A DC requires 1.61A secondary rating)
Unless the 10VA rating is conservative, those transformers will be pretty toasty (quite warm).
Look at the bridge rectifier and cap input filter in the following link:
Design Guide For Rectifier Use - Hammond Mfg.
The 0.62 factor, means that 1A secondary can run 0.62A DC
1/0.62 = 1.61
(1A DC requires 1.61A AC secondary rating).
At least meet the minimum requirement, or get a hot transformer.
(1A DC requires 1.61A secondary rating)
10VA / 6.3V = 1.59A, not quite enough to run a bridge and cap input filter, to run the IC.
(1A DC requires 1.61A secondary rating)
Unless the 10VA rating is conservative, those transformers will be pretty toasty (quite warm).
Look at the bridge rectifier and cap input filter in the following link:
Design Guide For Rectifier Use - Hammond Mfg.
The 0.62 factor, means that 1A secondary can run 0.62A DC
1/0.62 = 1.61
(1A DC requires 1.61A AC secondary rating).
At least meet the minimum requirement, or get a hot transformer.
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LM338 certainly won't work but LT1084 has 1.3V dropout voltage..... You seem to forget to take ripple voltage into account in the calculations?! I have good results but my 6V transformers have almost 6.8V in reality (under load). Of course I do use Schottky diodes and a 10,000 µF cap. The output voltage is set at 6V and no issue whatsoever. With LT4320 ideal rectifiers things brighten up even more as dropout voltage is then way below 0.1V. Expensive but I decided to built a few to have a little more headroom.
Using an LDO for least loss of energy in heat and therefor higher reliability is advisable (mandatory if you ask me). I have repeatedly done the trick with 0.35V dropout regulators and a 5V transformer for 5V output...
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Ah - thanks for that. I also have a couple of 12VA/6v transformers. They would be better. Space is a bit limited. I can hook one up and try it on one tube. I might have a 25VA 6v toroid for both. The 12VA transformers are square PCB mount ones.
DC-DC LM2596 3A Buck Adjustable Step-down - no heatsink needed.
DC-DC LM2596 3A Buck Adjustable Step-down Power Supply Modules Converter | eBay
DC-DC LM2596 3A Buck Adjustable Step-down Power Supply Modules Converter | eBay
With switchers the challenge will be to keep RF radiation out of audio circuits. What one gains in less heat one looses in the other department. LM2596 is one of the bad ones. I know it well from many a class D amplifier and first thing to do is to remove that reg as it often emits high frequency radiation. These are the Chinese versions as found on cheap boards though.
BTW I see I made a mistake in the previous post. Of course I use LT4320 ideal rectifiers as it would be impossible to have 6V output otherwise. In 5V output PSU's with 5 or, preferably, 6V transformers and 0.35V dropout regulators I use Schottky diodes and so at moderate loads. Then the 5V transformers need to be selected as things are really tight. Not good design practice but lower dissipation in return.
BTW I see I made a mistake in the previous post. Of course I use LT4320 ideal rectifiers as it would be impossible to have 6V output otherwise. In 5V output PSU's with 5 or, preferably, 6V transformers and 0.35V dropout regulators I use Schottky diodes and so at moderate loads. Then the 5V transformers need to be selected as things are really tight. Not good design practice but lower dissipation in return.
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I wouldn't use it to power active circuits, but I've used it for indirect heating in a headphone amp and could hear/measure no difference or noise increase in the output.
I thought you always used a Rod Coleman regulated supply. Is there some reason why that won't work for this?