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Bas Horneman 17th January 2003 08:53 AM

LM317 current regulator
If I build a current regulated filament supply how do I calculate the resistor values to get a certain current say 1,5A?

The voltage regulator that precedes it is simple in that I turn a pot till I measure 5V...but the current reg?

UrSv 17th January 2003 09:01 AM

I don't know much about tubes but it seems reasonable to assume that the filament is not supposed to be fed by a constant current. If the filament is 5 V I think you supply the 5 V and then the current draw is simply around the 1.5 A changing a little after heating up. I donšt think it is necessary to limit the current. Also you can not set the voltage and the current at the same time for this type load as it will either need to use the full voltage when current draw at 5 V is lower than 1.5 A (thus using less than the stated) and less than full voltage if current draw at 5 V is higher than 1.5 A (and thus woule be using more than 1.5 A if voltage set to 5 V).

I suggest using 5 V supply capable of at least 2 A and then everything should be fine. Please however trust other sources on tube supplies before me...

If however you need to regulate the current then I am sure the app notes/datasheet describe that. Essentially you connect a resistor on the output of the reg and then Vadj directly to the output end of that resistor. The current regulated would be approximately Vref/R1 as the reg tries to maintain 1.25 V between those pins. So your current would be 1.25/R1 which for 1.5 A would mean (solved for R1) 1.25/1.5=0.83 Ohms.

jim 17th January 2003 09:10 AM


Take a look at

Current Reg


Bas Horneman 17th January 2003 09:20 AM

Thanks for the replies!

My plan is to use a voltage regulator first LM350, then add the LM317 stage...which would use a 0,83R resistor...1,25/1,5 from Jim's link see below. (for a 1,5A current supply)

"One three pin adjustable voltage regulator and one resistor are all that is needed to make a current source, a high current one at that. The desired current is set by the value of the resistor, which is found by dividing the base voltage (1.25 volts usually) by the desired current. For example, if we want 900 mA, we use a 1.39 resistor, as 1.39 = 1.25 / 0.9 ."


jim 17th January 2003 09:53 AM

Input Voltage

Be sure your unregulated (input) voltage is high enough.
= heater voltage + 1.25 + 3V(min. regulator working voltage).
Example : current reg for a 6.3V heater needs at least :
6.3+1.25+3=10.55V raw DC at its input.


Bas Horneman 17th January 2003 10:01 AM

Jim, you're a star ;)

Ofcourse the lm317 also drops voltage..(Never gave it a moment 's thought)

jan.didden 17th January 2003 10:18 AM

I reg

A tube with a heater specified in volts (there are of course also tubes with a heater specified in current) should see the recommended voltage for optimum performance, e.g. 6.3V for the E-types.
What you want to do I think is limit the current at start-up to prolong heater (& tube) life, but give them the recommended voltage after that.

So what I would recommend is a regulated 6.3V voltage regulator with a current limit set to say 120% or 130% of the expected current comsumption.

Of course, it gets more complicated with several heaters on the same regulator.

Jan Didden

Bas Horneman 17th January 2003 11:07 AM

2 Attachment(s)
I got some prints from a fellow diy tube builder (Doede Douma). That have been made for the attached schematic. And I thought it was a good idea to use them both for a 300b amp I'm gonna build one day.

So initially I though just set the voltage reg to 5v and the current reg to 1,5A.. But Jim's suggestion sounded logical..The current reg needs a higher voltage because it will also drop voltage??

dhaen 17th January 2003 11:50 AM


A few words about the LM317:

Making sure that there is sufficient input voltage is essential for all regulators in all modes, although you may have to allow extra for the voltage drop across the output resistor in current mode. I aim for 4 to 5 volts. Remember the regulator needs working volts, even in the "trough"of the input ripple.

The National datasheet says it all:

Also, consider the wattage of the variable resistor on the output of the current mode supply. If the pot is turned near to the max current end, there is considerable current flowing through it, and only a small amount of track to dissapate the power.


Bas Horneman 17th January 2003 12:23 PM

Hi John,

So a 15v/3A transformer should do the trick..


How do you supply your tubes? AC/DC?

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