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Old 24th June 2006, 01:24 AM   #1
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Default LT1085 Current Regulation for DHT filaments

I attach below a circuit (from internt) of using the first LT1085 as a voltage regulator and the second LT1085 as a current regulator for feeding the filaments of 2A3.

I know how to calculate the value of resistors for the LT1085 as a voltage regulator, but the datasheet of LT1085 does not contain any information of using it as a current regulator.

It appears that the R6 (520 ohm) shown in the circuit does not tally with the formula I = 1.2 / R for LM-338K. How to calculate its value when using LT1085 as a current regulator?

How about the R7 (1K ohm) shown in the circuit?

Is the formula of LT1085 (3A) the same as LT1083 (7.5A)?

Can anybody lead me the way? Thanks.
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Old 25th June 2006, 12:17 AM   #2
kevinkr is offline kevinkr  United States
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Basically they are very similar parts, I recommend looking them up on the Linear website and downloading their datasheets to compare.

IMHO you don't need the voltage regulator section at all in power tube applications, I have designed and built them both ways and the sound was indistinguishable - the important thing is not to run them too close to their drop out voltage, typically under low line conditions you want a minimum of 3V across the regulator ic.

The current in current regulator mode is vref/sampling resistor value.

For 1.2A something slightly over 1.0 ohm is close enough.

I actually use a precision 1 ohm power resistor (3W) with a 50 - 100 ohm pot across it , wiper to the ref pin on the regulator - if a good pot is used this allows me to set the current very precisely. Tie the wiper through a 1K or greater resistor to the end of the pot connected to the load, that way if the pot fails it will default to the minimum current. I use a large cap on the output and I don't usually bypass the ref pin in current mode. (If you bypass this point I doubt the ground is the right place, probably should be referred to the high side of the load - your filament. Note I haven't tried this.)
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