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#1 |
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diyAudio Member
Join Date: Dec 2003
Location: Hong Kong
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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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#2 |
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diyAudio Moderator
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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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