current boost lm317 prob.

[deepanger]

am trying to current boost the lm317 to make agood bench ps.
iused amedium power BJT (BD140) for biasing, idont know how
to calculate the appropraite resistor value for the base so iused a
variable 100k. , it just worked as theres no BJT atall, imean the normal
operation of the lm317,, how do i make it work and to calculate
the proper value for resistor so ican decide the exact value of current
inwhich the BJT start conducting the excess current..

thanks all.

[UrSv]

I think you need to understand how it works first.

The resistor you want to know doesn't even need to be there and the one that is needed is missing and shoud be between emitter and base which would provide the voltage drop needed to switch the transistor on when the current reaches a certain value. Perhaps use something else than the BD140 as it is unsuitable for the task.

I strongly suggest reading any of the datasheets available for this type regulator as these deal with the current boost.

[deepanger]

ok in the datasheet of the regulator ifound the 1st schematic,
why using 2 BJTs (PNP ,NPN) while ijust can use one?!
as the 2nd schematic it just uses one transistor ,
ididnt know yet how ican calculate the proper resistor value for the
biasing current?!

[deepanger]

heres the 2nd schematic (not in the datasheet)..

[UrSv]

Actually the datasheet contains the same type setup in the later pictures but along with current control (which is not a bad thing to have).

Check this for variations closer to yours:

http://ourworld.compuserve.com/homep...den/page12.htm

[deepanger]

itook the 2nd schematic already from that page u posted,,

iput about 20ohm resistor bet. emitter and base as u said , but ifound
astrange thing actually, when i measured the current passing from
collector to lm317 output pin, then measured the total load current
ifound out they're the same... am sure that the regulator is the working one as it dissipates much heat,, so?!
am confused..

[deepanger]

iknow there're knowledgeable ppl inhere so can u help?

[paulb]

Originally posted by deepanger
itook the 2nd schematic already from that page u posted,,

iput about 20ohm resistor bet. emitter and base as u said , but ifound
astrange thing actually, when i measured the current passing from
collector to lm317 output pin, then measured the total load current
ifound out they're the same... am sure that the regulator is the working one as it dissipates much heat,, so?!
am confused..

The second diagram shows two resistors, 0.3 and 0.7 ohms. Which of these is your 20 ohm resistor? And what is the other value that you are using? How much current did you measure, and how did you measure it? You can measure the voltage across these two resistors and calculate the current from that.
Answer these questions and someone out there will be able to help. Or draw a diagram of exactly what you've built and post it, even better.

[deepanger]

ididnt use any of those 2 schematics,, iused the 1st one i posted and
it didnt work as ive put the resistor in the wrong place, so itried
the right thing (aresistor bet. emitter & base) , and ive found that
the BJT isnt working although theres acurrent flowing from collector
to the output (used an ammeter bet. collector and output pin of the
regulator),, ican find that only the regulator is working in its normal
operation like as theres no BJT at all..

heres the current schematic i use:

[cunningham]

If boosting the current output of the LM317, while still keeping the regulated voltage is your goal, then you should use the LM317 to drive another transistor. The LM317 regulator sould be connected so that the OUTPUT of the LM317 goes to the base of an NPN transistor. The collector should be connected to V+, and the emitter is the new output, just like an emitter follower circuit. The LM317 adjustable voltage regulator now only has to drive enough current to operate the transistor, Ic/Beta. If you use a 2N3055, the LM317 shouldn't have any trouble driving this transistor without a heatsink. The transistor may require one though.