Electronics > Beginners
LM317AHVT and Heatsink
mike_mike:
Thanks for the replies.
I have not found 120R resistor at the local electronic components shop. So I used 100R in series with 22R. I calculated that the current for 120R resistor is 1.25V/120R=0.0104A, while using the 122R the current is 1.25V/122R=0.0102A. The difference of current is very small.
It should be ok if I use 100R in series with 22R ?
Zero999:
122R vs 120R should make no practical difference, since it's less than the tolerance of the LM317.
newbrain:
--- Quote from: JS on July 30, 2018, 07:14:14 pm ---I don't know about the 1.2Ω resistor, seems a tad high, if you lower that value more current will go by the transistor and less for the regulator to take.
--- End quote ---
Yes, I also have some doubt about the emitter resistor value; as it is, I expect the current through the transistor not to exceed about 1 A.
Reasoning:
We have two diodes drop (D1+D2) voltage across the resistor R6 and the BE junction of Q1.
Let's say it's about 1V (as per the DS, with 3A through the diode), so 2V total (excess rounding).
Across R6 we have 2 - 0.7 = 1.3V.
Ohm's law gives us a current of 1.3V/1.2 \$\Omega\$ = ~1.1A
All the rest will be carried by the regulator.
--- Quote from: JS on July 30, 2018, 07:14:14 pm ---...at 3A (taking 0.65V for VBE) the transistor will be taking 2.6A and the reg 0.4A. If you lower it to 0.47Ω the transistor will take 2.8A and the reg 0.2A...
--- End quote ---
Am I missing something? See above my calculations.
JS:
--- Quote from: newbrain on August 08, 2018, 06:54:01 pm ---
--- Quote from: JS on July 30, 2018, 07:14:14 pm ---...at 3A (taking 0.65V for VBE) the transistor will be taking 2.6A and the reg 0.4A. If you lower it to 0.47Ω the transistor will take 2.8A and the reg 0.2A...
--- End quote ---
Am I missing something? See above my calculations.
--- End quote ---
You are right, I'm ignoring the diodes, as if they were just to limit the maximum current as you said but still working on he linear mode, my mistake. Reducing the value of the emmiter resistor will send more current to the transistor even if the diodes are conducting. 1.4V at the diodes, -0.65 VBE, 0.75V/0.47Ω gives 1.6A to the tranny, but limiting the current to the reg at 140mA for resistive share between the transistorand reg. I used 0.2A at the reg resistor, which means 2V before conducting by the diodes, so using 3 diodes instead of 2 makes that work, to keep the 2 diodes the emmiter resistor needs to be even lower and the approximations of VBE and Vf are more on the limit, where you probably want to measure them at the limit conditions to know how much current will be shared.
JS
mike_mike:
It should be a good idea to use a 1K potentiometer in series with the 10K potentiometer for fine adjustment of the output voltage ?
Will the power supply work correctly if I use the 1K potentiometer in series with the 10K potentiometer ?
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