I built below voltage Regulator Circuit and noticed that after sometime 1K resistor is getting a lot of heat and also I connected the multi meter to Output of this circuit and adjusted preset to output 12V. after sometime voltage seems to be drop by 0.1V continuously like 11.99, 11.98, 11.97 ... etc.

What is causing the voltage drop and heating?

I provided Input to this circuit with a 15V 2A Transformer.

Source of this Circuit enter image description here

  • \$\begingroup\$ What's the power rating on your resistor? I needs to be at least 1/2W otherwise it will heat up. \$\endgroup\$ – Tom Carpenter Oct 11 '15 at 14:15
  • \$\begingroup\$ the output is 30Vac ( 42V dc). even losing a few volts for the diodes + led leave 40V being dropped by the 1k = 1.6W \$\endgroup\$ – JonRB Oct 11 '15 at 14:16
  • 3
    \$\begingroup\$ @JonRB in the circuit yes, but the question says a 15V transformer was used, so \$P=\frac{(15\times\sqrt{2}-V_f)^2}{R} =\frac{(21-V_f)^2}{R} \approx 0.338 \mathrm{W}\$ \$\endgroup\$ – Tom Carpenter Oct 11 '15 at 14:18
  • \$\begingroup\$ @TomCarpenter I used a 1/4W 1K resistor :( and what could be the reason to drop voltage? \$\endgroup\$ – lasita Oct 11 '15 at 14:19
  • \$\begingroup\$ Iasita, 0.33W will noticeably warm even a 1/2W resistor. Over 60C I would say, so it will feel hot to the touch, but not sizzle if you put a wet finger on it. \$\endgroup\$ – Fizz Oct 11 '15 at 14:36

If you're really feeding this 15 VAC (not 30 VAC as shown in the diagram), your input bus is around 19 VDC, which is forcing about 20 mA through R1, and causing it to dissipate about 400 mW. This is too much for a 1/4W resistor, and even a 1/2W resistor would get rather warm. Either get a bigger resistor (1W), or pick a higher value to bring the current down. 2 kΩ would still put about 10 mA through the LED, and the resistor would only dissipate about 200 mW.

I can't explain the droop of the output voltage without more information. How much current are you drawing at the output? Is the LM317 getting warm as well? If so, put it on a bigger heatsink.

Is the voltage at pin 2 of the LM317 also drooping? If so, perhaps R3 can't handle the current you're putting through it, although I would think that 1.25 V / 220 Ω = 5.7 mA should be fine for any potentiometer.

|improve this answer|||||
  • \$\begingroup\$ I only check output by connecting multi meter to Output of this circuit without any load. \$\endgroup\$ – lasita Oct 11 '15 at 14:26
  • 2
    \$\begingroup\$ What construction technique did you use? Are all of the connections soldered, or did you stick everything into a breadboard socket? If the latter, poor contacts in the socket may be causing issues. Remember, I can't help you unless you answer all of the questions I ask. \$\endgroup\$ – Dave Tweed Oct 11 '15 at 14:51
  • \$\begingroup\$ That's pretty miffing indeed. At this kind of voltage R1 burning out would normally fail open. Try without it (and the LED) since they're non-essential for the circuit. You should meausre the voltage drop between pins 2 and 3 of LM317. If it's not a constant 1.25V (or thereabout), the regulator is probably bad. \$\endgroup\$ – Fizz Oct 11 '15 at 14:51
  • \$\begingroup\$ @DaveTweed First i built this on a breadboard and did not noticed the heating problem. so i went ahead and soldered all the components and then noticed the heat and voltage drop. i did not put any load i just checked with multi meter and it shows 0.1V voltage drop continuously. LM317 did not get any heat \$\endgroup\$ – lasita Oct 11 '15 at 15:01
  • 1
    \$\begingroup\$ What about the measurement of pin 2? \$\endgroup\$ – Dave Tweed Oct 11 '15 at 15:03

Please check '317 pin numbering. Correct: 1 = Adj 2 = Out 3 = In

|improve this answer|||||

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.