I have done a DC variable power supply, and checked it using multimeter the output voltage ranges from 1.3v to 19v, but the problem is when I added a DC load(small DC motor) it was not working, why?
Circuit diagram
At No load, the output voltage ranges from 1.3v to 19v but, with load, voltage becomes zero.

DC motor specifications Operating voltage - 3-6v Current- 300 mA

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    \$\begingroup\$ By 'not working' what do you mean? What is it doing? Not outputting any voltage at all? Dropping voltage? What is the size of the DC load? How much current does your motor need, and how much can your power supply give? What was your output voltage when testing? What voltage does the motor need? What were you expecting to happen and what did happen? You need to add some more information to the question. \$\endgroup\$ – MCG Jan 7 '19 at 8:50
  • \$\begingroup\$ post a schematic of the DC variable power supply you 'have done'. If it uses a small package 317 for instance and a high input voltage, it might go into thermal limiting at quite low output current. Without details of what you have there, we just don't know, any guesswork will be totally uninformed. \$\endgroup\$ – Neil_UK Jan 7 '19 at 8:57
  • \$\begingroup\$ I have posted the circuit diagram with my query. Please do have a look at it and clarify me \$\endgroup\$ – al rizwan khan mehboob ali Jan 7 '19 at 12:13
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    \$\begingroup\$ At 2.2k, R1 is the reason your output becomes 0V with any load. Why do you have a 2.2k resistor in series with the input to your regulator? \$\endgroup\$ – brhans Jan 7 '19 at 12:27
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    \$\begingroup\$ Remove it. But it was a mistake for it to be there, and you should explain where it came from. \$\endgroup\$ – pjc50 Jan 7 '19 at 13:19

As pointed out in the comments, the reason for this is the 2k2 resistor before the regulator. Because this is in series, any load will cause a voltage drop over it. A multimeter will use barely any current, hence you will see what appears to be a working circuit, however, even drawing just 10 mA will result in a voltage drop of 10*2.2 = 22V.

Remove this and your circuit should work fine.

Why it was added in the first place, I am unsure, my guesses would be perhaps for current measurement (because of the stated 1W), but had the value wrong, although even that would still have it placed incorrectly, it should be post regulator. The other thing is it could have been fitted wrong in the schematic and it was supposed to be in parallel with the smoothing capacitor to discharge it when powered off.

Either way, remove R1 and you should find your circuit works.

  • \$\begingroup\$ Perhaps the reason for R1 was to protect the diode bridge from the inrush current caused by C1, or something similar. The value is definitely off, though. \$\endgroup\$ – marcelm Jan 7 '19 at 15:20
  • \$\begingroup\$ Yeah, that's another good theory. It could be any reason! \$\endgroup\$ – MCG Jan 7 '19 at 15:55
  • \$\begingroup\$ With your reference, I placed that 2.2k resistor in parallel to C1, the circuit work perfectly but after few minutes, the resistors dissipates too much of heat, Can you say me an alternate way or better shall I remove that resistor, if I remove that resistor will that circuit has longer life??? \$\endgroup\$ – al rizwan khan mehboob ali Jan 7 '19 at 17:07
  • \$\begingroup\$ Any old resistor won't do, probably hence the 1W rating on it. The circuit will work fine without it. I was merely speculating what it could possibly have been used for \$\endgroup\$ – MCG Jan 7 '19 at 17:14
  • \$\begingroup\$ 2.2 Ohm would be a little more reasonable, especially considering the 1 W rating. Then it would work as an inrush current limiter/filter. Still a bit unusual. \$\endgroup\$ – pipe Jan 7 '19 at 17:31

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