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enter image description here

I am still learning analog electronics and possibly making a mistake here.

I am trying to build this power supply circuit but I am getting 0V as output instead of 4V.

This is the schematicenter image description here

Both of them show output voltage as 0.0V and 0.06V for schematic design and breadboard design respectively.

I don't understand where am I going wrong. There is nothing wrong with the source voltage. I keep on checking it.

Is the circuit wrong?

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    \$\begingroup\$ Current can only flow in a completed circuit, C3 is preventing that. \$\endgroup\$ – PlasmaHH May 30 '17 at 11:17
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    \$\begingroup\$ Put c3 in parallel to the battery, not in series. \$\endgroup\$ – Colin May 30 '17 at 11:19
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    \$\begingroup\$ Why is C2 a variable capacitor? \$\endgroup\$ – JRE May 30 '17 at 11:20
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    \$\begingroup\$ @JRE: i really would love to see a 1mF variable cap \$\endgroup\$ – PlasmaHH May 30 '17 at 11:21
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    \$\begingroup\$ Those two circuits are not the same. In both cases current can't flow because the circuit is not complete; in the first case because the grounds aren't connected (or at least, they don't seem to be). In the second case because C3 is in series with the voltage source, while it should be in parallel (like it is in the first circuit). \$\endgroup\$ – marcelm May 30 '17 at 11:22
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There are a few issues here.

The first issue is that, as others have said, your circuits are not identical. An equivalent to the first circuit would be:

schematic

simulate this circuit – Schematic created using CircuitLab

Note that the battery is in parallel with the first capacitor (acting as a filter to smooth out ripple input and improve stability). A capacitor in a DC circuit acts as an open circuit (ideally), which prevents current from flowing in the configuration you have.

The second thing to note is that C2 is an electrolytic capacitor, and you have used the symbol for a variable capacitor. Electrolytics are constructed differently than normal ceramic and film capacitors, with two main differences:

  • They have a much higher capacitance by volume
  • They are polarized and must be installed in a specific direction relative to voltages

One final thing to note with the LM317 is that the schematic representation can be a bit misleading for the common TO-220 package. The schematic symbol lists the pins as In/Adj/Out going from left to right, while the device itself has the pins as Adj/In/Out in the same direction. Make sure you connect your pins correctly!

Edit: updated schematic to reflect the possibly-different grounding scheme used in the topmost reference schematic.

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  • \$\begingroup\$ An equivalent to the first circuit would be: may not be quite right. In the first the grounds are given two different labels implying that they are two different and unconnected nets rather than a single ground. I'm sure that wasn't the intention but that's how it was drawn. \$\endgroup\$ – Andrew May 30 '17 at 13:22
  • \$\begingroup\$ @Andrew A fair point. I will update my schematic to reflect what I assume to be the datasheet schematic (the topmost one). \$\endgroup\$ – Chris M. May 30 '17 at 13:24
  • \$\begingroup\$ The circuit is working now. Thank you very much for your response. \$\endgroup\$ – knowledgeispower Jun 1 '17 at 12:28
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Your cicuit isn't closed to ground since C3 is in the Way. You should get rid of C3. You could put a resistor between the +-line and GND to stabilize the voltage. As you can see in your first picture, the capacitor should be placed between the 2 poles of the voltage source.

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  • \$\begingroup\$ Well the first sentence does answer the question and is even correct. The resistor however, is probably not such a good idea. \$\endgroup\$ – ilkkachu May 30 '17 at 17:32
  • \$\begingroup\$ By the Way, I see my fail myself :D but still the question "Is the circuit wrong?" was answerd. \$\endgroup\$ – bjoekeldude May 31 '17 at 19:28

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