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I am trying to prototype this circuit:

PSBench.gif

However I do not have a 2K Ohm potentiometer for R2, the closest I have is a 100K. So I have used 100K for R2 and 10K for R1. Using the forumula at the bottom these should be valid values for the resistors, however my output voltage only range between 0.01V and 0.4V. Also the output voltage slowly creeps by 0.01V increments the longer I keep the circuit switched on.

I have checked the wiring many, many times and replaced the LM317T twice and checked every component individually but always get the same problem. Can someone tell me what is wrong please?

I rebuilt the circuit so it is easier to understand the connections:

breadboard1 breadboard2

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    \$\begingroup\$ please add a photograph of the assembled circuit. \$\endgroup\$
    – markrages
    Jun 12, 2013 at 22:26
  • \$\begingroup\$ I'm not sure a photograph would be helpful because of how close all the components are on the breadboard (it would be difficult to make out all the connections). I am sure the wiring is correct, I have checked it 20 times atleast... \$\endgroup\$
    – codefi
    Jun 12, 2013 at 22:48
  • \$\begingroup\$ What are the voltages on pins 1 and 3? \$\endgroup\$
    – markrages
    Jun 12, 2013 at 23:07
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    \$\begingroup\$ If I recall correctly, the recommended value for R1 is 240 ohms - you can probably go up to 1K or so, but 10K is much too far from the recommended value to expect proper operation. \$\endgroup\$ Jun 13, 2013 at 0:06
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    \$\begingroup\$ @Adamme your link for the transformer wiring diagram takes me to a website hosting site - no transformer details there. However, looking at the photo of the transformer, I suspect that it has two separate 12 volt secondary windings - if so, those two unconnected terminals on the bottom could be connected together to give 24 VAC, or (better) you could move the the right wire to the second terminal from the left - but before doing that, measure the AC voltage on each pair of terminals to see what is really connected to what. \$\endgroup\$ Jun 14, 2013 at 23:28

3 Answers 3

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Based on the information in your schematic the voltage across the 3300 uf capacitor should be about 25 VDC which is the peak voltage of the 18 VAC at the secondary of the transformer. Since you are only getting 0.752V, something is clearly amiss. Try disconnecting pin 3 of the LM317T and then measuring the voltage across the 3300 uf capacitor. If it is about 25 VDC, then something is wrong with either your wiring or the LM317T. If it is not, then you have a problem with either the transformer, rectifier, or capacitor. In any case, I believe that your choice of values for R1 and R2 will not work in this circuit. The LM317T contains a 1.25V reference that appears across pins 1 and 2. The voltage divider formed by R1 and R2 causes the LM317T to change its output voltage to keep the 1.25V across R1. That is the basis of the first term of the equation for the output voltage. The problem is the second term. The LM317T outputs a small current from pin 1, designated as IADJ, which is nominally about 50 microamperes. This current flows through R2 and increases the output voltage by IADJ X R2 which accounts for the second term in the equation. When the recommended value of 2K is used for R2, this second term becomes 50 microamperes X 2K or 0.1V which is small compared to the usual output voltage of the LM317T. Also, the LM317T is designed to keep IADJ constant with temperature so as to not affect the stability of the output. However, you are using 100K for R2, so that IADJ now causes the output voltage to increase by 50 microamperes X 100K or 5 volts. This is a large fraction of the expected output voltage and could go as high as 10 volts since the specified maximum for IADJ is 100 microamperes. The data sheet is not clear on this point but I believe the designers expect the current through R1 to be much greater than IADJ. With the recommended value of 220 ohms, the current through R1 is 1.25V/220 or 5.68 milliamperes which is 57 times higher than IADJ. With R1 equal to 10k, the current drops to 0.125 milliamperes or 125 microamperes which is only 1.25 times the maximum value of IADJ. I strongly suggest that you find a pot much closer to 2k, lower the value of R1 accordingly, and try the circuit again.

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  • \$\begingroup\$ Thanks for the clear explanation, I had no idea I ADJ was important and went off the guess that keeping the ratio R1/R2 would be sufficient. Went to town to buy some 2k pots but couldn't find any so I ordered some online, I will be back to let you know how it went. \$\endgroup\$
    – codefi
    Jun 13, 2013 at 17:28
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The size of R2 can skew the regulation point. The adjust pin actually sources 100 microamps of current, which combines with the current from R1 to set the R2 voltage and the ultimate set point for the regulator.

enter image description here

Beyond that, the circuit appears reasonable to me (from a schematic point of view). You should check the steady (DC) voltage at the 3300 microfarad cap and at the 10 microfarad cap on the adjust pin.

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Transformer was incorrectly wired, causing a severe voltage drop when a load was applied to it. (See Peter Bennet's comment under the question for more details.)

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