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A Keithley 2400 Source meter is used to provide +7V. Using Fluke 179 Multimeter, the potential across the source meter leads is 7.00V.

The source meter is then connected to a LM317T voltage regular as shown. According to the datasheet, Using R1=150ohm and R2=240ohm should get me a voltage of 2.03V. Why does the voltage regulator output measure 2.308V?

enter image description here

Next I connected the voltage regulator output to a potential divider as shown below. When I measured the regulator output again, it now reads 3.034V!

enter image description here

Why did the voltage regulator output change from 2.308V to 3.034V?

The intention of using the voltage regulator is to create a 2V rail so it can be divided at the 1V point using 2 identical resistors. Something like the following below. (its most likely incorrect, but i'm not sure of another way to represent what I need)

The +2V will go into AREF pin of arduino, and the 1V will be used to bias a signal that swings between -1V and +1V. Any suggestion what would be a better circuit?

enter image description here

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  • \$\begingroup\$ Where are the capacitors? (marked C1 and C2 in LM317 datasheet) \$\endgroup\$
    – Al Kepp
    Mar 24, 2014 at 2:59
  • \$\begingroup\$ @AlKepp I did not use any capacitors with LM317, just the 2 resistors R1 and R2. Are the capacitors necessary? \$\endgroup\$
    – Nyxynyx
    Mar 24, 2014 at 3:00
  • \$\begingroup\$ Did your input supply voltage remain constant? \$\endgroup\$
    – Andy aka
    Mar 24, 2014 at 8:35

3 Answers 3

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Sounds to me like you might have the regulator wired wrong. Double check that it is wired correctly:

enter image description here

Capacitors are optional according to most datasheets, but are good practice to include.

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Referring to any LM317 datasheet, you'll see the following spec:

enter image description here

When there is no load connected to the OUT pin, the only load to the output is R2. With the R2 value you have selected the current is $$ I= \frac {V_{ref}}{R2}= \frac{1.25V}{240 Ohm}= 5.2mA $$

That may or may not be enough to keep the output in the proper operating condition to maintain regulation (since it's below the MAX and your device may be anywhere from 3.5mA to 10mA), so this may be the reason you are getting strange results as you vary the load. Also note that the output load regulation specs are given for a range >10mA (10mA - 1500mA).

I suggest a value for R2 that can maintain a current higher than the Max value of the datasheet. Such a value that gives a current of 10mA+ is R2=120 Ohm, and of course you'll have to change R1 too, to get the output voltage you want.

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  • \$\begingroup\$ Would using a 120 ohm resistor still help to maintain voltage regulation if the regulator output is going to the top of a 470k+470k voltage divider, which I am guessing is going to reduce the current to much lower that 10mA? \$\endgroup\$
    – Nyxynyx
    Mar 24, 2014 at 13:57
  • \$\begingroup\$ @Nyxynyx The divider you refer to, acts as an additional load, so the total output current will increase even more. \$\endgroup\$
    – alexan_e
    Mar 24, 2014 at 14:06
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Your resistor divider may be using low tolerance resistors. Have you used a high quality ohm-meter to check the REAL resistance of these components? Most are like +- 10%, which can give you bad results, as you have noticed with 300mV error.

Secondly, the second resistor divider network you have there is forming a parallel-resistance network, thus changing the divider that you had previously set up. You must decouple the networks, perhaps if small signals are used maybe use a voltage buffer line driver (Opamp with very simple voltage buffer setup)

You should analyze the resistor set up that exists by finding the equivalent resistances seen for the ADJ pin of the LM317, and you will see why it has changed it's output.

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  • \$\begingroup\$ It appears that the easiest solution will be to change R1, R2, R3, R4, where R3 = R4 such that the effective R1 to oR2 ratio is still 0.6 (taking into account the parallel resistance network since I ignored) which gives a output voltage of 2.0? I'm not sure how decoupling with the op amp works, but it seems to be more complicated. \$\endgroup\$
    – Nyxynyx
    Mar 24, 2014 at 4:36
  • \$\begingroup\$ the opamp technique is merely to put a effectively infinite resistance (tens of megaohms, even gigaohms) in between the two circuits in order to not affect your two resistor dividers. As others have mentioned, you should probably put a capacitor of some form on the output too, but it should work without it. If I was to put a cap there, i'd do a 10 , 47, or 100uF electrolytic - just as examples. \$\endgroup\$
    – KyranF
    Mar 24, 2014 at 5:12
  • \$\begingroup\$ But yes a quick fix may be to just adjust your resistor values like you said. \$\endgroup\$
    – KyranF
    Mar 24, 2014 at 6:02
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    \$\begingroup\$ the second resistor divider network you have there is forming a parallel-resistance network, thus changing the divider that you had previously set up. The output voltage is set by the RATIO of R1 and R2 and as long as any additional resistor is not connected only to one of the two resistors (effectively changing the ratio) it doesn't make a difference. \$\endgroup\$
    – alexan_e
    Mar 24, 2014 at 10:19
  • \$\begingroup\$ @alexan_e yes indeed that is what I was trying to say. As long as the ratio is kept, but also the value of the resistors. If Nyxynyx wants to make another divider it will influence his previous one. I would use a reference voltage diode and high value resistor on the input OR output into ADJ pin, or use an opamp and potentiometer if he/she wanted to tune it. And a 200Ohm minimum load resistor (for 10mA minimum load condition) with a 10uF -> 100uF output capacitor for stability. \$\endgroup\$
    – KyranF
    Mar 24, 2014 at 23:51

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