I made a circuit using LM338 voltage regulator. Input voltage to circuit is 9V @5Amps using a power brick. I used 220 ohm resistor as R1 and R2 is 1K ohm potentiometer. When I assembled the circuit, I can see output variable voltage is coming out. I have set my output voltage 6v (maximum I want). But whenever I put load on it, voltage drops and servos attached to it misbehaves. I am not getting enough power out of it. My assumption is R1 resistor sets current and R2 sets voltage. is it right? Where is the problem.? Example circuit

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    \$\begingroup\$ My power brick is regulated (Sony power adapter). I did not add capacitors as I was suggested capacitor are not necessary if voltages are low. \$\endgroup\$ Mar 7, 2016 at 13:41
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    \$\begingroup\$ Input capacitor depends on the distance from the power brick to your LM338 (more than an inch). Output depends on load. But its easier to just add them. Also, if you are adding a large load like a motor, try adding a 100µF capacitor or so. \$\endgroup\$
    – Passerby
    Mar 7, 2016 at 13:43
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    \$\begingroup\$ The LM338 can deliver 5 A when cooled sufficiently. If you draw 5 A at 6 V, the LM338 has to "burn-off" 9 V - 6 V = 3 V at 5 A = 15 Watt. You need a large heatsink for that. Also 3 V voltage drop is the absolute minimum the LM338 can manage. \$\endgroup\$ Mar 7, 2016 at 13:44
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    \$\begingroup\$ I did not add capacitors as I was suggested capacitor are not necessary if voltages are low. Don't listen to such advise as it is utter nonsense. Always place the capacitors as recommended in the datasheet. The person who advised you not to place them because the voltages are low has no clue obviously. \$\endgroup\$ Mar 7, 2016 at 13:45
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    \$\begingroup\$ "R1 sets current and R2 sets voltage": Wrong, R1 and R2 form a voltage divider: a fixed fraction of the output (R2/(R1+R2)*Vout) is compared to a fixed internal voltage reference (1.25V in your case), and the regulator has an internal control loop to cancel out the voltage difference. The series transistor (called "ballast") which serves as a variable resistor in the control loop is responsible for the minimum voltage drop for a given load current. \$\endgroup\$ Mar 7, 2016 at 14:08

1 Answer 1


Never assume.

Read the data sheet. R2 and R1 set the voltage - and that ONLY in the diagram you posted.

There is an equation in the data sheet (on the same page you took the diagram from) that explains how to calculate the output voltage from R2 and R1. The only tricky part is Iadj, which you have to find elsewhere in the datasheet.

The LM338 is a voltage regulator. You can use an LM338 as a current regulator, but that is a different circuit from the one you gave. If you need to regulate voltage and current, then you need to use the circuits given later in the datasheet.

You can only leave out the capacitors if the power conenctions are less than 6 inches (15cm.) In any case, you are better off including them. They don't cost much, and are much cheaper than replacing the hairs you ripped out while searching for a problem that could have been avoided by including them (the capacitors.)

The datasheet mentions the 6 inch limit for whether or not you need capacitors. It doesn't say anything about the voltage playing into that decision, so whatever advice you got isn't backed by the datasheet.

As for the poor regulation: You must cool the LM338 - it will need a good heatsink when operating at 5A. If it gets hot, it will shut down.

The LM338 needs a good 3Volt difference between input and output voltage to work properly. Your 9V input is too close to the required 6V output to work correctly and reliably.

  • \$\begingroup\$ Thanks for replying.. I checked all the circuits on datasheet but I couldn't understand them as my background is not electronics. Howevrr there was a circuit for 12v battery charger. Question is if R1 and R2 are not affecting current then it should supply input current to output port. Isn't correct? But I am not getting enough current near to input one \$\endgroup\$ Mar 7, 2016 at 14:21
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    \$\begingroup\$ The last paragraph in my answer is important: You need a higher voltage on the input in order to get 6 Volts on the output. You have 9V going in and expect 6V going out, but the LM338 "loses" 3Volts in the regulator. It can't regulate the 6V properly. Supply it with 10 or even better 12V and see how it behaves. \$\endgroup\$
    – JRE
    Mar 7, 2016 at 14:23
  • \$\begingroup\$ Also, add the larger output capacitor as Passerby suggested (100µF.) \$\endgroup\$
    – JRE
    Mar 7, 2016 at 14:24
  • \$\begingroup\$ i got 12 V battery, i will give a try with it. quick question. Does regulator chnage output current or not? When used as voltage regulator. Thanks \$\endgroup\$ Mar 7, 2016 at 14:25
  • \$\begingroup\$ i got 100uf and 470 uf capacitor I will add them as well. \$\endgroup\$ Mar 7, 2016 at 14:25

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