Pot on LM317 doesnt seem to be linear...

Discussion in 'The Projects Forum' started by Al Vince, Mar 1, 2015.

1. Al Vince Thread Starter New Member

Feb 24, 2015
27
1
I have a 5k linear pot on a simple LM317 circuit outputting to 12V ... the pot doesn't seem to have a smooth transition for adjustment... seems like a good half turn adjust like 1-2 volts.. then the other half adjusts the other 1o volts.

Circuit used...

2. Michael.Riley New Member

Jan 3, 2015
15
2
Do you have a potentiometer or a variable resistor? If you're using a pot as a variable resistor, check the label. If it's a "B" label it's a linear pot, so the resistance changes smoothly from position to position. If it's an "A" label, it's logarithmic, meaning the resistance will change dramatically across the turns at one end of the adjustment. (From instructables.com - "Wire a potentiometer as a variable resistor")

3. GopherT AAC Fanatic!

Nov 23, 2012
6,312
4,031
@Al Vince
Measure resistance at 1/4, 1/2 and 3/4 turn. Is it linear or not. 1200, 2500, 3700 ohms?

4. crutschow Expert

Mar 14, 2008
13,501
3,375
Are you using the potentiometer as a variable resistor (wiper connected to one end of pot) and not as a pot (wiper connected to the ADJ pin)?
Connecting the ADJ pin to the wiper will give a very non-linear output voltage versus pot position.

Last edited: Mar 1, 2015
5. Al Vince Thread Starter New Member

Feb 24, 2015
27
1
I think I have it figured out... I am using the wrong ohm rating for my pot... my pot is too high, so there is barely any voltage drop when pot is half way. I need around a 2k pot for 12v output. I am new to this stuff.

Thanks!

6. Al Vince Thread Starter New Member

Feb 24, 2015
27
1
Are you referring to connecting the pot as in this circuit?

7. k7elp60 Senior Member

Nov 4, 2008
479
69
R2 can also be a problem. If you have no load on the regulator the voltage may be inaccurate because the LM317 requires a minimum of 10mA of load current.. 1.25V/R2 determines this if no external load is connected. As a side note making R2 120 ohms causes R1 to be a smaller value to adjust the current. R2 of 240 ohms is fine for the LM117,or LM217.

8. crutschow Expert

Mar 14, 2008
13,501
3,375
Yes, that's the correct way to connect the pot for a linear voltage adjustment.

9. RamaD Active Member

Dec 4, 2009
254
33
I was wondering how Vout can be linear with R1 in the denominator. I am not saying about starting from 1.25V;
But current thro' R1 is constant, kR1 (k - constant) contributing to Vo has to be linear in the circuit. Where am I going wrong?

Ok, just figured out that the eqn is 1.25V(1+(R1/R2)) + (Iadj x R1)
Just the R1 and R2 are interchanged!

10. Al Vince Thread Starter New Member

Feb 24, 2015
27
1
Thanks!

Is that middle connection to the pot from R2 going to the wiper? And the ground and LM317 adj pin going to the other two ends on the pot?

11. GopherT AAC Fanatic!

Nov 23, 2012
6,312
4,031
The middle pin of the POT is the wiper. Connect that wiper to one other pin on the pot. Now connect the wiper+fixed end to the adj and connect the other fixed end of the pot to ground.

(If I don't say it here, someone will tell you you can also connect the wiper+fixed to ground and the other fixed end to adj of the LM317 if you want to do that - it is fine).