convert 50K pot into a 2K pot?

[pc88]

Suppose I have a 50K linear pot, but I really need a 2K linear pot. Is there an easy way to convert it? (electronically, of course)

 

[crutschow]

Not likely. What is the pot for?

 

[pc88]

Not likely. What is the pot for?

It's for the voltage adjust on your typical LM317 linear regulator. The datasheet and all the designs I've seen use something like a 2K pot. I have a 5K pot which I'm sure will work just fine, but I just wanted to make sure that I hadn't overlooked something "obvious".

 

[Hero999]

What voltage rage do you require?

Plug the resistor values into the formula on the datasheet and see if the resistor gives the voltage range you desire.

If you don't need a variable voltage regulator, then just use a fixed resistor of the appropriate value in its place.

What's the part number? Is it an LM317 or an LM117?

The diagram on the datasheet is more expensive LM117 which requires 240R for R1. If you've got the LM317 then you require 120R for R1 and half the value for R2.

 

[crutschow]

Ok, so it's a 5k pot, not 50K?

The 5K should work for the LM317. The 317 has slightly more bias current flowing through the pot then the 117 so it's output voltage may change slighly more with temperature, but otherwise it should operate ok.

 

[pc88]

No - I have a 5K pot. I was just using 50K as an example in asking the hypothetical question of what if you had a really high-valued pot but needed a low resistance pot.
Thanks for the help!

 

[jpanhalt]

In a pinch, I have used a paralleled resistor, but the taper obviously is no longer linear. John

 

[audioguru]

The correct value for R1 on an LM317 is 120 ohms.
Then a 5k pot will create a max voltage of 53.3V which is much too high.
A 2k pot will create a max voltage of 22.0V if the supply is at least 25.0V.

The tolerance of the pot might be 20% so watch out.

 

[crutschow]

The correct value for R1 on an LM317 is 120 ohms.
Then a 5k pot will create a max voltage of 53.3V which is much too high.
A 2k pot will create a max voltage of 22.0V if the supply is at least 25.0V.

Yes, R1 needs to be 240 ohms to give about a 25V maximum output with a 5k ohm pot. 120 ohms may be recommended in some data sheets but using 240 ohms should have little effect on the regulator output characteristics for most applications.

 

[audioguru]

using 240 ohms should have little effect on the regulator output characteristics for most applications.

National Semi invented the LM117/LM317 and other manufacturers copy it.
National clearly states in the datasheet that if R1 does not have a low enough resistance (120 ohms for the LM317) then some LM317 ICs will cause the output voltage to rise when the load is removed. A pretty serious problem if your load is a very low current circuit and it blows up because the power supply voltage decided to rise by itself.

The more expensive LM117 can use 240 ohms for R1 as is shown many times on the datasheet.

 

[crutschow]

National clearly states in the datasheet that if R1 does not have a low enough resistance (120 ohms for the LM317) then some LM317 ICs will cause the output voltage to rise when the load is removed. A pretty serious problem if your load is a very low current circuit and it blows up because the power supply voltage decided to rise by itself.

The more expensive LM117 can use 240 ohms for R1 as is shown many times on the datasheet.

I didn't see that on my National data sheet dated August 1999. But after some thought I understand the reason for it.

The pertinent difference between the 117 and the 317 is that the minimum load current for the 117 is 5mA and the 317 is 10mA (3.5mA typ. for both).

For the 117, the normal current through R1 is 1.2V/240 ohms = 5mA, thus the device stays in regulation with no load. Obviously the 317 would need 120 ohms to stay in regulation for 10mA minimum load.

One way around this is to add a 5mA dummy load to the 317 if you want to use R1 = 240 ohms and a 5k pot. You would add a 240 ohm load if you wanted to adjust down to the minimum 1.2V output. Of course, if you went up to 25V output, the power dissipated in the resistor would be 2.6W which may not be tolerable.

Another way is to use 120 ohms for R1 and place a 5k ohm resistor in parallel with the 5k pot. This would thus give you an approximate 1.2V to 26V adjustment range. The disadvantage, of course, is that the output voltage is now not linear with pot movement.

 

[audioguru]

I am glad that you understand that there are two ways to do anything:
1) The correct way.
2) The other way.

Or you are a gambler.

 

[Roff]

If you're desperate, you can add an approximately constant 5mA load. It wastes more power than just using 120 ohms from out to adj, but at 20V out, it's still only about 100mW.
Warning - the pin numbers are wrong (but correct for the sim).

 

[Hero999]

If you are using the power supply to power a circuit that always draws more than 5mA, there's nothing wrong with using 240R. Indeed, if the circuit uses a minimum current of 10mA then it's alright to use even higher value.

 

[Roff]

If you are using the power supply to power a circuit that always draws more than 5mA, there's nothing wrong with using 240R. Indeed, if the circuit uses a minimum current of 10mA then it's alright to use even higher value.

Except if you make it too high, the Adj pin bias current will interfere with output voltage predictability (not a problem when using a pot), and with stability vs temperature.

 

[crutschow]

I am glad that you understand that there are two ways to do anything:
1) The correct way.
2) The other way.

Or you are a gambler.

The trick is to determine whether your way is number 1 or number 2.

 

[gerty]

Ouch !!............

 

[audioguru]

The trick is to determine whether your way is number 1 or number 2.

I always do it the way that is in the datasheet. It is almost always correct.
Sometimes they make a mistake but not often.