0
\$\begingroup\$

I'm working on a project where I'm replacing the trimpots on a buck converter and a boost converter with proper potentiometers. The new potentiometers will be wired to the converters instead of soldered directly on, so this poses a problem: What wire gauge should I use?

Let's say my loads will be maximally drawing 10A. Does that mean I need 16-14AWG wire to attach the potentiomters? I don't think so but I don't totally understand these things. My current understanding is that the potentiometer acts as a voltage divider and it is the center pin which tells the chip how to limit voltage or current by being at a certain voltage.

The answer may be that it depends on the converter but I think there is a general pattern here. To give an example though, I'll be working on this Chinese converter (Google translate) Update: This post is the most information I can find on it.

As I understand it, it's based on the UC3842A SMPS controller driving at least one power MOSFET (I also see a big N-channel and then the smaller BU806 NPN.) I don't know if that's enough information to answer my question though. It seems like, if it works the way I think it does, it's pins 2 and 3 on the UC3842A which read the potentiometers.

Another example is the LM2596-based boards. I suspect they work in a similar way (and are a bit easier to understand.) There's one more board/chip that I'm interested in but I don't have any information on it yet since it's not in my possession and there's very little details online. If it's okay, I'll leave the ebay link here.

\$\endgroup\$
7
  • 2
    \$\begingroup\$ To be sure what you're talking about, you should add schematics (or parts of it as image) and links to datasheets. \$\endgroup\$
    – try-catch-finally
    Apr 17, 2016 at 10:12
  • \$\begingroup\$ Good point. I added datasheets for the two chips that I know and a link to a post with an unofficial schematic and some explanation. \$\endgroup\$
    – Anthony
    Apr 17, 2016 at 10:19
  • 1
    \$\begingroup\$ A trimpot cannot take much current let alone 10A which would release the magic blue smoke instantly, so why would swapping them out for external pots require big wire? \$\endgroup\$
    – JIm Dearden
    Apr 17, 2016 at 10:35
  • \$\begingroup\$ That's clever; it would not. \$\endgroup\$
    – Anthony
    Apr 17, 2016 at 11:32
  • \$\begingroup\$ Variable resistors obey Ohm's law, more or less, as you'd expect, more or less. \$\endgroup\$
    – Russell McMahon ♦
    Apr 17, 2016 at 12:40

1 Answer 1

Reset to default
2
\$\begingroup\$

Yes the potentiometers act as a variable voltage divider, so you can change the output voltage of the converter whenever you like. This voltage divider is connected to the Feedback (usually this pin is named as "Feedback") pin of converter IC, providing feedback for the regulation loop.

enter image description here

The datasheet usually contains information about this pin's bias current. In case of the LM2596, it is typically 10nA.

enter image description here

The feedback resistors are also recommended in the \$ k\Omega \$ range, reducing current flow.

\$\endgroup\$
1
  • \$\begingroup\$ Thank you. I will compare this with the UC3842A. And yes, the pot that I have on one of these LM2596 boards is 10Kohm. On the UC3842A, it's 100Kohm. \$\endgroup\$
    – Anthony
    Apr 17, 2016 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.