1
\$\begingroup\$

I am trying to make an adjustable power supply with XL4016 module and 24 volt 6,5A SMPS.

enter image description here

I can adjust volt with two pot (fine-course) because it needs two pins. But I couldn't think how to design a coarse / fine replacement for the current pot because it needs three pins for pot. I need middle pot pin too.

How can I connect two pots in place of a normal pot. The module wants to 10k pots.

\$\endgroup\$
5
  • 4
    \$\begingroup\$ Please edit your question and add a proper hyperlink to the datasheet for the XK4016 module. "This schematic doesn't work." (1) Where did you get it? (2) Why do you think it should work? (3) Doesn't work in what way? \$\endgroup\$
    – Transistor
    Commented Aug 23, 2020 at 21:59
  • 3
    \$\begingroup\$ what is a two current pot? \$\endgroup\$
    – jsotola
    Commented Aug 23, 2020 at 22:19
  • \$\begingroup\$ He wants to have fine and coarse adjust on the current limit setting. \$\endgroup\$ Commented Aug 24, 2020 at 5:46
  • \$\begingroup\$ @Jasen That may be a good guess, but there's nothing like that in the schematic. \$\endgroup\$ Commented Aug 30, 2020 at 18:05
  • \$\begingroup\$ he wants to modify the schematic. \$\endgroup\$ Commented Aug 30, 2020 at 22:03

3 Answers 3

1
\$\begingroup\$

Voltage adjust

The schematic shows that. R9 is a 3-terminal pot with the wiper connected to one end. This makes it a variable resistor.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Voltage coarse and fine.

If you want the fine adjustment control then add a 1k pot in series with R9 and with the wiper wired in the same configuration. The 10K part will then give you coarse control and the 1k pot will give you fine control.

Current adjust

schematic

simulate this circuit

Figure 2. Current coarse and fine.

Note that this adds an extra 1k into the R2 - R3 potential divider but if R2 is a 1% type it's resistance tolerance is ±2.2 kΩ.

\$\endgroup\$
2
  • \$\begingroup\$ But i need to set R3 pot. I've already arranged the volt pot(R9) with two pots \$\endgroup\$ Commented Aug 24, 2020 at 9:40
  • \$\begingroup\$ The wording of your question confused many of us. See Figure 2. \$\endgroup\$
    – Transistor
    Commented Aug 24, 2020 at 15:08
0
\$\begingroup\$

Put 1k pot in series with 10k pot... close enough This gives full range but not the foldback at CV cutoff when I drops <5%CC for Lithium but ok for 14.2 Lead acid

If numbered 123 the wiper going CW moves from 1 to 3 so if you to increase Vfb resistance short 12 both to increase Vout. Then R increases CW. Then a short goes to Vref =Vmin=1,5V ?

If you just want 12 to 15 why fine just use fixed 9k+ 1k pot or whatever fixed nom. you want

\$\endgroup\$
1
  • \$\begingroup\$ Now I see I gave same answer as @Tranny but with added detail. Coincidence. \$\endgroup\$
    – D.A.S.
    Commented Aug 24, 2020 at 2:58
0
\$\begingroup\$

You can get pots with two tracks:

enter image description here

Then you can put two-terminal resistors for fine adjust at each end of the coarse pot.

schematic

simulate this circuit – Schematic created using CircuitLab

But seeing as the 10K pot has much larger resistance (220K) in series at the top you could just put the fine adjust at the bottom of the 10K. there'd be a 5% difference in the fine adjust scale from top to bottom of the 10K but few people will even notice that.

\$\endgroup\$
2
  • \$\begingroup\$ Why have two ganged variable resistors when one will do the job? \$\endgroup\$
    – Chu
    Commented Aug 24, 2020 at 6:15
  • \$\begingroup\$ I was answering the general question of how to add fine-adjust to a potentiometer. in this specific case the full solution is not needed. \$\endgroup\$ Commented Aug 30, 2020 at 22:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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