Potentiometer alternatives

[Mzzed]

So a friend and I are building a power supply using buck converter. The buck converter is going to be using the adjustable version of the LM2678 switching controller that will regulate the output voltage by varying the duty cycle. The chip uses a voltage divider connected to the output voltage which the feedback voltage is measured and the chip uses this to adjust duty cycle to keep the voltage at the same value. By varying R2 in the schematic attached bellow, the output voltage can be selected.

Our problem is that this voltage divider is susceptible to interference from the inductor of the buck converter so we would like to make the voltage divider physically as small as possible to avoid this. That means we cannot run wires from the pcb to a potentiometer on the front panel of the power supply. We also tried looking at digital pots but they seem to only handle small currents and voltages around 5V where as our supply will be providingup to 20V on the output. The feedback pin requires 1.21V so the voltage drop across this variable resistor at R2 will be anywhere from about 1V to roughly 19V.

We would prefer to avoid placing the entire pcb vertically against the front panel. This would allow access to a pot on the pcb without increasing the amount of noise it receives from the inductor but we would prefer having the pcb positioned horizontally on the base of the power supply.

Are there any alternatives to digital pots or any methods to vary the resistance without also increasing it's susceptibility to noise?

 

[berkeman]

Would a potentiometer with a long shaft do the trick?

https://www.elliottelectronicsupply...d6e5fb8d27136e95/0/0/0000000018761_Def_13.jpg

 

[Averagesupernova]

Can you shield the inductor?

 

[Averagesupernova]

You also may want to consider running a drain wire along with the wires that go to the pot. I suspect with some clever layout you could get by with a front panel pot.

 

[donpacino]

Our problem is that this voltage divider is susceptible to interference from the inductor of the buck converter so we would like to make the voltage divider physically as small as possible to avoid this. That means we cannot run wires from the pcb to a potentiometer on the front panel of the power supply. We also tried looking at digital pots but they seem to only handle small currents and voltages around 5V where as our supply will be providingup to 20V on the output. The feedback pin requires 1.21V so the voltage drop across this variable resistor at R2 will be anywhere from about 1V to roughly 19V.

Why don't you make R1 the variable resistor? It would change the math, but allow the voltage drop to be significantly less and you can use a digital pot

 

[Merlin3189]

Cf. #3 use torroidal inductor?

 

[berkeman]

The chip uses a voltage divider connected to the output voltage which the feedback voltage is measured

BTW, be sure to check the input current specified by the Buck IC at the Feedback input pin, and compare that to your potentiometer's Minimum Wiper Current. If the Wiper Current is too low, the pot circuit will go unreliable over time (due to contact corrosion).

Here is the Bourns Trimmer Primer for more information on how best to use potentiometers:

http://www.bourns.com/docs/default-document-library/bourns_trimmer_primer.pdf

 

[NTL2009]

Could you use a Light Dependent Resistor (LDR) for R1 and/or R2? Then have a pot drive an LED to light the LDR. Or an LED and a mechanical 'shade/shutter' between them. You mentioned 19 V across R2, but what currents are required for R1 and R2?

https://www.digikey.ca/products/en/sensors-transducers/optical-sensors-photo-detectors-cds-cells/540?FV=ffe0021c&mnonly=0&ColumnSort=-732&page=1&stock=0&pbfree=0&rohs=0&cad=0&datasheet=0&nstock=0&photo=0&nonrohs=0&newproducts=0&quantity=&ptm=0&fid=0&pageSize=500

You can find some with the LED in the same package, but they may be diodes or transistors rather than R, but I think that could be made to work for you.

 

[Mzzed]

I'm not exactly sure what current is required right now but the data sheet should tell me. That's a great idea though thankyou! I didn't realize you could get the led and resistor in the same package.

 

[Mzzed]

Thanks all for your replies, I didn't expect so many answers. Thankyou!

 

[NTL2009]

I'm not exactly sure what current is required right now but the data sheet should tell me. That's a great idea though thankyou! I didn't realize you could get the led and resistor in the same package.

If you can't find one pre-made:

http://www.instructables.com/id/Make-a-simple-optocoupler/

 

[Merlin3189]

Can I just query the premise of this problem? The warning on the TI design example diagram seems to relate to the wire running from the output to back to the potential divider. This wire must exist whether you use two resistors or a potentiometer. The only extra wiring for the pot is from the output terminals (on the front panel?) to the pot. The wiper connection can simply follow whatever path you chose for the original feedback wire.

If the output and pot are on the front panel, there should not be any wiring closer to the inductor than in the fixed R design.
If the output is perhaps at the back, then the wires to the pot would have to run all the way to your front panel. These could be routed as circuitously as you like to stay well away from the inductor, because they carry negligible current.
==============================
If, despite the above notion, you still worry about magnetic pickup on the feedback wire, can I suggest adding an emitter follower to the potential divider, whether fixed or pot. The eg. circuit is for 3.5A supply, so using a few mA for this should not matter. (Not that I know anything about this: it just seems a good way of getting a clean signal back there. )

 

[Baluncore]

Is there really a problem here, or are you just fearful ?

There are a few things you can do.
1. Reduce voltage noise pickup by lowering the resistance values of R1 and R2 by a factor of 10.
2. Fix the value of R1 and keep it close to the chip. Use a variable resistor on the front panel for R2.
3. To reduce magnetic pickup, run a tightly twisted pair between the feedback divider at the chip and R2.
4. Put a capacitor, (10usec/R1), across the twisted pair at the chip end where R2 would have been on the PCB.

 

[Mzzed]

Thankyou, I think we've decided on using a toroidal inductor and wire the pot to the front panel as originally planned. If there is still noise we will then possibly just attempt to add shielding or maybe even simply go with a pot with a long knob. The simpler the better, but I'm sure it's likely we will use some combination of any number of these suggestions if this proves to be harder than we think it is in practice.

 

[NTL2009]

Makes sense to go simple first and see if there are issues. But I would still do the twisted pair and shielding of the wires to the pot as was recommended by others, that is very little effort during construction, and more work after. And if you retro that, and it still has issues, you are taking it apart again to try the next 'fix'. Might as well start there (and I'll bet that is all you'll need - although the LDR would be 'fun', I stripped one out of an old night light for a recent project). Good luck!