Replace potentiometer with 2 resistors.

Thread Starter

Bod

Joined Sep 18, 2016
301
I bought these a little while ago (MP1584 step down modules). The issue I had with these is they are extremely sensitive and also have a horrible voltage range (it goes sort of 1V, 2V, 3, 4, 4.5, 15V... as you turn the potentiometer.. and guess what voltage I needed - 5!). Anyway I dealt with it for a while until I blew my up micro-controller because I must have jogged the board and the potentiometer moved.

So obviously, the next best thing would be to get a constant 5V output board instead. While looking for one, I noticed the circuits are almost identical with just different values of resistors (and probably caps) and a different value inductor. That then got me thinking surely I can just replace the potentiometer with 2 resistors to get a fixed output. So that is what I did: (excuse the drawing)

Inkedfrontside_marked_LI.jpg
I did exactly as shown and put two 2.7K resistors across the two pins (NOTE: I have two of these boards - this is not the one I put the resistors on - the other one has the pot removed)
Unfortunately, I still got 23V on the output (max output = input - 1V). I would think this should be working because of diagrams like this: 1590140369497.png
The next thing I noticed was that the pin with the green arrow pointing at it doesn't seem to be connected to anything:
backside.JPG
If it was connected there should at least be some vias directly in line with the '+' (the board is flipped so the pot is on the left side now)
The pin may well be connected to ground but it just doesn't seem to be.

What am I doing wrong here? - maybe I have the resistors the wrong value but they should still at least affect the output a bit right?

Thanks,
Bod


EDIT: I just had a look over the board again. The left pin (green arrow) is indeed connected to ground. There's a trace hidden underneath pot that seems to got to one (if not both) of the vias on the underside, connected to ground (the ones between the 'U' and 'T' in 'OUT')
 
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Thread Starter

Bod

Joined Sep 18, 2016
301
hi bod,
Have you considered fitting a 20 turn Trim pot in place of that single turn pot.?
E
Unfortunately I don't have any pots on hand. Not sure why I even got one with and adjustable output in the first place. All I need is a constant 5V output.
If there is no connection to one outside leg then the pot is used as a variable resistor, not a voltage divider.
Does that mean I can just replace it with one resistor instead?

Also, isn't that how a pot works? One pin is connected to ground, one to VIN and the wiper terminal is the output.
 

jpanhalt

Joined Jan 18, 2008
9,686
You can probably get by with what you are doing, but I prefer a voltage regulator for my supplies, particularly for voltage sensitive devices. It is important to have a relatively constant voltage independent of load. Here's an example of what I built into the connector:

1590142781191.png
 

MrChips

Joined Oct 2, 2009
21,309
If it is not wired as a pot then one resistor will do.
Edit: measure conductivity between pin in question to ground.
 

Thread Starter

Bod

Joined Sep 18, 2016
301
If it is not wired as a pot then one resistor will do.
Edit: measure conductivity between pin in question to ground.
Oops... I'm wrong again. The pin is not connected to ground but rather it looks like the left pin (arrow) is connected to the wiper pin directly:
InkedIMG_0967_LI.jpg

Measuring between the left pin and ground (on the board with the pot) gives a reading of 52K ohms. Adjusting the potentiometer changes this reading. I turned it and the reading came up as about 200K ohms.
 

MrChips

Joined Oct 2, 2009
21,309
Things to note:

1) If you are designing a circuit to use a pot as a variable resistor it is a good idea to connect the unused pin to the centre pin.
The reason for this is as you turn the pot there is a chance that the wiper loses contact with the track. This results in an open circuit. With the unused pin wired in, the maximum resistance will be the nominal resistance of the pot even if the wiper does lose contact.

2) What Eric is saying, look at the circuit in TYPICAL APPLICATION. R1 and R2 indeed constitute a voltage divider from VOUT back to the FB pin (Feed Back). If your circuit board has the equivalent of R2 installed connecting FB to GND (I see 8.2kΩ), then all you need is one resistor R1. That is the purpose of your pot.
 

Thread Starter

Bod

Joined Sep 18, 2016
301
hi,
The resistors set the output voltage.
E
Things to note:

1) If you are designing a circuit to use a pot as a variable resistor it is a good idea to connect the unused pin to the centre pin.
The reason for this is as you turn the pot there is a chance that the wiper loses contact with the track. This results in an open circuit. With the unused pin wired in, the maximum resistance will be the nominal resistance of the pot even if the wiper does lose contact.

2) What Eric is saying, look at the circuit in TYPICAL APPLICATION. R1 and R2 indeed constitute a voltage divider from VOUT back to the FB pin (Feed Back). If your circuit board has the equivalent of R2 installed connecting FB to GND (I see 8.2kΩ), then all you need is one resistor R1. That is the purpose of your pot.
Thanks for the replies. I now understand how it works, I what I would need to do to get a constant output voltage, however, what I still don't understand is about the FB pin. Like the datasheet says, it is supposed to be connected to the output voltage through a voltage divider. In my circuit it seems the FB pin is connected straight to the VIN pin, through no resistors.
1590156762628.png
The FB is pin 4, bottom left:
Inkedfrontside_normal_LI.jpg
The FB on mine goes straight to VIN. It's connected to a resistor but that doesn't feed it nor is it a voltage divider.
Am I missing something here?

EDIT: I think I understand it now. The FB pin is connected on the right hand side of the SMD resistor which is connected through to ground. There's a via (hidden by my green line) which goes to the right hand pin on the pot. The wiper pin is connected to the output. Since FB is connected in the middle of those two, it is acting as a voltage divider feeding FB.
I was getting confused because I though it was connected to VIN. It's not it just looks like it since the silkscreen covers it partially.

EDIT 2: I've bodged in a 150K and 100K resistor to make 250K which I calculated as the Value of R1. Sadly I'm still reading 23V on the output.
If I know R1 is 250K and I know R2 is 8.2K, putting that into the Vfb equation:
Vfb = 5*(8200/(8200+250000)) = 0.159
Meaning Vout should be 5:
Vout = 1.59*((8200+250000)/8200) =~ 5 (exactly 5 if I use all the decimal points.
So technically I should see 5V on the output.
 

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ericgibbs

Joined Jan 29, 2010
10,184
hi bod,
Have been searching the web, those Vin are not connected to the FB components, it looks close but please recheck and measure with a meter.
E
AAA 345 15.58.gif
 

Thread Starter

Bod

Joined Sep 18, 2016
301
hi bod,
Have been searching the web, those Vin are not connected to the FB components, it looks close but please recheck and measure with a meter.
E
View attachment 207840
I actually did manage to figure out that it was connected after just tracing the tracks a little. It would have definitely been easier if I used a meter!
Thanks for the reply though.
 

MrChips

Joined Oct 2, 2009
21,309
The voltage at FB pin has to be 0.8V.
Calculate R1 and R2 to make VFB = 0.8V at whatever output voltage you need.
i.e.
R1 = 1.25 x R2 x (VOUT - 0.8)
 

ericgibbs

Joined Jan 29, 2010
10,184
hi,
Assuming that the designer has used a 40k from pin #4 to 0V. I would attach a 220k from Vout to pin #4, then check Vout.

E
 

Thread Starter

Bod

Joined Sep 18, 2016
301
The voltage at FB pin has to be 0.8V.
Calculate R1 and R2 to make VFB = 0.8V at whatever output voltage you need.
i.e.
R1 = 1.25 x R2 x (VOUT - 0.8)
Thanks for the reply. I think I have worked out the value of R1 which should approx. 43K ohms. I just rearranged the Vfb equation to give me R1 and when I use R1 in the equation, I get 0.8 for Vfb.

In your equation what is 1.25? Is it just some constant? In the data sheet it says 50.25 so why are you using 1.25?
 

MrChips

Joined Oct 2, 2009
21,309
1.25 is 1/0.8

In the example given R2 is 40.2k
1.25 x R2 = 1.25 x 40.2k = 50.25k

Can you find a 40.2kΩ resistor on your board? I don't see one. I see 8.2kΩ.
 
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