100k to 50k

[Noumenon]

So.

This might be pretty basic, but I have no idea.
I have this potentiometer that's at 100k, I can't replace this with a 50k potentiometer so how can I make a circuit where the 100k potentiometer controls a 50k potentiometer?

 

[Joel Rainville]

how can I make a circuit where the 100k potentiometer controls a 50k potentiometer?

What do you mean?

How are your pots wired?

 

[hantto]

Paralelconnect a 100k ohm resistor with it (or 50k from each pin to wiper) this will be almost like a 50kohm pot

EDITED - typo

 

[Noumenon]

Hm. I'm not really sure you understand me.

I want a 100k pot to control a circuit that goes from 0-50k. Basically so the 100k pot makes the circuit act like a 50k pot. where 0 on the pot is 0 in the circuit and 100k on the pot is 50k in the circuit.

 

[Joel Rainville]

Paralelconnect a 100k ohm resistor with it (or 50k from each pin to wiper) this will be almost like a 100kohm pot

Glad to see you guys understand each other :lol:

But still, I don't see how you can assume he's trying to make a 50k pot act like a 100k one. There's nothing in the question that sounds like it :

how can I make a circuit where the 100k potentiometer controls a 50k potentiometer?

 

[Noumenon]

Glad to see you guys understand each other :lol:

it's hard sometimes

But still, I don't see how you can assume he's trying to make a 50k pot act like a 100k one.

Me neither, I want the reverse. 100k acting like 50k

 

[Joel Rainville]

You could connect a 100k resistor between the wiper and a pin.

Or you could join both pins together with a 100k resistor.

Both methods give you 50k to ~0k, but the tapper won't be linear. I.E. middle position won't be 25k, more like 33k...

 

[Noumenon]

Okay. Thanx, I'll have to try it and see how it works out

 

[Joel Rainville]

Paralelconnect a 100k ohm resistor with it (or 50k from each pin to wiper) this will be almost like a 100kohm pot

By the way, this is plain wrong.

Assuming you're talking about a 50k pot :

A 100k resistor in parallel gives you a ~33k pot.

50k resistors from wiper to each pins gives you a ~25k pot.

 

[Joel Rainville]

With your typo fixed, this is right :

Paralelconnect a 100k ohm resistor with it [...] this will be almost like a 50kohm pot

...but this is still wrong :

50k from each pin to wiper) this will be almost like a 50kohm pot

EDITED - typo

 

[hantto]

Motivate please. Show your calculations so that I may understand, opposed to "BLEEEP! WROOONG!"

 

[Joel Rainville]

Motivate please. Show your calculations so that I may understand, opposed to "BLEEEP! WROOONG!"

Sorry, I thought it was obvious since you took the time to answer the original poster's question and that the explanation is actually in your own post! I thought it was another typo from your part...

If you agree that a 100k resistor in parallel with the 100k pot gives a max 50k resistance, why does the resistor suddenly become 50k between a pin and the wiper?

Take a 100k pot, with 50k resistors from pins to wiper, you get :

avg(100+50)/2 == 33k to ~0 on pin 1, 0 to 33k on pin 3...

The correct value for the resistor is again 100k, and in most applications you'd only need one, leavin one pin on the pot floating...

 

[Joel Rainville]

Wait, now, bleeep, my calculations are wrong! :lol:

In the above example, the total resistors values are 150k ohms. The 50k being in parallel : 50k/150k == 33.3k

The correct resistor value then is obviously 100k because 100k/200k == 50k.

But don't trust my calculations. Take a pot and a resistor. I just tried it with a 500k pot and a 470k resistor. I get ~250k to 0.

 

[checkmate]

The net resistance of any 2 resistors in parallel is
xy/(x+y)=y/(1+(y/x))
Let's say you fix y, then there is no way you can get a linear variable resistance by adjusting x. If linearity is not important, just go ahead and let the net resistance be 50k, max(x)=100k and find the corresponding y.

Or alternatively, maybe you can show us your schematics, and we'll see if anything else can be done to get the same effect. But of course, the easiest method is to just replace that potentiometer.

 

[hantto]

Hm, do you know how to calculate parallel resistors? the formula would be

1/R = 1/R1 + 1/R2 + 1/R3 ... and so on

this comes to if you have a 100k over a 100k pot:

1/100 + 1/100 = 1/R
2/100 = 1/R
2R = 100
R = 50kohms

And if you put 50k resistors from high and low terminals to the wiper, you would have 50k resistors in parallel with the pot itself. When the pot is in middle position it would equal 25k from high or low to the wiper. Totally from high to low the resistanse would equal 50k, because there is two 50k resistors on series which equal 100k which is in parallel with the 100k pot, thus equaling 50k. If you use the two resistor method the pot's response will not be fully linear.

Hope I got this right :lol:

 

[Joel Rainville]

The left circuit is still wrong... Man you're starting to confuse me. I don't think you understand how a pot works... Let me figure it out, I'll be back in a minute

 

[checkmate]

It's the same principle. Putting 2 non-linear variable resistances in series is not going to make them linear. All the configurations shown are non-linear.

 

[Joel Rainville]

It's the same principle. Putting 2 non-linear variable resistances in series is not going to make them linear. All the configurations shown are non-linear.

Checkmate, the debate is not about linearity. What I'm saying is that using those 2 50k resistors like he wires them is not going to give you a high value of 50k resistance, but 33k.

 

[Joel Rainville]

Totally from high to low the resistanse would equal 50k, because there is two 50k resistors on series which equal 100k

I think that's the part confusing me. Those 50k resistors are never in series. They would be if the wiper disconnected from the pot, but in reality current always flows through the wiper, lo or hi.

Does it make sense? Can someone confirm it makes sense?

 

[hantto]

Yes, you are also correct after further analysing. My starting point was that the pot would have 50k pot carasteristics at the middle, while yours was at high and low. Your solution is indeed better :lol:. If the output voltage is the only factor of intrest, both will suffice.