# Help with potentiometer needed

Discussion in 'General Electronics Chat' started by Voidugu, Jul 11, 2013.

1. ### Voidugu Thread Starter New Member

Jun 25, 2013
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0
I got a 10 K pot connected to 5v and ground. The potentiometer will be used as a voltage divider. I want to be able to utilize as much as possible of the potentiometer's rotation to vary the voltage output of the potentiometer between two voltages (Assuming that the rotor of the potentiometer can go from 0 to 180 degrees, i would like 0 degrees to be 2.5 volts and 180 degrees to be 2.8 volts).

Can you people provide me with a circuit that will allow me to do that and also allow me to change those boundaries easily?

Cheers

2. ### wayneh Expert

Sep 9, 2010
12,151
3,058
Just put the pot in series with resistors above and below it. Use ohm's law to select the proper size of those external resistors. You want 2.5V when the wiper is at 0Ω. That means the lower resistor must equal the top +10K. You also want 2.8V at 10KΩ. That means 2.8/5 = bottom+10K/(top+10K+bottom). Nothing but algebra between you and the answer.

It gets harder if there is a load on on your wiper, but if it's supplying a high impedance op-amp for instance, the math is not so bad.

3. ### #12 Expert

Nov 30, 2010
16,338
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It's all done with Ohm's Law.

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4. ### ScottWang Moderator

Aug 23, 2012
4,855
767
If you using the potentiometer of rotary type, it may bring you trouble, because the degrees of rotary type is not easy to get the precisely value of potentiometer.

If you can change to the slide type, maybe it's more easier to get the value and calculating the rest resistors.

The calculation formular as this way:

V_10K = 2.8V - 2.5V = 0.3V
I_10K = V_10K/10K = 0.3V/10K = 30uA

The leftmost resistor R2:
R2 = 2.5V/30uA = 83.333K

The rightmost resistor R1:
R1 = (5V-2.8V)/30uA = 2.2V/30uA = 73.333K

GND ← 83.333K(R2) ← 2.5V ← VR10K ← 2.8V ← 73.333K(R1) ← +5V

Because the current is too small, so you better adding a voltage follower using Op Amp.

Ok, now you already got the all values, so how to get the precisely values for R1 and R2, i will let you to think about it.

You may change the value of potentiometer from 10K to 1K or less and re-calculate R1 and R2.

Slide potentiometer 1.

Slide potentiometer 2.

Last edited: Jul 11, 2013
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5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Since pot tolerances can be ±30%, you might want to reduce the effect of the pot by shunting it with a more predictable resistor which takes most of the current.
See attachment.

This can be taken to any extreme you want, so long as you can afford the extra current. For example, if R3=100Ω, then ≈99% of the current flows through it, and the actual value of the pot becomes pretty much irrelevant, as long as R3<<Rp.
Bear in mind that if R3=100Ω, total current≈3mA. Still pretty low, unless you are using batteries. Even then, it ain't bad.

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6. ### tubeguy Well-Known Member

Nov 3, 2012
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You could use a couple op amps configured as voltage followers to supply the min and max voltages to the ends of the pot. These would each be set by a trim pot.

Last edited: Jul 11, 2013
7. ### WBahn Moderator

Mar 31, 2012
17,768
4,804
Since the pot may not actually get all the way to the end, you may want to include some calibration pots, as well. Also, that will allow you to compensate for component tolerances. If you just want a simple solution that uses three pots and two resistors, use the basic configuration initially described with a singe pot between two resistors. Then take the value of each of the resistors and reduce it by about 10% and use the nearest standard value. Then multiply each resistor by 20% and get potentiometers that are closest to that size, but larger (unless it is smaller by a very small amount). You now put each of the additional pots, wired as a rheostat, in series with the correpsonding resistor with it set to about the midpoint.

Adjusting them is an iterative process, but it should converge very quickly. Measure the voltage at the ends of the center pots travel and set it to the end that is the furthest away from where it should be. Then turn the pot on that side to get the desired voltage. Now do the same for the other side. This will mess up the first side a little bit, so go back and do it. Keep repeating until both sides are dialed in. If you think about it a little bit, you can reduce the number of cycles by anticipating the error that will be introduced by the next adjustment and compensating where you trim the current adjustment accordingly. But even if you don't do this, it shouldn't take more than a handful of cycles to get close enough.

8. ### Voidugu Thread Starter New Member

Jun 25, 2013
8
0
Thank you very very much for the help people.
By the way do you think i should use a smaller value pot in order to pass more current through it an minimise the effect of the current drawn by the comparator (KA393 to be precise) which will be connected to the pot?

Last edited: Jul 12, 2013
9. ### wayneh Expert

Sep 9, 2010
12,151
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In my opinion, the error introduced by the comparator input impedance will be small compared to imperfections in the pot itself. It might cause you to need tweaks to the surrounding resistors.

10. ### WBahn Moderator

Mar 31, 2012
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4,804
The comparator should draw negligible current. Look at the input bias current spec, use the max value, and then make sure that your pot current is at least, say, 100 or even 1000 times this. Since you are probably talking about a max input bias current of well under 1μA, using a pot with 1mA of current in it should make this a non-issue. Since you want a total of 300mV across it, you would be looking for a 300Ω pot. You would probably find that a anything form 100Ω to 1000Ω will work just fine. If you use a 10kΩ pot then your pot current will be 30μA. That will probably be about 1000x the typical bias current, but only about 100x the max bias current. Depending on how accurate you need the endpoints (and, remember, most of the suggestions that have been made allow you to calibrate that), that will probably work fine. I wouldn't recommend going too much more than that (again, depending on what your accuracy needs are).

Part of what you need to be doing depends on things you haven't told us. Which is more important, that the limits not exceed 2.5V to 2.8V (meaning that slightly less is tolerable) or that the limits include 2.5V to 2.8V (meaning that slightly more is tolerable). How accurate and stable do these limits need to be? How variable is the voltage that will be driving the pot? A simple two-resistor and one-pot solution might be all you need, or perhaps you need to use a zener or a bandgap voltage reference and temperature compensation. No way we can tell that from what you've provided.

11. ### Voidugu Thread Starter New Member

Jun 25, 2013
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In a synopsis. I am building a temperature controlled soldering station around the hakko 907 pencil. The pencil has got a 50 ohm RTD (at room temperature) which can vary up to 150 ohms (at max temperature). One voltage divider is created by connecting a 100 ohm resistor in series with the RTD. This gives me the temperature feedback i need. The divider i asked about is what will be used to vary the temperature. Both voltage dividers are connected to a comparator (KA393) which drives a MOC3023 which controls the main power triac which controls the current going to the element of the hakko 907 pencil. The pencil is powered by 24VAC from an 80watt transformer. The 24VAC are also passed through a full wave rectifier, smoothed by a capacitor and then through a 5volt linear regulator powering the two voltage dividers, the comparator and the led of the MOC3023. I tested the arrangement and it turns out that at 2.5 volts (minimum setting) from the RTD-resistor arrangement the heat generated is adequate for SMD soldering. Also at 2.8 volts (maximum setting) the heat generated allows for soldering connections with large thermal mass (ie soldering a heatsink on a pcb). I haven't though fully figured out the max and min setpoints therefore i would like to be flexible with them. ie make a voltage divider arrangement that would allow me to vary the min from say 2.5 to 2.7 volts and the max from 2.8 to 3 volts. (As you must have already understood, the higher the voltage on the RTD-resistor divider, the higher the temperature of the heater)

12. ### BillB3857 Senior Member

Feb 28, 2009
2,400
348
If you need a high degree of accuracy for your 0 degree and 180 degree points, use the circuit #12 posted but place trim pots in place of the fixed resistors he shows. Also note that most single turn pots are more in the 270 degree range rather than 180 degree.

Sep 9, 2010
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14. ### Voidugu Thread Starter New Member

Jun 25, 2013
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Thank you very very much for the help people. Have a great great day