Back2Basics Using a Trim Pot

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StudentSA

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I need to have an adjustable voltage "signal". This is fed into a IC that uses it as a referance for comaprison. What has confused me is that I have biased my trim pot (50k) with 12V.

I have 12V and ground at the end terminals. I then use the wiper pin as an output and adjust the output signal level using my multimeter. Lets say I set it to 8.5V.

However when I connect the "wiper" pin to my IC (Input Pin) all of a sudden the voltage drops to like 3V and I need to readjust the wiper to get it to 8.5V again.

Why does this happen? Any advice is appreciated...
 
Post a circuit. I'll bet that you are exceeding an input common-mode range, or you are connecting the wiper to something other than an opamp input.
 
Well basically, Im trying to use a trimpot to set the upper voltage limit of an LM3914 LED VU Display Driver IC.

The IC Pin "Rlo" is tied to ground and "Rhi" is fed from the trim pot as shown:



I need to set the "Rhi" Pin between 7V - 10V and hence use a trimpot.

Is there a more simple way or is my trimpot value just wrong?
 
Reduce the trimpot value to 5K, or raise the voltage setting resistors inside the LM.... circuit. You cheated, you didnot post an adequate schematic.
 
The minimum equivalent resistance of a 50k ohm pot at the wiper output occurs at midscale, where it is 12.5k ohms. It becomes less on either side of the center. It is this resistance which is acting as a voltage divider with the input resistance of the circuit the pot is connected to. Ohm's law explains it all.

As MikeMl said you need to reduce the value of the pot or increase the value of the circuit resistance to reduce the effect you have noticed.
 
One way to reduce the impedance of the source is to add a voltage follower to the wiper of the pot, but that seems like a silly thing to do when all you really need is less resistance in the pot.
 
This shows you what is going on. A 50K pot is set to 8V Open Circuit when powered with 12V. V(out) shows what happens as a function of the load resistance connected between wiper and ground. The X-axis in the plot is V(R), but the load resistance R1 is R=V(R), so is varied from 1K to 1meg. Note that for load > 1meg, V(out) approaches the theoretical 8V, but V(out) drops for lower values of R1.
 

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Thanks, I think I understand what you getting at.

The load impedance is required to be large compared to the source for maximum power transfer.

So if my load is fixed to 10k I am required to have my source impediance very small to compensate. However lets not forget about efficency! I do not want a "simple" voltage level signal consuming too much power.

If I use a 1K pot, theoretical power consumption (of the pot) would be P = V*I = 12*(12/1000) = 0.14W.

Seriously is this the amount of power required?
 
StudentSA said:
If I use a 1K pot, theoretical power consumption (of the pot) would be P = V*I = 12*(12/1000) = 0.14W.

Seriously is this the amount of power required?
Click to expand...
That is correct.

To reduce the power, you can use a follower follower (as #12 suggested), such as a op amp follower (gain = +1) between the pot output and your load. Then you could still use your 50k ohm (or even somewhat higher value) pot. For that you would need a low-power, rail-to-rail type op amp.
 

Yes, if you are using only resistors to create a "voltage divider". There is a way of making a voltage divider that utilizes active circuits which will consume less power, and have a much lower output impedance than the resistive divider. It was mentioned several posts ago; make a divider (pot) that has high resistance, say 1megΩ. Connect the wiper to the non-inv input of a modern CMOS rail to rail opamp. Connect the opamp as a voltage follower. Connect Vdd to the top of the pot and the Vdd pin of the opamp. Connect the bottom of the pot to Gnd, connect Vss of the opamp to Gnd.

That will have a low (<0.001Ω) output impedance (if the load current is less than a few tens of mA). At zero load current, the right CMOS opamp will consume much less power than your 1K pot....
 
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