# Why is this load resistor connected this way and what could be its value?

There is a potentiometer vane and the diagram includes protection components and looks messy, I simplified it by re-drawing as below: The circle R1, R2, R3 and the gap represents the vane's 10k potentiometer section. The datasheet says "Wiper Contact Resistance" is 400 Ohm, so I modelled it as Rwp. Rw resistances above represents the wire resistances for the red and blue wires which passes the supply current. Rs is the possible supply output resistance.

It is clear to me that Ch1 and Ch3 measures the potential difference between the point X and point Y. This is called I guess Kelvin connection. Ch2 is the poti output so the ratio Ch2/(Ch1-Ch3) can be obtained which corresponds to angle. I want to calibrate this by using an encoder versus this ratio.

The vane has a dead band. Now for the this dead zone the datasheet suggests a pull down/up resistor RL between Ch3 and Ch2 but it doesn't mention or suggest any value. My question is what could be the reason RL is tied between Ch2 and Ch3 but not between Ch2 and the ground? And what value for RL can be reasonable?

My guess:

The potentiometer is not a full 360° to prevent the wiper short-circuiting X to Y. This means that when the wind is from the west the wiper will not be in contact with the resistance track and so will be floating. This could lead to false readings so the designers have a pull-down resistor so that the wiper will appear to be at Y while in the dead zone.

Channel 3 is measuring the voltage at Y and this will be slightly higher than GND (> 0 V). If RL were to be connected to GND then that would pull Channel 2's input below the Channel 3 zero reference and the subtraction would result in a negative value for the position (which is always positive with the arrangement shown).

And what value for RL can be reasonable?

RL is going to pull the wiper voltage away from the potentiometer's divider voltage. This effect will be lower at either end of the wiper because the low impedance of that end of the track will win over the value of RL. The maximum loading effect will be at mid position. You can run some calculations in a spreadsheet for this to see the effect. 50 × the pot resistance seems good enough to me (without doing any calculations) to get you within a couple of percent. With wind variations I doubt that accuracy is all that important but I'd be inclined to put the dead-zone opposing the prevailing wind.

• Thanks for the answer so between that dead sonze the ratio should be flat as Ch2/(Ch1-Ch3) correct?(II ask because I will calibrate this). And secondly to minimize the nonlinearity caused by RL if I set it 4.7Meg do you think it is fine? (since you said times 50 is adequate). Too high RL woul be an issue? Feb 23, 2020 at 22:51
• (1) Your formula looks correct. (2) I suggested that you calculate the maximum error at midpoint. Did you? :^) Feb 24, 2020 at 17:38

The datasheet says to keep wiper current below 10 microamps (max 10 milliamps though). The load resistor, if present, must be large enough that the wiper current does not exceed 10 microamps, and large enough that it does not change the measurement too much, as it will form a resistor divide network with the 10k pot inside the device and the return currents will flow in the long wires back to the far end. The reason it is between wiper CH2 and negative reference CH3 is that basically CH3 is what should be considered as most negative voltage you can get out from the pot. Reasonable value depends of course what device is reading the voltages and at what accuracy. 5V at 5Mohms would make max 1 microamp flow, and output impedance of 10k pot would be max 2.5k so error would be about 1/2000 or good to about 11 bits of resolution.

• Max output impedance of 10k pot is 5k or 2.5k? Feb 23, 2020 at 23:15
• Good catch; a 10k pot would have 5k from wiper to both ends, so max impedance is 2.5k. Feb 24, 2020 at 0:06