Calculation OF Potentiometer PDF

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CALCULATION OF POTENTIOMETER

Potentiometer Example No1 A resistor of 250 ohms is connected in series with a second resistor of 750 ohms so that the 250 ohm resistor is connected to a supply of 12 volts and the 750 ohm resistor is connected to ground (0v). Calculate the total series resistance, the current flowing through the series circuit and the voltage drop across the 750 ohm resistor.

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Potentiometer Example No2

A 270o single-turn 1.5kΩ carbon track rotary potentiometer is required to provide a 6 volt supply from a 9 volt battery. Calculate, 1. the angular position of the wiper on the track in degrees and, 2. the values of the resistances either side of the wiper. 1. Angular position of pots wiper:

Then the wipers angular position is 180o or 2/3rds rotation. 2. Potentiometer Resistance Values:

Then the resistive values either side of the wiper are R1 = 500Ω and R2 = 1000Ω. We can also confirm that these values are correct by using the voltage divider formula from above:

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Loading the Wiper In the simple voltage divider example above, we have calculated the values for R1 and R2as 500Ω and 1000Ω respectively, to produce a voltage at the wiper terminal (pin 2) of 6 volts with a wiper angular position of 180o. We have assumed here that the potentiometer is unloaded and producing a linear straight line output, so VOUT = θVIN. However, if we were to load the wiper terminal by connecting a resistive load, RL, the output voltage would no longer be 6 volts as the load resistor, RL is effectively in parallel with R2, the lower 1000Ω part, and thus affects the total resistive value of the load part of the voltage divider network. Consider what would happen if we connected a 3kΩ load resistance to the wipers output terminals.

Loaded Potentiometer Wiper

So we can see that by connecting a load across the terminals of the potentiometers output, the voltage has decreased in this example, from the required 6 volts to just 5.4 volts as the loading effect of the 3kΩ resistor gives a parallel equivalent resistance, RP of 750Ω instead of the original 1kΩ. Obviously, the higher or lower the resistance of the connected load the greater or lesser the loading effect on the wiper. So a load resistance in the mega-ohms range would have very little effect compared to one that was just a few ohms in value. Thus, to return the output voltage back to the original 6 volts would require a small adjustment of the potentiometer wiper position (18o in this case) as now RT is equal to 1250Ω (500 + 750).

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