Answer:
One of the ways to accomplish this is to use the resistance as a part of Wheatstone bridge and op-amp in difference amplifier configuration. This circuit is shown below:
First about the bridge:
In the following circuit R1, R2 and R4 are fixed at 10K. R2 is the resistance that varies from 10K to 3475ohms.
When R2 is equal to 10000 ohms, the voltage Va and Vb are both 2.5V. Therefore the difference amplifier gives an output voltage equal to
Vout = (RF/R5) (Vb - Va) = 0V
When R2 is equal to 3475 ohms then Va = 2.5V but Vb changes to
Vb = 10K * 5/(10K + 3.475K) = 3.71 volts …………......using potential divider theorem
Therefore Vout = (RF/R5)*(3.71 - 2.5) = (RF/R5)*1.21 = 4.13*1.21 =5V
Hope this helps.
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