Jan 16, 2012

Voltage Divider Quiz Solution

I was lucky enough to get a few responses on my quiz question this time around. While there were many failed attempts at finding the answer among my friends, someone did come through and restore my faith in humanity. Let’s look at how he did it -- or at least how I assume it was done.

Step 1: Since I told you in the hints that this was a comparator circuit, you should remember that a comparator compares two input voltages and outputs a high or low signal if one is above the other, which can be controlled via the circuit configuration. Since this opamp has a relatively high input impedance, you can ignore any loading effects and solve the two voltage dividers very easily.

Step 2: Find the reference voltage going into the inverting input (-) of the comparator using the voltage divider equation. At this point you are only concerned with the bold portion of the circuit in the picture above so the equation should look like:

The units cancel on the ohms range and your answer ends up in volts. You now know that the LED will turn on when the input voltage on the non-inverting input (+) goes above 3 volts.

Step 3: Apply the voltage divider equation again to the 4.7kΩ  and ??? Ω resistor network in bold in the picture above. This time you must rearrange the terms to solve for the mystery resistance using a known voltage. The equation should now look like:
This equation requires a few more steps to solve, but it is not complicated if you know basic algebra. Dividing both sides by 5 volts you end up with:
Now get rid of the denominator by multiplying both sides by its value (4.7kΩ + ???Ω) and end up with:
Therefore, the LED should turn on when the mystery resistance is above 7.05kΩ because it will cause the output of the comparator to go high.

Step 4: Here on To the Rails we care about numbers as much as the end result so I decided to build this circuit and do some comparisons between our predicted outcome and the practical implementation. I used a variable resistor known as a potentiometer (the blue thing) to vary the resistance around the point that turns on the LED. Without knowing the resistance, I turned the pot to a point where the LED barely turns on in order to get the most accurate resistance reading possible. Once I felt I was as close as I could get to the resistance that causes the LED to turn on, I removed the pot from the circuit and measured its value. **NOTE** It is very important to remove the pot to measure the value. Leaving it in will cause you to measure the resistance seen by the pot within the circuit itself. And as always, NEVER MEASURE THE RESISTANCE OF A LIVE CIRCUIT!!

As you can see from the pictures, even with all the uncertainties associated with the manufacturing of the resistors, the accuracy of the potentiometer, and the voltage of the power supply, we still managed to get a reading of 7.09kΩ as our observed answer. For those keeping score at home, that makes our percent error only 0.57%...not bad.


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