How to make a voltage testing circuit

Question

I want to make a very cheap tester that I could use on cables that are used in the field at my company. We usually use multimeters but not everyone knows how to use one. It would basically be like a power supply tester on a computer.

How could I do this if the voltage is 14 V dc?

Answer 1

If the multimeter is too difficult you want a device which just has a good/bad indication. You need a circuit which compares your voltage with a 14V reference. Some margin is needed, like everything between 13.5V and 14.5V is OK. This is a tailor-made job for a window comparator:

You choose the resistor values of the input resistor ladder so that V1 = 14.5V and V2 = 13.5V. The table shows what happens for different input voltage situations.

Case 1: \$V_{IN} > 14.5V\$
The inverting input of CMP1 is higher than the non-inverting input, so that output will be low. The LM339 has open-collector outputs, so you can wire-OR the outputs. The combined output will be low if at least one of the outputs is low. This is already the case here, so we don't even have to look at the state of CMP2.

Case 2: \$13.5V < V_{IN} < 14.5V\$
The non-inverting input of CMP1 is higher than the inverting input, so that output will be high. The output transistor is off, and \$V_{OUT}\$ will be pulled high by R3.
The non-inverting input of CMP2 is higher than the inverting input, so here we get the same: output is high. Since none of the outputs pulls \$V_{OUT}\$ low, it will be high due to R3.

Case 3: \$V_{IN} < 13.5V\$
The non-inverting input of CMP1 is higher than the inverting input, so that output will be high. The output transistor is off, and \$V_{OUT}\$ will be pulled high by R3.
But the inverting input of CMP2 is higher than the non-inverting input, so this output will be low. And despite CMP1's high output the output transistor of CMP2 will pull \$V_{OUT}\$ low.

If we now place a green LED between \$V_{OUT}\$ and ground we get a good indication if the input voltage is 14V +/- 0.5V. Choose R3 so that you have 20 mA through the LED.

note: for a portable tester you probably don't have a power supply of 15V, maybe a 9V battery. Scale the input voltage down to for instance 5V, and choose the values for the input resistor ladder accordingly. The circuit consumes rather a bit of power, so you may want to use a pushbutton to switch power. Push the button and the LED lights = voltage OK.

edit: improved version
While I wrote this I was aware of the power consumption drawback (hence the pushbutton), but that's a compromise for having a circuit this simple. An additional driver will reduce the power when the LED is off.
The LM339 has a few more comparators left, so we use one of these. Remove the LED and use a 10k resistor for R3. \$V_{OUT}\$ will now switch between V+ and ground without consuming much power.
Enter CMP3. Use a resistor divider to get \$\frac{V+}{2}\$ on the non-inverting input, and connect our old \$V_{OUT}\$ to the inverting input. The output will now be low if \$13.5V < V_{IN} < 14.5V\$. Connect the LED in series with a pull-up resistor to the output. Choose the resistor so that you have 20 mA through the LED when the output is low.

Answer 2

Do you just need a power/no power test? If so, then you can use two LEDs and a diode in series and parallel with each to handle the reverse voltage. Then a current limiting resistor. Either way to hook it up, it will light one LED for one polarity and another for the other polarity. No LED means no voltage.

If you have the possibility of using a higher voltage, you will need to put a zener shunt or something else to limit the voltage and not damage the LEDs.

An even simpler solution would be to pick up some LED based voltage testers for car electrical systems at an auto parts store. Car DC goes up to 14V.