I'm trying to figure out what I'm doing wrong here. When I have a cadmium sulfide photocell unconnected I can get a resistance reading across it. When I connect it to a voltage source, I'm not getting any resistance reading whatsoever, but I can measure a voltage drop across it that is variable with light, so I know its still working. I've set my multimeter from the 2K ohms setting all the way up to 2 mega ohms and I still get the "1 . " reading when its connected to any voltage source whatsoever. If I disconnect the voltage source, I can get a resistance reading again.

What gives? Does the photocell have some property that I don't understand?

 

It's not the photocell, it's the ohmmeter function. To read a resistance, the meter has to run some current through it from its internal battery. A live component that already has current passing through it will be unmeasurable.

 

If you were using the old-style D'Arsonval meters with the needle set to read Ohms, applying the test leads to the powered circuit would result in the needle slamming hard to one of the ends of the scale. A fellow engineer jokingly called this effect "fast Ohmies"

Always measure the voltage across a circuit before you attempt to measure resistance. If you read any voltage at all, you should understand that you cannot measure resistance in at least that part of the circuit.

 

Presumably it is in series with a known fixed resistor. Measure voltage across that resistor, then calculate current using I=V/R.
The same current is flowing through the photocell so you can then measure voltage across photocell and use R=V/I to determine the resistance of the photocell.
Current will change with the amount of light and if you have anything else connected up that will affect the reading.

 

I guess I am a total newb. Thanks for the help guys.

What I'm not understanding, however, is why I could pull out the photocell, replace it with a fixed resistor, and then get an ohm reading WITH the voltage source attached.

 

You won't get the correct value for the resistor and there is a good chance that you will blow the fuse in your meter if there is power to the circuit.

 

Good to know. Thanks.

 

If you were able to measure resistance, then you did not have a complete current path through the two resistors. Current flow would've caused readings on Ohms to be out of limits.

The meter itself uses the internal battery to send a small current through the device being measured for resistance. That way, the resistance can be calculated due to the voltage drop/current flow.

 

Just for reference, there are some meters than can measure resistance in the presence of small DC voltages across a component. The one I use is the HP 3456 digital voltmeter -- it's called the offset compensated ohms measurement. The instrument measures the voltage across its terminals, then makes an ohm measurement, subtracting the first voltage reading. It was intended as a convenience when you wanted to know a resistance in the presence of e.g. small thermoelectric voltages. Anyone who has done troubleshooting with a regular digital multimeter and seen a measured negative resistance could make use of this feature.

I used one of these instruments in the early 1980's and fell in love with it -- but I knew I'd never cough up the thousands of dollars required to buy a new one. ebay changed that in the early 2000's and now I'm the happy owner of a used one in good condition. Except for its size, it makes a good bench multimeter, although I occasionally wish it also measured current.

 

I would like an all in one, bench top unit.

o'scope, logic analyzer, multimeter, and while im at it, function generator and counter, and if it could cook, even better.

Might as well have an isolated power supply thrown in for good measure.

All for under $400 would be a prize for any budding EE or hobbyist.