How much current through an LDR?

[blandoon]

I'm doing a mod to a Morley wah that involves putting +9v through an LDR (CdS cell) - I don't have any datasheet info on the ones I have, but for this application I imagine it would have to handle about half an amp.

I'm wondering if it would be better to run the LDR to the base of a transistor or something like that, to essentially create a voltage-controlled resistor. Anyone have any ideas on the simplest way to implement that?

[rubberlips]

generally LDR's have a resistence of >1Mohm when in darkness. When they're in light about 3K-9K and 50K- 140K depending on which LDR you have. Even with the lowest resistence - say 3K, the current through it will be 3mA. This isn't determined by the current handling but because of the resistence of the LDR.

Now if it's to control current - using a transistor to vary the current would be the go

Pete

[blandoon]

What I'm trying to do is to add a CV output function to the pedal, for use with some modular synth gear... it's one of the older Pro-Series wahs which came in several flavors, with and without volume or distortion. But they were all built on the same circuit board, so by using some of the unpopulated volume circuitry I can add another LED/LDR pair to the board without having to disrupt the wah.

When I tested it just now I got little or no change in the voltage by trying to "darken" the LDR by hand, but I'm guessing it will behave better once it's back in the pedal and enclosed in darkness.