LED/LDR question

[zener]

http://members.shaw.ca/roma/wah.html

It says there in the modified version that the RV1 sets the gradual lighting of the LED while the RV2 sets the decay time.

The gradual lighting of the LED is like the pressing of the pedal in a wah and the decay is the releasing of the pedal, right?

Isn't the RV1 will just make the LED more dimmer or brighter depending on the setting? It's just in series with another variale resistor?

I breadboarded the circuit and used two 10ohm resistors for the RV1 and RV1a respectively. I did so because of that thought. As i engaged the switch, there's no gradual lighting of the LED, it just lit on immediately. But there is the gradual going off of the LED and it depends on the RV2 setting. I haven't tried putting a higher resistance for the RV1 and RV1a because I'm away from home until tomorrow.

Am I correct with my thought on RV1 and RV1a?

Thanks for any help.

Zener

[brett]

You're right - the LED lights to full brightness once the switch is closed.  The reason for that is that without a 5k or so resistor in RV1, the rise in voltage is too fast for you to see.  By the time the output of the op-amp is 2V, the LED is well-lit.  

You are right that RV1 and RV1a set the brightness, but with your small value resistor, the brightness is set so high that the LED hits max brightness really fast.  Try a minimum of 1k of resistance in series with the 1N4148 and LED.  (I don't know why the 1N4148 is there, but I suppose it can't hurt.  Maybe it's saving your LED from death by excess current?).

hope that helps a bit