# How to derive an equation for the delay time to LED shutoff?

The question asks to create an equation for the time delay for the LED to turn off in terms of capacitance(CT), reverse current (IL) of the 1N4148 and the threshold voltage (Vt) of the MOSFET for the circuit below.

The hint given was that we can treat the 1N4148 as a current source with a value of 2 nA.

From the experiment that I performed, Threshold voltage = 1.824 V, Capacitance = 0.01uF, resistance = 600Ω, yield the LED turn off after 8.71 seconds.

I feel like I would use the capacitor equation, i=C dv/dt, solving for dt, and plug in the values above. Is this a step in the right direction to solve this question.

• If you do that, what do you get? Surely it would be quicker than waiting some unknown time for an answer on this site. Commented Sep 17, 2017 at 0:19
• Random stranger agrees vgs will decay as linear ramp if diode reverse current is constant 2nA Commented Sep 17, 2017 at 0:21
• That's true, I did perform the calculations, dt = Ct * (Vgs - Vth)/(IL) and I managed to get 13.47 seconds, which is 35% off from 8.71 seconds. My only concern is this the correct way of solving this problem? Commented Sep 17, 2017 at 0:35
• Maybe you are working with the wrong threshold. Where did you get the 1.824V number anyway? Also have you considered that the gate of the MOSFET may contribute to the discharge of the capacitor as well and shorten the overall LED ON period?? Commented Sep 17, 2017 at 0:55
• The 35% error could be entirely due to the difference between the estimated and actual values of the diode leakage current -- or leakage elsewhere in the circuit. Commented Sep 17, 2017 at 1:20