0
\$\begingroup\$

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.

Automatic Shutoff Switch

\$\endgroup\$
6
  • \$\begingroup\$ If you do that, what do you get? Surely it would be quicker than waiting some unknown time for an answer on this site. \$\endgroup\$
    – WhatRoughBeast
    Commented Sep 17, 2017 at 0:19
  • \$\begingroup\$ Random stranger agrees vgs will decay as linear ramp if diode reverse current is constant 2nA \$\endgroup\$
    – sstobbe
    Commented Sep 17, 2017 at 0:21
  • \$\begingroup\$ 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? \$\endgroup\$
    – Delta117
    Commented Sep 17, 2017 at 0:35
  • \$\begingroup\$ 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?? \$\endgroup\$
    – Michael Karas
    Commented Sep 17, 2017 at 0:55
  • \$\begingroup\$ 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. \$\endgroup\$
    – Dave Tweed
    Commented Sep 17, 2017 at 1:20

1 Answer 1

Reset to default
0
\$\begingroup\$

I'd have said your formula would be:

dt = (Ct * (Vbatt - Vth)) / IL

This gives a time value of 15.88 seconds

The circuit starts with a Vgs = Vbatt --> 5V and will discharge linearly from there to the 1.824 V, the FET Vgs(threshold) value. Note that the LED may appear off long before this point is reached, since at very low currents there may be no visible light emitted to the human eye.

If you set up an experiment with real components, who knows what time value you would actually measure. The capacitor may be =/-10 - 20% and the diode reverse current would be highly variable too.
Even if the Vgs(threshold) was very accurate, as said earlier, the LED would appear off while there was still some current flowing, even if you used a sensor to measure the light output.

In an experimental configuration it might be worth adding a high value resistor from Drain to Vbatt (perhaps 100 kOhm) and use an oscilloscope to measure when the V(Drain) rises to Vbatt - LED forward voltage level.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.