Electronic – Calculate the turn-off delay of an LED in parallel with a capacitor

capacitancecapacitorelectricityled

I am trying to create a Custom turn off delay for an LED. For this I take an Capacitor in parallel to an LEDenter image description here

My Idea so far was:
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I have the General Equation for an Capacitor. And the General Equation for Discharge. My U0=5V. The current is "controlled" by the LED. For the LED I assumed the Equation for an Diode Current. Now is the part where I assume, I did something Wrong. Because the Current Equation depends on the Voltage I inserted the Discharge Equation. Now I Put everything in the General Capacitor Equation. Because even WolframAlpha couldn't solve this Integral I assume I messed it up earlier. Can anyone help me?For this example -> I assume an Vf von 2V

//EDIT: Added Equation in Latex
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Best Answer

You have your differential equation wrong.

In your equation (1) you state \$U_c(t) = U_c(\infty) - \left( U_{c}(\infty) - U_c(0)\right) e^{-\frac{t}{\tau}}\$. This is the solution to a first-order linear differential equation. It's probably one you found for a simple RC circuit. Then you try to plug that equation into another differential equation, and realize you fouled it all up.

The correct relation is

  • \$i_d = -i_c = -C \frac{dv_c}{dt}\$,
  • \$i_d = I_{ss} \left( e^\frac{v_c}{n V_T} - 1\right)\$

Now you can do a simple algebraic substitution and get $$ C \frac{dv_c}{dt} = -I_{ss} \left( e^\frac{v_c}{n V_T} - 1\right) $$ or $$ \frac{dv_c}{dt} = -\frac{I_{ss}}{C} \left( e^\frac{v_c}{n V_T} - 1\right) $$

This is a \$1^{st}\$-order nonlinear differential equation. If you've forgotten your diff-eqs (or haven't studied them yet), it's not easily solvable (this is why we have simulation tools, after all).

We can persevere, however. It's separable, by doing some algebraic shenanigans:

$$ -\frac{C}{I_{ss}}\frac{dv_c}{e^\frac{v_c}{n V_T} - 1} = dt $$

Integrate both sides:

$$ -\frac{C}{I_{ss}}\int{\frac{dv_c}{e^\frac{v_c}{n V_T} - 1}} = \int dt $$

Now, I see that \$-1\$ in the denominator, and I know two things: it's going to be trouble, and \$I_{ss}\$ is so small that \$e^\frac{v_c}{n V_T} \gg 1\$. So I'm just going to approximate it away:

$$ -\frac{C}{I_{ss}}\int{\frac{dv_c}{e^\frac{v_c}{n V_T}}} = \int dt $$

This is easy to solve. I integrate both sides:

$$ -\frac{C}{I_{ss}}\left(-n V_T e^{-\frac{v_c}{n V_T}} \right) = t + t_C $$

Do a bit of algebra, and take the log of both sides:

$$ -\frac{v_c}{n V_T} = \ln\left(\frac{I_{ss}}{n V_T C}(t + t_C)\right) $$

And, finally, after skipping enough steps that I am not going to get 100% on my final exam:

$$ v_c = -{n V_T}\ln\left(\frac{I_{ss}}{n V_T C}(t + t_C)\right) $$

This will hold true-ish until \$v_c\$ drops low enough that \$e^\frac{v_c}{n V_T} \gg 1\$ no longer holds -- but that's going to be long after the LED is no longer emitting light, and the diode equation isn't holding up anyway. You just need to guess at \$n\$, find \$I_{ss}\$ from the LED current vs. voltage, and \$t_C\$ from the initial voltage. And then do it all over again when the temperature changes, or you change LEDs.