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This question was asked before, but I am struggling with the same question and I can't figure out "how to deal with it in the real world" as someone put it at the time.

Olin's approximation is the first idea I came up with: assume the LED is a voltage source, assume the capacitor initially charges up to 5V - Vled_drop, and assume a simple RC discharge. However simulating the circuit seems to show that the approximation is nowhere close to being valid. With this circuit: Circuit

If I assume a simple RC discharge model the simulator gives

RC = - 3s / Ln(V(3s) / (5V - Vs)) = 4.5

after 3s, when the calculated value would be

RC = 1.8mF * 1k = 1.8

(the simulator uses Vs = 1.64V for a red LED).

t = 0s t = 3s

In other terms the capacitor discharges much slower than the RC=1.8 would indicate. Is there any better way to predict what will roughly happen without a simulator, or without solving the actual differential equations?

Franck

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  • \$\begingroup\$ Why would anyone not use a simulator? \$\endgroup\$ – Andy aka Apr 9 '20 at 8:22
  • \$\begingroup\$ I tried that yesterday to tune a MosFET based astable multivibrator, and my conclusion is that when you don't understand the basics that doesn't really get you anywhere :) \$\endgroup\$ – Franck Apr 9 '20 at 13:16
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Your first RC formula is not right. You need to consider Vf=1.64V of the LED as an offset for the input voltages. This is the correct formula:

RC = -t/ln((Vt-Vf)/(V0-Vf)) = -3/ln((2-1.64)/(3.87-1.64)) = 1.65 Seconds

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  • \$\begingroup\$ That was my mistake indeed, thanks! \$\endgroup\$ – Franck Apr 9 '20 at 11:50
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The test circuit.

I haven't time to try to figure out your equations which seem to contain both variables and SI unit symbols (which makes it very difficult to read). I offer the following comments:

  • Assume that the LED Vf = 2 V.
  • When SW1 is closed C1 will charge up to about 2/3 of the way from VD1 to 5 V due to the R1 / R2 ratio - so about 4 V.
  • During discharge you can assume that Vf will remain fairly constant although it will decrease with current.
  • Your RC discharge is now calculated from 4 V to 2 V. So your RC time constant is τ = R2C1 = 1.8 s so you should see a discharge of 63% from 4 V to 2 V (1.26 V) in 1.8 s.

Try that out and see if it clarifies anything.

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  • \$\begingroup\$ Indeed, I had missed that the discharge is down to Vd, not to 0... thanks! \$\endgroup\$ – Franck Apr 9 '20 at 11:49

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