If i have 20uF capacitor with charging 3V . Now i m gonna attach 1 ohm resistor and 2 V led in series with capacitor . Now current will flow through the led is 1Amp max For a while , but if 1 Amp current will flow through the led it must be burn out theoretically so why it will not burn out practically ?
closed as unclear what you're asking by pipe, Daniel Grillo, Voltage Spike, pjc50, dim Oct 20 '16 at 12:49
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You may not know this but a 5mm LED rated for 75mW has an internal ESR of about 15 Ω so adding 1 Ω is not significant. ( but at 3V ESR is probably closer to 10 Ω then rises above 20 Ω below 1.9V )
Recalulating without 1 OHm
- Assuming Vf=2.1 (Red)
- Vi= 3V on 20uF cap
- Imax= (3-2.1V)/15Ω =60mA
calculating light pulse Decay time to 1.8V (dim)
- If T = 20uF * 15 Ω = 300 µs decay time to < 40% of 3V
- even though LED turns off below 1.8V
- this will not be visible
You're assuming a simplified model of the forward biased diode (constant Vf forward voltage in series with constant "on resistance"). This model leads you to believe that the initial current would be (3V - 2V) / 1ohm = 1Amp. Your maths are correct but the simplified diode model is giving you an incorrect result.
A general procedure for estimating how a source and a load will interact, is to plot a graph of the source's V/I characteristics and the load's V/I characteristics, and see where the two plots intersect.
A real diode follows an exponential curve. Check the datasheet for your "2V LED" (which I assume must be a green LED). Most likely the datasheet specifies Vf=2V at a current of maybe 10mA or 20mA. From this Vf,If point you can determine the exponential curve values. (This is assuming the Shockley diode equation and a constant temperature.)
Draw an exponential curve through that given Vf,If point, and you have the characteristic V/I curve of that diode as a load.
Next, draw another line on the graph, representing the source characteristic of the 3V source. You're only asking about what happens initially, so for now let's not worry about how fast the capacitor discharges. The source V/I curve for a 3V source in series with 1 ohm, is a straight line that passes through (3V,0A) and also passes through (0V, 3A). In other words, if you shorted out the 3V capacitor through 1ohm without the load, then the initial short-circuit current would be 3V/1ohm = 3Amp.
The place where the diode's V/I curve intersects the 3V source's V/I curve, is the point that represents the initial voltage and current that will flow when the 3V 1ohm source is connected to that diode.
As a minor addition to Tony's answer. The energy stored in the capacitor is probably also not sufficient to generate enough heat to destroy the diode.
The energy stored in the electric field of the capacitor is:
W = 1/2 * C * V²
W = 1/2 * 20uF * 3V² = 0,0009J
That's not much, but in the end it also matters how fast this energy is discharged.
A current of 1 A will not directly destroy your LED attached to the capacitor. But it is still recommended to stay around 20 mA (if it is a red one of 5 mm).
It is not that because it works for a while that it will keep working.
Besides, there is no interest to keep a LED at 1 A. The luminous intensity (cd) does not upgrade as much as from 0 to 20 mA.