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I don't understand our professor's explanation for the answer to this problem:

An LED is connected in series to a 150ohms pull up resistor and a MOSFET with drain-source resistance of 3ohms. The Pull up resistor is connected to a 5V source and the MOSFET is connected to ground (say MOSFET is turned on i.e. the gate). Now if there is a 2V drop across the LED, what is the current flowing through the LED.

I calculated the LED's resistance to be 102ohms giving the current through the LED to be 19.6 mA (which is correct).

But my professor says that the LED has no resistance at all and that my method of calculation is wrong (despite there being a voltage drop). And he said something about thevenin resistance which I did not understand.

Do you know what he is talking about?

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What he says is that basically your LED has resistance 0 because in his model, if you drain 100 kA it will still drop 2V...but it's false: it will shine like the sun before burning :). –  clabacchio Mar 30 '12 at 14:30
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2 Answers

The idea is this: suppose that in our over-simplified model, the led drops constantly 2 V no matter what, so it has resistance =0, because dV/dI = 0.

Then, you know that the other resistors must drop 3 V, which gives 3/180 = 16.6667 mA.

The parallelism with Thevenin is that if you represent your LED with the equivalent Thevenin circuit, you have Vth = 2V and Rth = 0 Ohm.

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The Thevenin resistance is the amount by which something's voltage will change if the current changes by a certain amount, and he is assuming that the voltage drop across the LED will be 2.0 volts whether one puts one microamp or one mega-amp through it. If the voltage across the LED doesn't change at all with respect to current, then the effective resistance is zero.

In practice, the voltage drop on real LEDs isn't constant. A somewhat more realistic LED model might drop 1.8 volts plus 0.02 volts/milliamp (so at 10mA it would drop 2.0 volts, at 20mA it would drop 2.2 volts, etc.) If one side of an LED is connected to a fixed voltage source, the other side would have a Thevenin resistance of 20 ohms (20 volts/amp). Note that with any sort of real LED, the Thevenin resistance will change with current. If it were fixed, an LED whose behavior was as described above (assuming infinite precision on the above numbers) would drop 1.80000000002 volts with one picoamp flowing through it, and would drop precisely 20,001.8 volts with one mega-amp flowing through it. In practice an LED with only a picoamp flowing through it would drop almost no voltage, and an LED with a mega-amp flowing through it would have an essentially unmeasurable voltage drop (since it would be an unenclosed cloud of plasma).

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Your answer is correct except for the picoamp part: diodes and transistors have logarithmic voltage vs. current curves for small currents. –  Jason S Mar 30 '12 at 19:38
@JasonS: For moderately small currents I would expect that. But by the time one gets to the picoampere range, I'd expect some secondary linear effects to show up again. For example, if contamination creates a parallel resistance of 100 gigohms, the voltage drop could be at most 100 millivolts per picoamp. Perhaps some LEDs are so well made that there's less leakage than that, but I'd be surprised (note that 100 gigohm is a very high resistance--100,000,000,000 volts per amp; very few insulators are that good). –  supercat Mar 30 '12 at 20:36
ah, good point. sometimes I forget about things like that. :-) –  Jason S Mar 31 '12 at 2:44
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