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I am new to electronics. What would be the loop equation for the following circuit?

How would I find the current through the LED and the silicon diode in the circuit?

Any help would be appreciated.

enter image description here

JRE
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nattweu
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    Diodes are nonlinear, so you'll end up with nonlinear equations. See [this](https://electronics.stackexchange.com/a/565393/95619), for example. – a concerned citizen Aug 05 '22 at 13:22
  • @aconcernedcitizen You could still just use KVL and ideal diode approximation which would result in one diode not conducting. You would need to guess which diode, calculate the numbers out and see if they are consistent. If not, you guessed wrong. – DKNguyen Aug 05 '22 at 13:53
  • @DKNguyen: Depending on the LED, one or both diodes will be conducting. If Vf for the LED is < 3V, it will conduct. LED Vf sets the voltage across the Si diode the 220 ohm resistor. LEDs always have Vf > Vf for a standard Si diode, because of the higher required band gap to give photons of the desired "color". – John R. Strohm Aug 05 '22 at 15:13
  • @DKNguyen True, you could use approximations, the same way you could use real diodes. Or just simulate it. A pity OP hasn't specified. – a concerned citizen Aug 05 '22 at 20:48

3 Answers3

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The simplest way is to just use KVL and ideal diode approximation then guess between one of the four combinations of diode operating conditions: [conducting, blocking], [blocking, conducting], [conducting,conducting], [blocking,blocking].

You then calculate the numbers out with that assumption and see if they are consistent with the operating conditions you guessed (i.e. is the diodes you guessed to be blocking actually reverse biased?) If not, you guessed wrong and you repeat the ENTIRE process with another guess. Do not reuse previously calculated values from other guesses. They are not valid.

You are allowed to eliminate obvious possibilities by inspection rather than going through the entire process.

[blocking,blocking] can be eliminated by looking to see if the voltage supply exceeds the Vf of at least one diode. If it doesn't, the other three possibilities are eliminated and you have your answer.

You can also check if the supply exceeds the Vf of one diode but not the other. That will narrow down your possibilities real fast.

If we continue and the supply exceeds both diodes' Vf, then you need to guess and check. KVL of [conducting, blocking] or [blocking, conducting] may show the case where neither case being wrong nor correct which means it falls in between at [conducting, conducting]. And if it does come to this then the use of the ideal diode model may not be as accurate as you need it to be, depending on the circuit.

So if you use your brain you can save a lot of work. If the circuit is too complicated or you simply do not know enough to eliminate possibilities, you can just run through every possibility until you find one that works. This method also works for transistors and their mode of operation too. But much messier and eliminations are tougher, but with 4 modes of operation per transistor you really want to put effort into eliminating things without math since just one extra transistor changes the possibilities from 4 to 16. Also works with mixed diodes and transistors.

DKNguyen
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    Because of the resistor in series with one diode, it's quite possible for both diodes to conduct at the same time in this circuit. – user253751 Aug 05 '22 at 15:07
  • @user253751 Hmmm, yes I suppose that is true. – DKNguyen Aug 05 '22 at 15:13
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    Given the LED's Vf is likely to be 0.5-1 V higher than the silicon diode's, conducting/conducting is almost certainly the intended operating mode of this circuit. – The Photon Aug 05 '22 at 15:31
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There is a approximately answer: there is no current between diode and LED.

LED voltage is approximately 1.5 V.

V(r100)=3-1.5=1.5 (V).

I=1.5/100=15 (mA)

Normally 15 mA is more than LEDs saturation current.. then answer is correct.

---You need LED V-I curve to solve this problem accurately.

winny
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mohsen Negaresh
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    A forward voltage of 1.5 V is not a typical value for a LED and the other current path is ignored here. – Jens Aug 05 '22 at 15:43
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This is one of those you can just eyeball, rather than writing and solving equations.

Assuming a Bog-standard red LED, Vf = 1.7V.

That leaves 1.3V across the 100 ohm resistor, for 13 mA through the whole thing.

A Bog-standard Si diode has Vf = 0.7V. That puts 1V across the 220 ohm resistor, or about 4.5 mA through it.

13 mA - 4.5 mA gives 8.5 mA through the LED. Red LED drive networks are usually designed for 10 mA through the diode, so it may be a bit dim.

John R. Strohm
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