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I have this circuit:

enter image description here

If D2 was a resistance, I could just combine it with R2 to get some composite resistance that I could put in series with R1. That way, I could calculate the resistance of the entire circuit and, thus, the total current.

Now, this is not the case here but I know a few things here:

  • whatever voltage drops at D2, also drops at R2
  • the current through R1 is the same as through R2+D2

But I do not know the current from the beginning, so I cannot calculate the voltage drop at R1. And the diode is non-ohmic, so I cannot find out anything about it.

In this way, I do not see how I could apply Kirchhoff's laws or any other methods because I seem to be missing information... where do I start here?

IceFire
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  • Although the diode is not ohmic, it still has a characteristics describing its voltage-current relation. – peterh Apr 28 '19 at 22:12

1 Answers1

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One good place is to start is LTL-307EE datasheet. From the datasheet take look at current profile for the device. Below is the forward current and voltage characteristics.

LED forward current and voltage graphs

Also take look at absolute maximum rating for the device.

Absolute Max ratings

Now you have to size R1 resistor such that no more than in this case 30mA of current flows through the LED. R1 is called a current limiting resistor. I would size the R1 to make ensure 25mA of current, with will provide Relative Luminous Intensity of around 3.0. Like wise R2 can be used to manage current LED current flow.

I hope this help you get started.

Mahendra Gunawardena
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  • Well, I know the usual (V_in-V_led)/I formula where V_led and I are the desired values. But does R2 not change everything? It seems to get more complicated that way since current splits according to resistance – IceFire Apr 29 '19 at 04:32
  • Yes, R2 can also be used to manage the current flow in the LED. Can you please give use the purpose of the circuit? What is V2? Is it battery or constant current source? – Mahendra Gunawardena Apr 29 '19 at 11:44
  • V2 is a constant voltage source. Purpose of the circuit is to learn electrical engineering with details – IceFire Apr 29 '19 at 12:46
  • I suggest selecting a forward voltage of 2V. The representative current is around 20mA . Ohm's laws is V=IR. so you can calculate the representative resistance. Replace the LED with a resistor. Also you know the voltage across the resistor. – Mahendra Gunawardena Apr 29 '19 at 16:23
  • I would like to know how to calculate exact values and what influence R2 has. Also, Ohm’s law does not apply to an LED, so what is the point in replacing it? – IceFire Apr 29 '19 at 16:25
  • @IceFire Replacing doesn't mean actually replacing the component, you have to find the work-load. That LED then, behaves instantaneously Ohmic, look at the graph, the Shockley equation is very stiff, during turn-on time, you can replace the exponent with a line or look at comments, replacing it with a resistor. – Sam Farjamirad Apr 29 '19 at 17:24
  • @SamFarjamirad ok, say, I use (12-2)/.002=5kΩ for R1, so that 2V is left for D2. U/I (ohmic) gives then 1kΩ for D2. Now, R2 is at 200Ω, so that R(D2) || R(R2) = 200kΩ/(200Ω+1000Ω)=167Ω. Total resistance of the circuit is then 167Ω + 5000Ω (R1) and applying Ohm's law now gives 23mA... now, I can analyze voltage across R1 again which is 5000*.0023=11.5V. But then, only 0.5 is left for D2, so R1 was a bad value. What is wrong in my process here or how can I come up with correct values? – IceFire Apr 30 '19 at 08:00
  • @IceFire I order to force 20mA through the R(D2) resistor and large resistor of something like 5K needs to be selected. Now R2 becomes a blocking resistor. From a practical sense you can use a potentiometer to control the intensity of the LED. – Mahendra Gunawardena Apr 30 '19 at 22:14
  • @MahendraGunawardena okay, I see! Then, I get the intuition. Practical is relative, in a complex circuit, I would not decide on resistances with a potentiometer. Some equation system with a Shockley equation should do the math here! Enough sub-questions for this question, thank you! – IceFire May 01 '19 at 13:52