# LED Current Calculation

Discussion in 'LEDs and Optoelectronics' started by krishna42099, Oct 10, 2012.

1. ### krishna42099

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Oct 1, 2012
Hi there, I've seen a green led, and its forward voltage drop is 3.2 to 3.8 volts given in the datasheet, current is 20 mA (typically) and max of 30 mAmps...now, the question, I've got zener diode of 3.3 volts, I would like to connect this led paralled to the zener diode...Do I still need a resistor in series to the LED and how can I calculate or predict the current through the LED???

2. ### duke37

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772
Jan 9, 2011
You need to define the current through the led not the voltage across it. A led acts as a poor Zener diode so why add another?

Decide on a current through the led, say 20mA. Subtract 3.5 from the power supply voltage and calculate the resistor, R=V/I

3. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
... Why would you like to do that? Is the zener diode already present in an existing circuit? If so, no, you shouldn't add the LED in parallel with it. Doing so will most likely affect the circuit.

It would help a lot if you would post your design as it stands, and explain what you want to do.

Last edited: Oct 11, 2012
4. ### krishna42099

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Oct 1, 2012
Hi duke, sorry I guess you didn't get my question...., this tym i'll present my question more clearly, ...I've been to a website looking for a green led, then I've got one, given that its forward voltage drop is 3.2 to 3.8 volts and max current is 30 m Amps.

Now say for example I've got A 3.5 volt battery, I assume that I can connect up this battery without any resistor (correct me if I'm wrong) to the LED...now if I can then how can I calculate the current through the LED??

5. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
You can't.

It will vary between LEDs, as temperature changes, and possibly over time (as the junction heats up)

You can look at the specs and take a guess at what the current will be and you may get close.

The current will also vary as the voltage varies. It will likely vary (as a percentage) by a greater amount than the variation in voltage (as a percentage)

6. ### krishna42099

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Oct 1, 2012
hi, this is just a part of the circuit, the led in this circuit is just for indication whether load is connected or not, so, if I add a resistor instead of a zener diode, the voltage drop on the resistor changes with the current in the circuit..but for a zener diode it stays constand...so there is always same amount of voltage drop on the zener diode no matter how much current flows through it...., then thats a green led with a forward voltage of 3.2 to 3.8 volts and zener diode is 3.3 volts..., do i still need a resistor in series with led or how can i do resistor calculation if its necessary..

I know generally to calculate resistor value..., (available voltage - forward voltage)/current through led, but in this case, available voltage and forward voltage is same, so comes 0/current which is equal to 0, in that case if I connect this led directly to the zener diode, then how can i predict the current through it???

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7. ### KrisBlueNZSadly passed away in 2015

8,393
1,272
Nov 28, 2011
You can't really do what you're trying to do. The zener voltage and the LED's forward voltage are close together, and will vary depending on various factors, so how much of the current flows through each of them will be poorly defined.

You haven't explained the purpose of your circuit. It looks like you're using a zener to drop the voltage applied to the load from 9V to around 6V, but that's just a guess.

If that's all you're doing, and you want to illuminate the LED when the load is connected but not otherwise, you can do this using an NPN transistor that receives base bias from the voltage across the zener, and supplies current to the LED in its collector circuit using a separate path to the battery positive.

Use a 2N3904, BC547B or similar. Emitter to battery negative. Base through a 10k resistor to cathode of zener. Collector through 1k resistor to cathode of LED; anode of LED to battery positive. The 1k resistor sets the LED current. Remove the LED that's connected across the zener.

8. ### duke37

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Jan 9, 2011
You now have a much more reasonable supply voltage.

Get rid of the zener, it only consumes power for no reason.
If the led voltage is 3.5 ( somewhere in the middle of the range) you need to drop 5.5V. A 270 ohm resistor would give about 20mA.

In your circuit, with a 3.3V zener (+/- 5%?) you will be dropping 5.7V across the 10 ohm resistor so needing over half an amp. Power will be over 5W so something will have to give, possibly all three components.

9. ### krishna42099

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Oct 1, 2012
thx guys, I did some research on the web and found out, the problem here with an led parallel to the zener diode which got the same forward voltage drop as led, the actual reason is at the same voltage, the led resistance is more than the zener, so current tend to flow through zener instead through led,..in this case I guess, if one connecting led over the zener diode, the led forward voltage should be less than the zener's voltage...then it works..

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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2,852
Jan 21, 2010
With a zener in parallel with a LED the voltage across both of them will be the same (by definition).

The current through them will vary in a fairly complex way according to the V vs I curve for each.

There is no current limiting, so without it, you risk destroying both devices.

If you have a current limiting device (say, a resistor), then the zener diode is doing nothing useful and should be removed.

11. ### krishna42099

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Oct 1, 2012
yup!   