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I have connected 10 LEDs in series, the LEDs have a current rating: 20 mA which is 0.02 A and forward Voltage: 2V.

My power supply provides me with: 5V DC 2A

By using the formula down below:

$$R = \frac{vs-vf}{i} = 5 - \frac{2}{20} = 150 \Omega$$

I can see that LEDs need 150 ohm.

Which is in total \$10*150 = 1500\Omega\$

I connected: 1 k resistor, 2 x 150 ohm resistor and 2 x 100 ohm resistor which is a total of: 1500 ohm.

But the 10 LEDs dont turn on, what's wrong?

I can still turn 3-4 LEDs with or without the resistors.

Try to explain this down below please:

I tried to connect a single green led by itself without a resistor. It turned orange which is a sign that it needs a resistor.

Then I connected 150 ohm resistor to the led and it became green. I then changed the resistor value from 150 ohm to 1 k ohm, but it did not change the brightness of the led, it still the same green, no fading in brightness.

I tried to change the value to 33k ohm, it faded in brightness, but I can still see some greenish brightness in the LED.

What's wrong? Why is the LED responding like that to the resistor's value? I mean 1k ohm is much more then 150 ohm but it worked the same.

P.S.: I bought the resistors from China via eBay.

JYelton
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user1022734
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  • 10 led in series would require a voltage of at least 20v (10*2v), that is not available in your system. You should use 5 parallel sets with two led in series each. Please refer to this reply for resistor calculation – alexan_e Jan 22 '14 at 10:07
  • But i have seen once small 5mm light led about 8 leds connected that use only 2 x aa battery which have 1.5v ech, how is that possible then ?? – user1022734 Jan 22 '14 at 10:10
  • A forward voltage drop of 2v at 20mA means that the voltage across each led will be 2v (for 8 leds that is 16v), connecting 8 leds in series and expecting them to light with 3v doesn't make sense. Using a lower current with a slightly lower Vf is possible but this will not help for 8 leds. – alexan_e Jan 22 '14 at 10:13
  • What could they have done?? I still have the light with me. What could the trick behind it be ?? – user1022734 Jan 22 '14 at 10:15
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    I don't think they are connected in series or there is a circuit that multiplies the voltage of the batteries. Make sure that you distinguish correctly between series and parallel connection – alexan_e Jan 22 '14 at 10:17
  • I Think Down in the circuit the leds is wired in parallel. But that dont explain why the leds behave like that to the resistor values?? – user1022734 Jan 22 '14 at 10:22
  • I think you are confusing series & parallel. If you have 10 LED's in series and you calculate your current limiting resistor is 150 Ohms then you do not need one per LED, you just need one. However, you do need enough volts across each LED, which will be 1/10th of the total voltage plus the drop across your resistor. – John U Jan 22 '14 at 10:23
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    The led should be used with either a resistor to limit the current or a driver that drives the led with constant current. i tried to connect a single green led by it self without a resistor it turn orange which is a sign that it need resistor. using the led like this will damage it very fast – alexan_e Jan 22 '14 at 10:59
  • I know, i was just explaining that and when you Read more you Will see i did end using resistor and tried 2 values, but they worked the same.. – user1022734 Jan 22 '14 at 11:02
  • 1k ohm is much more then 150 ohm but it Worked as same if you use an ammeter you will see that the led current is not the same, your eye can't judge the current – alexan_e Jan 22 '14 at 11:05
  • But isnet 1k ohm supossed to change the led brightness??? – user1022734 Jan 22 '14 at 11:15
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    The brightness was definitely different (as was the current) but they eye can't quantify the difference easily. If you use two leds side by side with 150 ohm and 1K you should see the difference. In any case the proper way to measure the current is by using an ammeter. You can also measure the voltage drop across the resistor and calulate the led current by using Ohms law – alexan_e Jan 22 '14 at 11:25

4 Answers4

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Here is the forward conduction characteristic of a typical LED: -

enter image description here

At 2V forward voltage the LED consumes 20mA and is quite bright. At 1.8V forward voltage the current has fallen to 5mA and the LED will be somewhat dimmer than the 20mA scenario. At 1.6V the LED is taking almost zero current and below this voltage it is doubtful that it illuminates at all.

If you have 10 similar LEDs in series, to obtain a forward current of 20mA requires a forward voltage of 20V.

If you are happy with a forward current of 5mA then 18 volts is all that is needed. And, if your power supply is (say) 20V, a series resistor is needed that drops 2 volts at 5mA therefore its resistance will be about \$\dfrac{2V}{5mA}\$ = 400 ohms.

If you have a limited supply voltage, you can wire the LEDs in parallel banks or use a boost converter to lift your 5V to about 20 volts. The CAT4238 device from Motorola can do this: -

enter image description here

Andy aka
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Your calculation of \$ R = \frac {vs-vf}{i} = \frac {5v-2v}{0.02A} = 150 ohm \$ is for 10 individual resistors that should be connected as shown below (leds in parallel).

enter image description here

alexan_e
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Forward voltages add when you wire LEDs in series. You need a 20 volt supply to run your 10 in series. With a 5 volt supply you can run the LEDs in 5 parallel pairs, using either a single 1/2 watt 250 ohm resistor at one end of the circuit, or smaller wattage resistors in each circuit branch.

Wayfaring Stranger
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I will address the part of your question about the different resistance values having no or little apparent effect:

"Ultra bright" LEDs today are remarkably efficient and can still appear quite bright at currents well below the typical or maximum stated on the data sheet.

So, let's consider just one green LED which has a forward voltage \$V_f\$ of 2V and a nominal forward current \$I_f\$ of 20mA.

With a power supply voltage of 5V, we could calculate the resistor value needed for 20mA as:

$$R = \frac{3}{0.02} = 150\Omega$$

But what if we wanted to drive the LED with only 1mA?

$$R = \frac{3}{0.001} = 3k\Omega$$

The LED might not illuminate with so little current, but many LEDs (for example, these inexpensive Panasonic indicator LEDs) will start to illuminate with as little as 2mA or less. For 2mA, the calculations above would give us a resistor value of 1.5kΩ.

If you exceed the maximum forward current (20mA), the LED will be damaged. One sign of impending damage is a green LED illuminating orange or yellow. That color is an indication that the LED is being driven too hard. (Too long at such currents and the LED will stop working altogether.)

(Check the data sheet for whether you can exceed the nominal forward current for various duty cycles. For example, many LEDs which have a nominal \$I_f\$ of 20mA can be driven at 40mA or more with duty cycles of 10% or so.)

If you connect the LED to a variable resistor (a potentiometer) you should be able to see a nonlinear brightness increase as you decrease resistance from 1.5kΩ to 150Ω To ensure that you don't damage the LED, connect a 150Ω resistor in series with your potentiometer. When your potentiometer is at 0Ω, your LED will be at 20mA.

JYelton
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