LED lights behave strangely, why?

Question

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.

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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.

Answer 1

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

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: -

Answer 2

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).

Answer 3

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.

Answer 4

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.