I want to ask a question about a project I am making, I am using a seven segment. Here is Information about the seven segment: http://www.4project.co.il/documents/doc_188_54.pdf

I want to use this seven segment but when I read about it, it said that the maximum voltage allowed is 2.2 (V) also maximum current is 20 (MA) and I only have the ability to give it 5 (V), my question is : Can I give 5 voltage and use a resistor to decrease the current to 20 (MA), will it work or will it burn it? and is it OK to go pass the max voltage but not pass and max current,and the device will still work?

Please help, I have already burned two seven segment....yeah.

  • \$\begingroup\$ Hi Abed, glad you found the info really useful. I've just rolled back your edit because the question section is just for questions. You can always thank people by accepting answers (as you've done) and when you have a reputation of 15 you'll be able to up-vote the answers as well. \$\endgroup\$ – PeterJ Feb 11 '14 at 7:37

On the below schematic, D1 and D2 are segments of your (common anode) display.

When the switch S2 is closed, let's say you want 10mA through D2. The voltage across D2 will be around 2V (it's shown as 1.8 to 2.2V at 20mA, and it won't be much different at 10mA). So, VM1 will read 2V, give or take. To make the current 10mA, we make the value of R2 such that the current is 10mA. The voltage across R2 (VM2) is 5V - 2V = 3V.

Therefore, the resistor should be R = \$ E\over I\$ (Ohm's Law) = 300\$\Omega\$. You might pick standard 5% values such as 300\$\Omega\$ or 330\$\Omega\$.

The current is something you as a designer have to choose. Too high (too low a resistor value) and LED life will be reduced (sometimes drastically, as you have found). Too low and the LED will be too dim.


simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ The phrase you're looking for is "E24 values". \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 9 '14 at 18:23
  • \$\begingroup\$ Yeah, I know Ignacio, but that's less likely to be useful to someone who doesn't know to even use a resistor. Unless you're building millions, E96 values might make just as much sense, at least in 0603 resistors. With networks, I'd stick to E12 values. \$\endgroup\$ – Spehro Pefhany Feb 9 '14 at 18:30
  • \$\begingroup\$ Sure. But that's why you link to a page that explains them. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 9 '14 at 18:43
  • \$\begingroup\$ Thanks for the additional information. How about the power dissipation? \$\endgroup\$ – Spehro Pefhany Feb 9 '14 at 19:11
  • \$\begingroup\$ Power capability of a resistor is independent of its value. \$\endgroup\$ – Ignacio Vazquez-Abrams Feb 9 '14 at 19:53

Choosing the resistor correctly will stop your LED burning out and, by the looks of the specification this display is very lightweight when it comes to current. It says: -

  • Absolute maximum current is 20mA
  • At 20mA the LED segment may develop 1.8 volts across it

If you worked to these limits, and assumed your 5V power supply might be 5.5V, the value of resistor that protects the LED will be: -

\$ \dfrac{5.5V - 1.8V}{0.02A}\$ = 185 ohms.

This means your resistor should not be less than 185 ohms to produce 20mA through the LED with a little headroom on power supply variations and LED forward voltage drop variations BUT this is still only guaranteeing the current is 20mA and THIS value is an absolute maximum rating. Based on this, I'd probably add 50% to the resistor value and pick a 270 ohm resistor.

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  • \$\begingroup\$ Thanks, it really helped and also I want to thank you for taking the time to explaine this subject to me. I really appreciate it. \$\endgroup\$ – Abed Feb 11 '14 at 7:28

Just like any other LED, the individual segments of the 7-segment display also needs current limiting.

That LED you linked to in the PDF is a common-anode, which means the positive end is common to all the LEDs, and the negative end is individual.

You should use ohms law to figure out what resistor you need.

If you have a 5.0 Volt supply, and the LED is 2.2V @ 20mA, then the resistor must drop the remainder (5 - 2.2 = 2.8V) @ 20ma. Ohm's Law says that $$ R = {V \over I } = {2.8V \over .020A } = 140 \Omega $$

So you can use a 140 ohm resistor for each segment, and this will limit the current to 20 mA.

If you are turning all the LEDS on at one time, you need one resistor in each individual cathode. If you are multiplexing the LEDs with a microcontroller, then you can use one resistor only in the common anode of the 7-segment display.

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