0
\$\begingroup\$

I'm a beginner in electronics and I was studying when I had a question, how to calculate the voltage after passing by a resistor. Ex: there is a circuit of 10mA with a 5V voltage, what is the voltage in an LED, after the current pass by a resistor of 100 ohms?

\$\endgroup\$
2
  • \$\begingroup\$ Precisely what LED is it - is there a data sheet. Also, can you be clear, is the maximum current that can be taken from the 5V supply (short circuit current) only 10 mA? \$\endgroup\$ – Andy aka Jul 18 '16 at 11:21
  • \$\begingroup\$ Where does this 10mA come from? The actual current really flowing through the resistor and the LED will be determined by the characteristics of the LED (its voltage drop). Even if the LED is rated at 10mA, it doesn't mean that, with your circuit, 10mA will flow. \$\endgroup\$ – dim Jul 18 '16 at 11:25
1
\$\begingroup\$

According to Ohm's law, if 10 milliamperes passes through a 100 ohm resistor, then the voltage dropped across that resistor will be :

$$E = IR = 10mA \times 100\Omega = 1 \text{ volt.}$$

So, if you started with 5 volts, that means the remaining 4 volts will be dropped across the LED.

However, in the real world there's usually a little more to it than that.

To start with, if you want to drive an LED it's best if you get its data sheet and look at the forward voltage and forward current specifications.

What you'll find for a typical red LED is that when there's 20 milliamperes through it there'll be about a 2 volt drop across it. Be careful here because that doesn't mean that if you put 2 volts across the LED 20 milliamperes will flow through it, it means that if you force 20 milliamperes through the LED it'll drop about 2 volts.

So, your circuit will look like this:

enter image description here

and, from Ohm's law,

$$ R1 = \frac{3V}{0.02A} = 150 \text{ ohms}$$

\$\endgroup\$
3
  • \$\begingroup\$ Thanks, I was looking for this infotlrmation everywhere and I couldn't find it \$\endgroup\$ – jfinizolas Jul 18 '16 at 11:39
  • \$\begingroup\$ @jfinizolas be careful here - the answer given is strictly speaking true (in a sense) but when you connect an LED you might find that the voltage drops to more like 2 volts. To be sure, answer the comment I made below your question. \$\endgroup\$ – Andy aka Jul 18 '16 at 12:35
  • \$\begingroup\$ @Andyaka: You're right; I'll go back and edit my post for the real world. \$\endgroup\$ – EM Fields Jul 18 '16 at 13:10
0
\$\begingroup\$

Notice that LEDs do not behave like resistors but more like, well, diodes.

That's why it's harder to do calculations on voltage and current on them and why people recommend using constant current sources to drive them.

For learning Ohm's law, LEDs are not a good example. However, you can have a look at the datasheet of the LED. It might say something like

typical forward voltage Vf: 2V and typical forward current: 20mA.

From that information, you can say that the LED typically behaves like a resistor of 2V/0.02A = 100Ohm at that specific voltage and current. (Actual characteristics of a device may deviate from the typical values from the datasheet!)

You can expect the LED to draw about 20mA when you apply 2V to it. So if you have a 5V source, and you want to get 20mA out of it and you know that the LED wants 2V you know that you need to drop 5V-2V=3V via a resistor to reduce the 5V to 2V.

Hence, the resistor should be 3V/0.02A = 150Ohm. Ideally then, the voltage across the resistor is 3V, the voltage across the LED is 2V, and the current through both is 20mA.

This should work in experiments. For longevity of the LED however this is not ideal. As said above, LEDs behave like the diodes they are. This includes that the current through them strongly goes up as voltage rises. A simple resistor with 2.2V across it will pass 10% more current than it would at 2V. An LED @ 2.2V may pass 50 or 100% more current than it would at 2V which will destroy the LED quickly.

Because of that, and because an LED only has a typical Vf specified, you cannot rely on the actual Vf of it. For instance, a device with 2V typical might actually have 1.9V Vf. Driving it with 2V is already 5% over what it should really get and will be detrimental to the device's health.

\$\endgroup\$
1
  • \$\begingroup\$ "You can expect the LED to draw about 20mA when you apply 2V to it." is backwards. What you can expect is for the LED to drop about about 2 volts when 20 milliamperes is forced through it. It is, after all, a lossy diode. \$\endgroup\$ – EM Fields Jul 18 '16 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.