5
\$\begingroup\$

I apologize for the very basic question but I seem to be lacking some fundamental knowledge.

I am using an infra red LED in a circuit and it seems like it is drawing around 15mA at 5v. I was told today that I am running a chance of burning it since infra red LEDs require 1 to 2 volts as well as I should consider adding a resistor to protect it from higher currents. I got confused because my understanding of current is that the consumer draws as much as it needs and the only requirement for the power source regarding current is that it must supply at least that. For example if my power source is a 5v 1Ah battery the current flowing would still be 15mA regardless of the higher capability of the battery. I got further more confused by the second part. I thought resistors affect the EMF, not current, how can I calculate what resistance I need in order to limit current to 15mA?

\$\endgroup\$
  • 6
    \$\begingroup\$ There are literally hundreds of posts here of how to calculate a series resistor for a LED. Looks these up. IR or not IR does not matter. \$\endgroup\$ – Eugene Sh. May 27 at 17:14
  • 6
    \$\begingroup\$ @EugeneSh. The point of the question is more about understanding whats going on, than just about calculating a series resistor for a LED. And the question is a good real world example why beginners sometimes get confused. \$\endgroup\$ – Rev1.0 May 27 at 17:49
  • 5
    \$\begingroup\$ I would rephrase like this: A consumer will draw the amount of current it demands at a particular voltage. Most consumers of power, especially engineered devices with internal power supply circuitry, are designed to be relatively insensitive to small changes in voltage. A "5V" USB device is engineered to draw similar amounts of current at 4.8 or 5.2 volts. Diodes (including) LEDs, are an extreme exception. A diode's current-voltage relation is exponential, so a small change in voltage results in a HUGE change in current. (This kills the LED.) \$\endgroup\$ – Glenn Willen May 27 at 19:10
  • 3
    \$\begingroup\$ In principle if you had a perfect LED and a perfect power supply, you could power an LED with just a regulated voltage. We never do this, because (1) LEDs are not perfect (each one is slightly different); (2) power supplies are not perfect (it's hard to regulate to the precision required); (3) all your components change slightly with temperature. So regulating voltage accurately enough to control LED brightness without destroying the LED is impractical, and instead we have to focus on regulating the current. The cheapest way to do this is to add a resistor. \$\endgroup\$ – Glenn Willen May 27 at 19:12
  • 4
    \$\begingroup\$ Possible duplicate of How can I efficiently drive an LED? \$\endgroup\$ – Dmitry Grigoryev May 28 at 11:21
11
\$\begingroup\$

I am using an infrared LED in a circuit and it seems like it is drawing around 15 mA at 5 V.

That is unlikely unless it has a built-in resistor.

enter image description here

Figure 1. The I versus V curves for a range of typical LEDs shows that the forward voltage of an IR LED is about 1.25 V at 20 mA. Source: LED IV curves.*

Note that if you were to apply even 2 V to the IR LED that the current would exceed 100 mA and it might not last long. At 5 V a bare IR LED would almost certain die instantaneously.

I was told today that I am running a chance of burning it since infra red LEDs require 1 to 2 volts as well as I should consider adding a resistor to protect it from higher currents.

A better explanation would be that you need to limit the current through the LED to a safe value. A resistor is the simplest way of doing this.

I got confused because my understanding of current is that the consumer draws as much as it needs ...

Yes, but the graph is showing us that at 5 V it "needs" a very high current.

... and the only requirement for the power source regarding current is that it must supply at least that.

That's correct (subject to my previous comment).

For example if my power source is a 5 V 1Ah battery the current flowing would still be 15 mA regardless of the higher capability of the battery.

Only if there is a current limiter in the circuit. Again, a resistor would be the normal solution.

I got further more confused by the second part. I thought resistors affect the EMF, not current, ...

The resistor will both limit the current and create a voltage drop.

How can I calculate what resistance I need in order to limit current to 15 mA?

There are many, many sites on the Internet that explain this. The load-line tool below may be of interest.

enter image description here enter image description here

Figure 2. A loadline graphic calculator for various LEDs on a 5 V supply. Source: Loadline resistance graphic tool.

To use the tool:

  • Select your LED colour: IR in your case.
  • Select the current you want: 15 mA in your case.
  • On the If axis move up to the 15 mA line and move across to the IR curve.
  • Select the nearest loadline curve: 270 Ω.

A 270 Ω resistor will limit the current through your IR LED to about 15 mA on a 5 V supply. There will be 1.25 V across your LED and about 3.75 V across your 270 Ω resistor.

\$\endgroup\$
  • \$\begingroup\$ So if I had a 1V power source even if it was capable of supplying 100A theoretically it would be safe to connect the LED without a resistor? \$\endgroup\$ – php_nub_qq May 27 at 18:25
  • 1
    \$\begingroup\$ Yes, it would be safe but pretty useless. Can you see from the graph how much current would flow? I show how to measure it on the video at the bottom of the page. You might also find What is an LED useful. \$\endgroup\$ – Transistor May 27 at 18:39
  • \$\begingroup\$ Yes I think I understand now. I meant more like 1.25V but I typed 1, my bad. \$\endgroup\$ – php_nub_qq May 27 at 18:45
  • \$\begingroup\$ It can work but you can run into problems with variations in the forward voltage, \$ V_f \$. I've written on this too in Variations in Vf and binning. \$\endgroup\$ – Transistor May 27 at 22:15
  • \$\begingroup\$ I have one final uncertainty. On a raspberry pi I can use a GPIO pin to power the LED, the voltage will be 3.3V but the specs say that the maximum current a single pin can provide is 16mA. What will happen in this case? \$\endgroup\$ – php_nub_qq May 28 at 6:24
12
\$\begingroup\$

Instead of saying the consumer (circuit) draws as much as it needs, it is better to say current will flow as much as is allowed by the circuit.

How much current a circuit will allow is based on its resistance (or impedance). An LED has a voltage drop, but very little resistance. This small resistance means that more current will flow through it than is healthy.

A resistor is an easy and common way to limit this current. There are many many examples on this site alone on how to do that. Search for "LED current limiting".

To get you started, use Ohm's Law to calculate the resistor you want:

Subtract the LED's voltage drop from your source voltage. This is how much voltage you need to drop across your resistor. Now use the amount of current you want to pass through your LED to calculate (using Ohm's Law) the size of resistor that will do this.

\$\endgroup\$
2
\$\begingroup\$

I got confused because my understanding of current is that the consumer draws as much as it needs and the only requirement for the power source regarding current is that it must supply at least that.

This is an extremely common misunderstanding. There is nothing fundamental about electronics that makes this case. It just happens to be a property of some devices.

For example you can buy two 120V light bulbs and be assured that ever if they are different brightnesses, they will both run correctly on 120V. But this is because you bought 120V light bulbs.

You could imagine an alternate universe where you bought two 1A light bulbs. You could be assured that they would both run correctly on 1A even if they were different brightnesses, but they would operate at different voltages. In this world, we might have 1A power supplies that provide whatever voltage the device needs, say up to 500V. In this world, so long your supply was the right current and its maximum voltage was sufficient, the device would work safely.

So "the consumer draws as much as it needs and the only requirement for the power source regarding current is that it must supply at least that" is an attribute of devices that are specified to work with a specific input voltage. An infrared LED is not such a device. In fact, it has a specified forward current that you are expected to supply and the voltage will be whatever is required to drive that current.

\$\endgroup\$
0
\$\begingroup\$

my understanding of current is that the consumer draws as much as it needs and the only requirement for the power source regarding current is that it must supply at least that.

No. That is a convention/standard to which you are expected to build, for most applications.

The convention is a constant-voltage system in which your equipment must self-limit current. That is the standard (instead of constant-current where you self-limit voltage drop) because:

  • Due to transformer and battery technology, it is easier to build constant-voltage distribution than constant-current. (Though honestly, transformers could go either way).
  • Transformer based constant-current systems develop scary-dangerous voltages when driving into insufficient load, and that creates a lot of problems, least, insulation.
  • Items are added to constant-voltage systems in parallel, which is a heck of a lot easier to implement than series connections.

However, constant-current mode is perfectly respectable now that electronic sources are able to limit open-circuit voltage... and highly appropriate for LEDs particularly. In fact, it is preferred to drive LEDs in constant-current mode.

However, it's not new. For instance, arc-discharge lights (arc, fluorescent, neon, HID, Jacobs Ladder) are basically a dead short after they strike, and they have always had to be driven in constant-current mode to keep them from immolating.

So you would run in constant-current mode inside the equipment you are building. That equipment will ultimately draw from a power source such as battery or mains, and that will be a constant-voltage system most likely.

\$\endgroup\$

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.