My plan is to use a Teensy logic output pin to switch a transistor to switch an infrared LED

The LED handles up to 1A peak current. I will use it as a computerized IR remote transmitter so the duty cycle will be very low.

The entire circuit will be powered by the USB connection. I don't want to add a second USB connection for more power, nor an external non-USB power supply.

Q: how much current should I drive the LED with?

I don't have a target LED current, I just want it to be as bright as is reasonable given the power supply.

I won't have any other outputs or big loads, just a SPI input from a temp sensor.

  • 1
    \$\begingroup\$ You can double the available current from USB (max 500mA) if you have 2 connectors connected to two USB ports (max 1A). That's what some external hard disk drives do. \$\endgroup\$ – avra Oct 3 '11 at 14:05
  • \$\begingroup\$ @Avra, your comment is the closest answer to my actual question. Pls make it an "Answer" so I can upvote it. \$\endgroup\$ – Larry K Oct 3 '11 at 14:11
  • \$\begingroup\$ @avra - He only needs 100 mA. He cannot use more than 100 mA. Even 100 mA is more than he should sensibly use in practice. (1 Amp is only allowed for 100 uS max with 1% duty cycle) \$\endgroup\$ – Russell McMahon Oct 4 '11 at 0:41


  • The maximum continuous current that the LED may draw is 100 mA.

  • If you use a linear power supply from the USB circuit you will not need more than 100 mA. At low duty cycles you will need proportionately less.

  • Operation at 100 mA continuous is liable to be risky.

  • If pulsing to 1A duty cycle of <= 1% is required = mean of 10 mA.

From data sheet:

  • The rated LED continuous current is 20 mA

  • The absolute maximum continuous LED current is 100 mA.
    Lifetime can be expected to suffer at this level - plus see below re thermal ratings.

  • The maximum pulsed current is 1A but
    this is at 1% duty cycle and with not more than 100 uS oulse.
    As 1A x 1% = 10 mA mean the max continuous rating is higher.

Note that the seem to be somewhat serious about the absolute maximum continuous rating.
ie some data sheet writers put figures like that BUT they are inconsisent with other figures in the datasheet.

In this case the consistency of the other data suggests they mean what they say OR that they have cut and pasted the datasheet from elsewhere and changed some figures. Less likely here. The datasheet IS cut and pasted between their other IR LED products, but that's legitimate **. There is however some overlap in thinking between various parameters suggesting that walking to near the data sheet limits will get you (or the LED burnt. (eg see below - dissipation at 100 mA is for 1% duty cycle but 100 mA is abs max continuous current.)

Absolute max current is at 25C ambient.
This derates at about 27 C/Watt (BOTE calculation from fig 1.)

Power dissipation in free air <= 25C is 150 mW.
Max Vf at 100 mA = 1.4V typical and 1.8V worst case.
Dissipation will be 140 mW and 180 mW in each case *.
This is around the 150 mW figure. The 100 mA dissipation figure is given for 1% pulsed output suggesting that they expect die heating to alter this adversely.


If the data sheet can be trusted, you may be able to operate this LED at say 80 mA continuous at 25C ambient in free air.
Thou may be able to pulse it at up to 1A at <= 1% duty cycle at <= 100 uS on time per pulse.

(* Ignoring the 1 mW or so that escapes as IR light)


Fig 8 - The word "relative" on the Y axis should be removed.

Fi6 6 - The radiant output / angle graph has been hand drawn and is not reliable. Nothing really looks like that. They were trying to fit the +/- 10 degrees half power points and then hand draw the rest.

** Cut and pastes:


This appears to be a cut and paste of Everlight ds..


  • \$\begingroup\$ I appreciate your time on this. But the question is not the minimum save current, but rather the max conservative amount given the USB pwr supply. \$\endgroup\$ – Larry K Oct 3 '11 at 18:01
  • \$\begingroup\$ @Larry K - The first line in my answer directly answers your requirement. The maximum mean current that this LED can ever legally draw is 100 mA continuous. Place a suitable smoothing capacitor at the 5V input (as you must do to make the design "safe") and you cannot ever "legally" draw more than 100 mA. That is the max conservative value and also the minimum safe value if you want to run this to its very limit. Answer: 100 mA. \$\endgroup\$ – Russell McMahon Oct 4 '11 at 0:39
  • \$\begingroup\$ Why can't I use more than 100ma? As I say in the Q, the purpose is IR control. Eg doc. The max steady on time is 9 uSec, if I read it right. The duty cycle is extremely low. (Send an IR cmd once an hour.) My thought is to go with 200ma. Doesn't the LED spec support this usage? \$\endgroup\$ – Larry K Oct 4 '11 at 2:29
  • \$\begingroup\$ @Larry K - Please reread my answer more carefully. It's well explained there, several times. eg "If pulsing to 1A duty cycle of <= 1% is required = mean of 10 mA." The USB power supply cares about the mean current as you MUST have a local filter cap to supply the surges (as I also note). The usage that the LED suppots is stated or implied by the data sheet. 200 mA is a grey area but you can be certain that it is not allowed for more than 50% and will actually be much less based on logical interpolation of provided data points. Thermal limit for "more than a moment" is 100 mA. \$\endgroup\$ – Russell McMahon Oct 4 '11 at 2:36

Since you will be pulsing the LED with limited duty cycle, your actual "power supply" for the pulses would be a capacitor on your board, not the USB, which would effectively "trickle charge" the cap in between pulses.

However, your micro isn't going to be able to drive more than a few tens of milliamps. To get higher drive currents you will need an outboard transistor or driver chip.

I'd be tempted to try breadboarding the whole thing and experimenting with what kind of drive current and circuit you need to get the desired range and performance (preferable while connected via a USB hub or to a computer you don't care that much about). If you have a scope, you can put a small resistor in the ground line of the LED supply/circuit and measure the voltage across that resistor to determine the instantaneous current.

  • \$\begingroup\$ Thanks. Sounds like you recommend a destructive test--trying different values until failure. I'd rather be much more conservative and just pick a safe value. As I said in the OP, I will be using a transistor to drive the LED. \$\endgroup\$ – Larry K Oct 3 '11 at 14:13
  • \$\begingroup\$ No, not at all. I'm recommending a functional test to figure out how much current you need. Since you'll be supplying the instantaneous current from a cap, you shouldn't challenge the USB port in the slightest if your circuit is working properly. The preference for a disposable test machine is on the off chance that at some point your circuit will not be working properly. \$\endgroup\$ – Chris Stratton Oct 3 '11 at 22:42
  • \$\begingroup\$ @Chris Stratton - No test needed. Data sheet provide all specs. 100 mA is the absolute max required. \$\endgroup\$ – Russell McMahon Oct 4 '11 at 0:42

You can double the available current from USB (max 500mA) if you have 2 connectors connected to two USB ports (max 1A). That's what some external hard disk drives do.

More info about USB power consumption is here.


You should be safe with 200ma pulses if you are using the linked RC5 protocol. In the course of one command transmission (25ms) the IR LED is on for 1/8 to 1/6 of the time (duty cycle), so if the max instantaneous on current is 200ma the average current is 25-33ma (depending on whether you use 25% or 33% duty cycle of the individual pulses; the other factor of 2 comes from the pulses being sent only 1/2 the time during a command). Commands repeat every 113 mSec so on longer time frames the duty cycle is even lower by another factor of 4.5 or so (~6-7mA average). That easily fits USB current drain as well as the LED's power dissipation.

As others have said, for brief 7 uS pulses it will mainly be your smoothing capacitor which provides the power, with the USB recharging the capacitor.

And yes, I'd test on a breadboard first. The proper resistor to get 200mA peak will depend on the specific diode (varying Vfwd depending on batch) and the particular transistor's voltage drop at that current when driven by your uC.


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