I've done some reverse engineering work on the protcol for a remote that... sucks. I'm replacing it with one of my own design, but while I've improved on the interface, the range and view angle of my system are terrible. I'd like to complete this as soon as possible... I, uh, fried my original remote while trying to debug my own. =]

So, like any good geek, I thought I'd borrow from someone else's successes and pulled up the circuit diagram for the TV B Gone:

TV B Gone Circuit

My question is, why have one resistor and transistor for each LED, instead of chaining the LEDs in series and controlling them with a single transistor which is controlled, in turn, by the arduino pin coming in through a single resistor?

I have no qualms about implementing the same way (frankly, I'm tempted to use about 32 LEDs, resistors, and transistors just for the hell of it,) but I'd like to understand why it was done this way.

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    \$\begingroup\$ I should ask a secondary question, though I think I'm just going to dig into their code to try to figure this out for myself: Why would there be 2 pins controlling the LEDs? The only difference between them is their viewable angle and range - they're all 940nm. \$\endgroup\$
    – user30997
    Aug 18, 2012 at 5:12
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    \$\begingroup\$ Another interesting design choice: they have no current-limiting resistors between Vcc, the LEDs, and ground. I looked at the data sheet, and the LEDs can handle a whopping 100mA of current, so I guess that they're simply not necessary if the board isn't going to provide that much power? \$\endgroup\$
    – user30997
    Aug 18, 2012 at 6:00

3 Answers 3


The forward voltage for an IR LED is much lower than for a visible light LED, typically around 1.3 V, but rising if you push real high currents through them, like > 100 mA. There seems to be no reason why you couldn't place two of them in series, especially if your Vcc would be 5 V. If your Vcc comes from a pair of AA batteries though, the voltage drop of two LEDs + the transistor's saturation voltage may come close to Vcc and that could limit the output current.

The two outputs to drive the four LEDs are to avoid overloading the microcontroller's output. Or better, should avoid overloading. A 120 Ω resistor means 35 mA base current per transistor, and that's too much already for the AVR, let alone the 70 mA which it will draw now.

The 2N3904 is not a good transistor for this either: it's only rated at 100 mA and the low hFE necessitates the high base current. A BC337-40 has an hFE of minimum 250 at 100 mA collector current, then 5 mA base current should be enough to drive it. A base resistor of 820 Ω will allow you to drive all four resistors from 1 pin. The BC817 is also rated at 500 mA.

Alternatively you could use a FET to drive the LEDs. A PMV20XN can handle several amperes and has an on-resistance of only 25 mΩ so it will dissipate hardly any power. 1.5 V gate voltage is sufficient for 2.5 A.

A note about current limiting. Usually we'll have a resistor in series with the LED for that, but if you look at the schematic of a commercial remote control that resistor is often missing, because they count on the batteries' internal resistance for that, and then they save another 0.001 dollar per remote controller.

This is not a good idea if you power from a mains powered voltage regulator. That will limit the current, but at a too high level, and if it doesn't destroy the LED immediately it will severely limit its lifetime. So a small series resistor is recommended. At 5 V supply and 2 LEDs in series you'll have a voltage drop around 2.9 - 3.0 V, so for 100 mA you need a 30 Ω resistor. Peak power will be 300 mW, but at a 50 % duty cycle average power is only 150 mW, then a 1/4 W resistor will do.

  • \$\begingroup\$ Wish I could upvote twice. Thanks. I'll be using your advice when I rebuild my own circuit. \$\endgroup\$
    – user30997
    Aug 18, 2012 at 7:31
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    \$\begingroup\$ @user30997 - Shall I post it again, so you can upvote a second time? ;-) Thanks for the accept! \$\endgroup\$
    – stevenvh
    Aug 18, 2012 at 7:34
  • \$\begingroup\$ simple emitter follower circuit would do the same thing. \$\endgroup\$ Aug 18, 2012 at 15:19
  • \$\begingroup\$ @sandun - Yes, but it has the same components, so there's not an advantage in it. The base current will vary with the collector current due to the change in forward voltage for the LED. In common emitter you don't have to watch out for that. \$\endgroup\$
    – stevenvh
    Aug 18, 2012 at 15:23
  • \$\begingroup\$ @stevenvh: Depending upon the V/I curve for the LED, I would think an emitter follower might eliminate the need for a resistor, but the 0.7-volt BE drop might make the circuit more sensitive to battery voltage. \$\endgroup\$
    – supercat
    Aug 18, 2012 at 18:51

Chaining the LEDs in series means you need a higher voltage supply to drive them all. And putting them in parallel may lead to issues if the LED characteristics are not nicely matched, or if your transistor can't handle the current of all LEDs at once.

They may have used multiple microcontroller pins for flexibility -- for example, this device now has the option of lighting fewer LEDs and therefore saving battery power.


It appears to me that the circuit is expecting the 3904's to limit the amount of current flowing through them to the correct amount for the LED. Since the transistor rather than a resistor is being used to limit current, and since each parallel-wired LED (or string of LEDs) requires its own current-limiting device, that implies the use of a separate transistor for each LED. I don't think I'd design a circuit that way, since it's sensitive to the beta of the 3904's, and transistor beta characteristics are normally not specified very tightly. Still, the circuit does have the advantage that current is somewhat less sensitive to VDD than it would be if it simply used a hard-switching transistor and then series resistors for the LEDs.

As for the use of two processor pins to control two separate LEDs, my guess would be that if the LEDs are pointed in substantially different directions the controller might be activating them at different times. Infrared remote signals typically alternate between 50% PWM and off. If during the "50% PWM" time one drives two sets of LEDs alternately, the required peak current would be cut by half. The one disadvantage would be that anything which only saw light from one LED would see a full-strength carrier wave, but something which saw some light from both LEDs would see a carrier wave whose strength was the difference in strength of the two LEDs' light. This factor could be mitigated by using e.g. a 25% PWM signal and having the two sets of lights operate on adjacent quarter-cycles. This would allow the use of higher LED currents, which would offset the receivers' reduced sensitivity to non-50% PWM waves. Further, a device which saw light from both LEDs would see a nice 50% carrier.

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    \$\begingroup\$ In most all battery powered remote controls the IR LED current is limited by the internal resistance of the pair of AA cells in series that provide the power. It is is not the transistor that intrinsically limits the current. This scheme has several advantages. It maximizes the available brightness to the IR LEDs without wasting battery power as heat in a resistor. Also it has the attribute to be able to continue to work as the battery voltage decays. In effect the remote control is able to more fully consume the battery capacity. \$\endgroup\$ Sep 11, 2012 at 21:43
  • \$\begingroup\$ That seems rather dangerous. A fresh pair of AA batteries has an internal resistance of under 0.3 ohms. If one were driving an IR LED which dropped 2.7 volts at one amp, fresh batteries would have no trouble putting that one amp through it. \$\endgroup\$
    – supercat
    Sep 11, 2012 at 21:49
  • \$\begingroup\$ What Michael Karas says is basically true. I have the PCB of an old Phillips TV remote in front of me right now. The [B]C33725 transistor driving the IR led is connected directly to the plus and minus poles of the two AA batteries (which are connected in series). There is actually a place on the PCB where a current limiting resistor could be mounted, but it's soldered straight through. \$\endgroup\$
    – Fizz
    Nov 2, 2014 at 8:57
  • \$\begingroup\$ @user3588161: Different LEDs and transistors have somewhat different voltage-drop characteristics. If a manufacturer is making a million of something, it may make sense to find parts which will behave suitably even if driven with a "rigid" 3.3 volt supply. Further, in many cases if one samples parts from a manufacturing lot, it may be possible to characterize their behavior more precisely than data sheets (behavior will likely vary more from lot to lot than within a lot). All that having been said.... \$\endgroup\$
    – supercat
    Nov 2, 2014 at 16:24
  • \$\begingroup\$ @user3588161: ...it may well be that there's enough of a margin between desired "on" current and maximum safe "on" current (given the pulse width) that merely limiting the size of the high-side output drivers would be sufficient to limit the amount of current passed through the transistors to a safe amount, even given a wide range of component variations. \$\endgroup\$
    – supercat
    Nov 2, 2014 at 16:28

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