# fastest way of doing on/off-modulation of a LED?

I need to modulate a LED quickly (multi-megahertz range) on/off. It's a high-power LED. I have had some trouble locating any well-known methods to do this. Simply switching the voltage with a FET turns on the LED quickly but the fall-time will suffer, and to solve that, I guess there are some different solutions, for example maybe switching in a reverse bias for a short while? Any ideas?

I think the underlying problem with the turn-off is that the charge-carriers make the p-n junction act a bit like an inductor in that the current will keep on going for a short while after turning off the voltage gradient, but I haven't found any reference on this.

I know that a laser-diode can be modulated much faster.

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lasers are by no means comparable to LEDs (except that both emit light). The fastest laser yet is more than $10^{9}$ times faster than a LED. – stevenvh Jun 23 '11 at 0:17
@stevenvh: yes, that qualifies as "much faster" right? ;) – Bjorn Wesen Jun 23 '11 at 8:04
Why can laser diodes be modulated faster than regular light-emitting diodes? – endolith Sep 1 '11 at 5:26

If you are trying to send data this way, don't try to modulate it 0%-100%. Go 10%-90%, this will be way faster.

To switch it off rapidly, you need 2 transistors in push-pull configuration, PNP+NPN or N-MOSFET + P-MOSFET, so that in 'off' state LED will be shorted to ground. Achieving high speed with BJT would be easier.

If you need to go over 1-5Mhz, you will need to add anti-saturation Schottky diodes.

Another thing to try is bridge circuit out of 4 BJT's - it will eliminate remaining charge in LED even faster (as LED will be reverse biased in off state), but I haven't tried that. Some LED's might die if reverse biased too much.

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thanks for your answer, it was something like this I was experimenting with! additional question: what do you mean with 0-100 being slower than 10-90%? obviously the total rise/fall time will be slower, but if the rise/fall edge speed is the same, the eye area increases in total, and helps my modulation. why is a BJT-setup faster by the way? the lack of FET gate charge modulation? – Bjorn Wesen Jun 23 '11 at 8:04
Yes, driving MOSFETs at high speed is hard. About the 10-90 thing - rise from switch-off is slower than from 10% of brightness. So due to that in DVD-RW drives for example this approach is used - laser is never switched off, but in "off" state it's power is significantly lower. – BarsMonster Jun 23 '11 at 8:29
Also, seems that I was wrong about 90% - disregard that, 2 cases are 0-100%, and 10-100%, not 0-100 and 10-90. – BarsMonster Jun 23 '11 at 8:32

LEDs themselves take a little time to turn off, but I think a few MHz is still possible.

It sounds like your problem is the off time of the transistor used to switch the LED. Try driving the LED from the emitter instead of the collector. Logic output drives base of NPN directly, collector connected to supply, emitter to resistor, then to LED, then to ground. Since the transistor never saturates, it should turn off quickly. The base is being actively forced to a low voltage, which should also help turn it off quickly.

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 yes I'm at least going to try with a FET with a faster specified off-time - although simply moving the gate-charge is a delimiting factor as well so buffers are needed. do you think a BJT would actually work better in this case? I'm pulling 2A... – Bjorn Wesen Jun 22 '11 at 21:14 Are you implying that because it doesn't saturate, the slew rate has less effect on the switching time since $\Delta t= \left ( slew rate \right ) * \Delta V$ ? – Nick Halden Jun 22 '11 at 23:17 would one of those emitter switched bipolar transistors we saw the other day help perphaps? I remember they seemed to have faster switching times than normal NPN bjts Nevermind I guess they were for high voltage applications – Nick Halden Jun 22 '11 at 23:21 @JGord: I wasn't really talking about dV/dt. In this case we are switching current, not voltage. A saturated BJT takes a while to turn off because there are lots of extra minority carriers in the base. They hang around after the base current is shut off, causing conduction until eventually used up. A BJT never saturates in emitter follower mode, so can turn off quickly. No, a emitter switched bipolar would not be appropriate here. – Olin Lathrop Jun 23 '11 at 12:27

Have you considered using a "transistor driver" to drive your LED? (Or perhaps considered using a "transistor driver" the way it was intended to be used, to drive a transistor -- that then drives your LED?)

I'm talking about devices such as the Microchip MCP14628, the Texas Instruments TPS28226, etc., available at my favorite electronic supply websites, all of which the datasheet claims can switch a highly capacitive load in 10 ns. (Hopefully your LED is much less capacitive and so those chips can switch it faster).

p.s.: the datasheet for each transistor drivers gives a big-sounding number for "peak power". That number is only valid for very short pulses. LEDs often have a similar "peak power" rating about 4 times the continuous-power rating. I hear that most optical communication systems are carefully designed such the system turns the LED or laser on for at most one or two bit times before turning it off and letting it cool -- such as one-of-two encoding aka Manchester code, and one-of-four encoding aka PPM.

I hear rumors that some IrDA devices can communicate at 16 Mbit/s, 96 Mbit/s, or 1 Gbit/s. Is this close enough to what you want to do that you can buy something off the shelf? Or perhaps buy something off the shelf, crack it open, and make relatively minor modifications?

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there is a simple circuit for fast LED switching on this website. http://www.fiber-optics.info/articles/light-emitting_diode_led Haven't tried it but I am working on the same problem. need fastest turn off time after continuous operation

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