Current-driving a powerful IR-illuminator array

Not at all, and this also lets you avoid unequal LED voltage drops while driving all the LEDs in parallel. You would have 40 resistor + LED branches all tied in parallel with a medium power MOSFET switching the whole thing ON/OFF, and a common voltage regulator with large output filter capacitor supplying the power.

You will be running only 500mA through the resistors at 3% duty cycle.

No- keep it simple.

Put the 40 LEDs in serie, so you only need one power supply, say

60..80V, with one resistor-in-serie (a few watts) and a MOSFET to switch ON and OFF.

Jacques

BW wrote:

I once used the following approach to drive 25 IR leds in sync with a camera frame sync: From a 12V power supply I drove 5 strings of leds, each string having 5 leds in series. Each led string was connected anode to +12V, cathode to the collector of an NPN transistor, the emitter was connected to ground through a 10ohm resistor. By driving the base from 5V, the circuit behaves like a current source of (5-0.7)/10 = 430mA. Need more current, increase the drive voltage. All bases were connected together though a 10ohm resistor. The total current draw is 2.2 A while each led gets 430mA.

Meindert

Thankyou, I guess the powerloss through the resistors is worth the gain in simplicity in this case. The other solution I was contemplating was to assume the hfe (current gain) of a darlington NPN was fairly stable, and I could feed a resistor-derived small known current into the darlington and get a fairly known high output current. But then again, I would have to contend with the unequal voltage drops in a LED series configuration and that could be worse (I hadn't thought about that) as well as temperature/batch varying hfe.

Any recommendation on a standard MOSFET suitable for this kind of power ?

Regards,

Bjorn

There is no such thing as a standard MOSFET. You want to select a MOSFET with RDS,ON such that RDS,ON*4 Amps< 0.05*(VDC,supply-VF,LED) supplying the circuit- i.e. contributes less than 5% error in your current calculation, it must achieve this RDS,ON with a gate-source voltage drive that you have available, and it's VDS ratings must be compatible with your circuit voltages. See Fairchild, International Rectifier, Philips and whomever selector guides to pick a candidate form the selector chart, and then iterate through availability from your suppliers.

I still toying with a system to flash Led's in sync with a camera. Edmund Optics makes a flasher, but its not heavy enough for my case.I don't see much of a problem making my circuit, even though its going to peak at 65 watts. I probably will use Fet switching to make the common connection at ground.

greg

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Transistor beta is notoriously variable and not suitable for that purpose. If you want to vary the current level, it would be better to make your DC power supply variable and adjust current that way. Put a big honking capacitor at its output so that most of the pulsed current derives from that and not the active elements of the supply. You could get away with a small 250mA capable supply that way. You want C so that

4Amps/C*1ms 80,000 uFarads-Volt of DC power; eg a 12 Volt supply makes from 80,000/12=6800uF capacitor or so- make it 10,000u at 16WVDC.

20 amps at 3% duty cycle is indeed 600 mA average current. But the RMS current, the thing wot fries resistors, is 3.5 amps.

John

JF,

the schoolmarm thing is annoying. His response was perfectly intelligible.

John

I designed a high efficiency LED flasher that has all the LEDs in series, so they all shared the same current. The pulse is formed (regulated and time limited) by an LC series circuit. The capacitor is precharged to a specific voltage, prior to each pulse by a flyback switching power supply. This capacitor is discharged through the LED string by a MOSFET, that turns full on for a little more than half of the duration of the pulse. But there is also an inductor in series with this discharge. A diode carries the inductor current through the LEDs, after the MOSFET turns off, till the inductor current runs down to zero.

The values of the capacitor and inductor determine the pulse duration, and the capacitor starting voltage sets the peak current.

You don't want to drive 40 LED/R strings in parallel - it's terribly wasteful. And, as you know, LEDs are current operated, and you can't depend much of Vf being matched. Do as others have suggested - use series strings, maybe 4 strings of 10 or 5 strings of 8, then drive each string with a current driver of some kind - for this, I prefer an ordinary NPN with a calibrated voltage on the base and an emitter resistor such that when (Vb - .7)/R = Iled. I've been known to use two series silicon diodes go get about 1.4V on the base, so then there will be about .7V across the emitter resistor - just use Ohm's law to give you the proper LED current (R = .7V/If). The NPN will regulate the current and the collector gives you the voltage compliance you need.

Like I said, for regulated current, I'd use an NPN, that can handle the voltage and current you need. I don't have a specific recommendation - I think the last one of those I did I used a TIP34C or so.

Have Fun! Rich

LED strobe lights should be readily available at much lower cost than building one yourself. For example in Australia, electronics hobbyist supplier Jaycar

has 2 white light models available ranging from AUD39.95 for a 24 LED model to AUD59.95 for a 50 LED model. To find the info on these units type SL2895 and SL2896 respectively into the product search box at the top left of the home page.

You might also find this project

interesting for photography.

He's not going to do 20Amps. That 500mA Ipk is for a 100us pulsewidth not exceeding 20% duty. The 50-100mA pulse is more realistic and operation can be continuous. I would not bother with RMS current computations. In the case of 40 LEDs in parallel he has 40*Iled*Vf total dissipation in the LEDs and 40*Iled*Vdc power delivered from the supply, leaving Iled*(Vdc-Vf) dissipation in each resistor, a peak multiplied by

1/30 for average power dissipation per resistor. If he goes to 8 strings of 5 LEDs then then each string looks like a single LED of 5*Vf at Iled making the peak power dissipation per string resistor Iled*(Vdc-5*Vf), or generally strings of N LEDs giving Iled*(Vdc-N*Vf)/30 average power dissipation per string resistor. Another misconception about these arrays is that somehow forcing identical current through each LED zeroes out the optical radiant power mismatch. It doesn't, and since the aging alone is -26% over the lifetime of the emitter, this error is much larger than the minuscule +/-8% 4-sigma spread in Vf, putting the 2-sigma at 4%, an insignificant number even for arrays as small as 40 LEDs.

Well, he said he was, about 20 lines up. I just wanted to make sure he didn't size the series resistors based on average current.

John

[...]

John, I'm missing something here. How does rms enter here, and how did you get an rms current of 3.5A?

Say we have a 40 volt PWM supply and a 2 ohm resistor. The current is 20 A.

When the duty cycle is 100%, the power dissipated in the resistor is 40^2 /

2 = 800 W. With 50% duty cycle, the power is 0.5 * 800 = 400 W, right?

So for 3% duty cycle, the power is 0.03 * 800 = 2.4 W.

P = E * I, so the average I = 2.4 / 40 = 0.6 A

But the duty cycle times the average current is also 0.03 * 20 = 0.6 A.

So where does rms enter into this?

Mike Monett

Ok I'm back after some actual lab-work ;)

I studied the current-source design with an NPN with an emitter resistor (Re) to GND and a string of 5 leds between Vcc (12V) and the collector. I tried to make the dimensions so that Ve is around 500 mV (to leave space for a Vce of over 1 V and the voltage drops across the LED's which is around 2.1 V per LED at the currents I'm interested in), so for a LED current of Ic=250mA, I chose Re=2.2 ohm. Given the drop over the base-emitter port of the NPN of 0.7V, I'd have to have a Vb of

1.2 V.

Somewhere here I realised that at a beta of say 30-40, I would have to support an Ib of around 10 mA (I actually measured this), which is quite much. The trigger source is 5V and I have difficulties of getting those 5V down to the required 1.2V. A resistor-based voltage-divider is not good enough since choosing small R forces a too heavy load on the trigger buffer, and choosing a high R makes the 15 mA create a too big voltage drop (pulling the NPN out of the "easy" feedback model). So I switched the transistor to a darlington NPN with beta > 750 giving an Ib of about 0.2 mA (measured). Still this did not create a sane configuration with the voltage divider model (the 0.2 mA creates a too big voltage drop nonetheless).

I also tried various combinations of 1N4148 strings to drop the voltage, but it does not work as I think it does (I get voltage drops of only 530mV over each 1N4148 can that be right?). I tried both simply putting the diodes in a series from the trig buffer (5V) down to the transistor base. I also tried a resistor from the trig buffer to the base, and a diode string from the base to GND, but this configuration didn't work as expected either :)

I'd be happy to have some thoughts on this seemingly simple circuit :) Perhaps adding a voltage follower (another buffer) at the input port to support the higher currents is enough ? I'll try that in the lab I guess...

Regards,

Bjorn

--
I wasn\'t commenting on the intelligibility, I was commenting on the
top-posting.  

I notice you don\'t, [top post] and have commented, on occasion, that
it\'s not a good practice, so what\'s your problem?

Simple solution would be: SMPS from your source voltage to 80V with 40 LEDs in series in line with switchable current source to the ground.