Current-driving a powerful IR-illuminator array

[BW]

Hi! I want to drive a fairly powerful IR-illuminator array using
high-efficiency IR-LED's (Agilent HSDL-4230 to be specific), which can
support continous currents of 100 mA and peak currents of up to 500 mA.
I want perhaps 40 of these.. and essentially I want to flash them all
in sync to an electronic camera shutter of around 1 ms width, with a
duty-cycle of perhaps 1-to-30. Now when googling around for suitable
circuits, most refer to relatively small power demands, with LEDs that
use a current of only a tenth of this.. both with resistors and with
MAX-circuits etc.

Would it be crazy to try to get the right current by the old
resistor-in-series trick ? Obviously running 20 amps continously
through some resistors would be crazy but here the duty-cycle is so
high that on average the current is only 20/30 amps..

Is there a better way of say switching the LED array with some
darlington transistors and having an additional circuit that monitors
the current and adjusts the current into the transistors to regulate ?

Regards,
Bjorn

 

[Fred Bloggs]

Hi! I want to drive a fairly powerful IR-illuminator array using
high-efficiency IR-LED's (Agilent HSDL-4230 to be specific), which can
support continous currents of 100 mA and peak currents of up to 500 mA.
I want perhaps 40 of these.. and essentially I want to flash them all
in sync to an electronic camera shutter of around 1 ms width, with a
duty-cycle of perhaps 1-to-30. Now when googling around for suitable
circuits, most refer to relatively small power demands, with LEDs that
use a current of only a tenth of this.. both with resistors and with
MAX-circuits etc.

Would it be crazy to try to get the right current by the old
resistor-in-series trick ?

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.

Obviously running 20 amps continously
through some resistors would be crazy but here the duty-cycle is so
high that on average the current is only 20/30 amps..

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

Is there a better way of say switching the LED array with some
darlington transistors and having an additional circuit that monitors
the current and adjusts the current into the transistors to regulate ?

No- keep it simple.

 

[Jacques]

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

 

[Meindert Sprang]

Hi! I want to drive a fairly powerful IR-illuminator array using
high-efficiency IR-LED's (Agilent HSDL-4230 to be specific), which can
support continous currents of 100 mA and peak currents of up to 500 mA.
I want perhaps 40 of these.. and essentially I want to flash them all
in sync to an electronic camera shutter of around 1 ms width, with a
duty-cycle of perhaps 1-to-30. Now when googling around for suitable
circuits, most refer to relatively small power demands, with LEDs that
use a current of only a tenth of this.. both with resistors and with
MAX-circuits etc.

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

 

[BW]

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.

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

 

[Fred Bloggs]

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.

 

[GregS]

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.

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

 

[John Fields]

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

---
From:

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comment, but won't have to re-read the entire article.
And if your reply appears on a site before the original article
does,
they'll get the gist of what you're talking about."

 

[Fred Bloggs]

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.

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 <0.05*VDC or C>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.

 

[John Larkin]

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.

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

 

[John Larkin]

---
From:

http://groups.google.com/support/bin/answer.py?answer=12348&topic=250

"Summarize what you're following up.

When you click "Reply" under "show options" to follow up an existing
article, Google Groups includes the full article in quotes, with the
cursor at the top of the article. Tempting though it is to just
start
typing your message, please STOP and do two things first.
Look at the quoted text and remove parts that are irrelevant.
Then, go to the BOTTOM of the article and start typing there.
Doing this makes it much easier for your readers to get through your
post. They'll have a reminder of the relevant text before your
comment, but won't have to re-read the entire article.
And if your reply appears on a site before the original article
does,
they'll get the gist of what you're talking about."

JF,

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

John

 

[John Popelish]

Hi! I want to drive a fairly powerful IR-illuminator array using
high-efficiency IR-LED's (Agilent HSDL-4230 to be specific), which can
support continous currents of 100 mA and peak currents of up to 500 mA.
I want perhaps 40 of these.. and essentially I want to flash them all
in sync to an electronic camera shutter of around 1 ms width, with a
duty-cycle of perhaps 1-to-30.

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.

 

[Rich Grise]

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.

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.

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

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

 

[Ross Herbert]

Hi! I want to drive a fairly powerful IR-illuminator array using
high-efficiency IR-LED's (Agilent HSDL-4230 to be specific), which can
support continous currents of 100 mA and peak currents of up to 500 mA.
I want perhaps 40 of these.. and essentially I want to flash them all
in sync to an electronic camera shutter of around 1 ms width, with a
duty-cycle of perhaps 1-to-30. Now when googling around for suitable
circuits, most refer to relatively small power demands, with LEDs that
use a current of only a tenth of this.. both with resistors and with
MAX-circuits etc.

Would it be crazy to try to get the right current by the old
resistor-in-series trick ? Obviously running 20 amps continously
through some resistors would be crazy but here the duty-cycle is so
high that on average the current is only 20/30 amps..

Is there a better way of say switching the LED array with some
darlington transistors and having an additional circuit that monitors
the current and adjusts the current into the transistors to regulate ?

Regards,
Bjorn

LED strobe lights should be readily available at much lower cost than
building one yourself. For example in Australia, electronics hobbyist
supplier Jaycar http://www.jaycar.com.au/index.asp 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
http://xray.bmc.uu.se/markh/cam_trig/cam_trig.html interesting for
photography.

 

[Fred Bloggs]

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

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.

 

[John Larkin]

He's not going to do 20Amps.

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

 

[mike Monett]

[...]

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

 

[BW]

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.

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

 

[John Fields]

JF,

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

---
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?

 

[Slavek]

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.