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I would like to pump 4-5 A to a high-power LED for 100 µs. My system has only a 3.3 V battery, and this 100 µs high-power event takes place once every 10 seconds.

What is the best way of doing this without upsetting the batteries?

The answer below is very good. However, I am looking for a schematic I can use and test out.

More exact requirements:

  • Battery: Li-ion
  • Current 5 A
  • Pulse Duration: 100 µs
  • Pulse Rise time <100 ns
  • Minimum time between pulses 10 ms
  • Pulses are controlled with a 3.3 V I/O GPIO from a controller
  • The voltage drop across the LED is 3.5 V. Ideally I like to be able to put three or more in series (10.5 V voltage drop)
  • Datasheet to the LED

Bonus question

If you have a better LED recommendation with a large angle that is in the invisible range, please let me know.


I have implemented this project, and it works well except the leakage current. No matter what I tried, I couldn't get rid of the leakage. I tried a few types of opamps add a pull down resistor to the output of the opamp, etc. I end up turning off the opamp to cut the leakage. It works, but it is not very neat. I would be happy to hear what experts think of the situation.

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    \$\begingroup\$ What type of battery? \$\endgroup\$
    – stevenvh
    Commented Apr 13, 2012 at 7:18
  • \$\begingroup\$ Just for curiosity: why do you need to pulse a high power LED for a so short time? \$\endgroup\$
    – clabacchio
    Commented Apr 13, 2012 at 7:46
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    \$\begingroup\$ @clabacchio I will take a picture in that time.. it is an infrared flash \$\endgroup\$
    – Ktc
    Commented Apr 13, 2012 at 9:20
  • \$\begingroup\$ @stevenvh li-ion rechargeable battery. energy capacity unknown at this time. \$\endgroup\$
    – Ktc
    Commented Apr 13, 2012 at 9:21
  • \$\begingroup\$ Awesome! I'm also curious about how you will take the pic :) \$\endgroup\$
    – clabacchio
    Commented Apr 13, 2012 at 10:01

4 Answers 4

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This is the most efficient way I can think to do it. There's a MAX1682 charge pump to give you 6.6v at the super capacitor. The voltage doubler is pretty efficient, probably more then 90%, but they can't supply huge currents. But what's the average current?

5A * 100us / 10s = 0.05mA.

That's well within the MAX1682's 45mA spec.

From a brief look at the datasheet, I couldn't see any reason it wouldn't work with such a large capacitor for C2.

Thanks to Russell McMahon for advice about charge pump efficiency. It looks like an inductor based solution would be more efficient, but would require more components. Take a look at something like MAX17067. This also has the benefit that it can produce the higher voltage required by three LEDs in series. I'll add it to the schematic tonight.

Flash 1

Now the important bit. You'll notice that there's no current limiting resistor. The current limiting will be performed dangerous open-loop style by the MCU. You'll have to get this right by calculation or trial and error (or both).

By supplying PWM to Q2's gate, you will be able to use the inductor as an efficient current limiter. But you won't get a very reliable current this way. It may not matter hugely, as long as 1) sufficient power is delivered to the led in 100us, and 2) the LED's current limit is not breached.

Here's a simulation I did in Altium. I used a 5uH inductor (not the 10mH shown in the schematic). And I supplied PWM with 12us on time, and 3us off time to the gate. I didn't use the 100uF capacitor, just a fixed voltage source instead. So you could expect some current droop.

Current control with inductor and PWM

Red is the current in amps, and blue is the PWM signal. You can see that you get close to 5A within 20us, and stay pretty close to it after that.


If you want better current regulation, then you can add a sense resistor, and use it to feed back to the MOSFET.

Flash 2

Here we have a 0.5ohm current sense resistor. At 5A, this should give us 2.5v to the comparator negative input. This is compared against the value from the pot. If the current is too high, the comparator switches off, and vice versa. The switching speed will vary depending on the hysteresis of the comparator. If the speed is too high, then you can increase the hysteresis (and decrease the switching speed) by adding a few hundred k resistor between the comparator output and its + input.

Note: You must use a high speed (<0.1us propagation delay) comparator with open drain output. You might look at the LMV7235 which is available from Farnell for about one pound.


Added:

The circuits above assume only one LED. If you still want to use 3 in series, you can use two MAX1682s to give you 13.2v.

Also, many thanks to Telaclavo for his advice on this.


Added:

OP has stated:

  • He wants a very fast rise time on the current
  • Not interested in efficiency
  • There will be a single pulse, or two pulses 80us apart, then a long pause
  • Wants a simple, robust circuit

Here is a circuit which is a linear current regulator. This is only feasible because the duty cycle is very low. This circuit will likely over-heat the transistor if the duty cycle is too large.

Flash 3

Thoughts:

  • A high voltage from the MCU or 555 will switch on the LED. A low voltage will switch it off.
  • Set the current using the voltage divider, or put in a pot so that it's adjustable. Or use a digital pot or DAC so the MCU can vary it.
  • In the schematic, the current is set to 3.3A. You can set it to whatever you want.
  • I drew only one LED, but it's meant to represent three LEDs.
  • If you use only a single LED, then set the boost regulator's output voltage lower accordingly.
  • I suggest a 555 based pulse generator for safety reasons, so it would be fairly hard to leave the current on
  • You could also make it safer by choosing a boost regulator which has a current limit. So, even if the flash is left switched on, the regulator would just limit the current anyway.
  • I cannot say what the switch on time will be. This will depend on the inductance of your wiring.
  • You should lay out the PCB carefully to avoid EMI.
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  • \$\begingroup\$ Be careful, the Q1 you drew has its source on the left (connected to C2). Real-world silicon MOSFETs have a parasitic diode that goes from S to D, so Q1 will always conduct. Even if you flip it horizontally, it will be difficult to turn it on, because none of S and D are ground. There are other issues, too. Too late here. I'll comment tomorrow. \$\endgroup\$
    – Telaclavo
    Commented Apr 23, 2012 at 0:07
  • \$\begingroup\$ @Telaclavo - I have updated the schematics. I hope they are correct now. Please let me know what you think. \$\endgroup\$ Commented Apr 23, 2012 at 9:12
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    \$\begingroup\$ A capacitor charging a capacitor will be relatively inefficient if the pump capacitor voltage drops by a substantial percentage during discharge or charge. (Work out 1/2CV^2 before and after discharge and be surprised). Best efficiency will be obtained with a switching regulator using an inductor. \$\endgroup\$
    – Russell McMahon
    Commented Apr 23, 2012 at 16:43
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    \$\begingroup\$ @Rocketmagnet how about this Exar (exar.com/power/led-lighting/regulators/step-up-down-regulators/…) solution with super cap? Unfortunately, the solution cannot time the pulses but perhaps that can be done with the external mosfet and secondary control mechanism. All I am proposing is inserting mosfets before the LEDs and control those mosfets based on the pulse timing. \$\endgroup\$
    – Ktc
    Commented Apr 25, 2012 at 11:21
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    \$\begingroup\$ @Ktc - The only thing is, with a FET, there are many more ways to choose the wrong one. A BJT really only needs to be NPN. Of course, the even simpler way to do this is just with a current limiting resistor. But then it might be a bit too sensitive to variation in LED forward voltage drop. \$\endgroup\$ Commented Apr 27, 2012 at 11:51
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That's an average power of

Power = 5 A \$\times\$ 10.5 V \$\times\$ 100 \$\mu\$s / 10 ms = 0.525 W.

Average power is easy for almost any battery. You just need a store to accomodate the pulse.

A capacitor that will "droop" say 0.5V in 100 \$\mu\$s needs to be

C = I \$\times\$ t / V = 5 A \$\times\$ 100 \$\mu\$s / 0.5 V= 1000 \$\mu\$F.

A supercap would do well here if voltage rating is OK.

E&OE

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    \$\begingroup\$ What is "E&OE"? :) \$\endgroup\$ Commented Apr 13, 2012 at 11:52
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    \$\begingroup\$ Why would a supercap be better than an ordinary electrolytic? \$\endgroup\$ Commented Apr 13, 2012 at 11:58
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    \$\begingroup\$ E&OE = errors and omissions excluded. (a generic disclaimer) \$\endgroup\$ Commented Apr 13, 2012 at 12:53
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    \$\begingroup\$ As @Madmanguruman says - error and omissions excepted BUT I tend to use it only where I have been raoidly throing powers of 10 around or writing equations out of my head and may have (stupidly) put I on top rather than bottom line of an equation or missed off or added a 0 or so. ie - the principle is what matters, check my arithmetic before you depend on my answer. \$\endgroup\$
    – Russell McMahon
    Commented Apr 13, 2012 at 14:13
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    \$\begingroup\$ @FedericoRusso - A supercap liable to have a better discharge capability for it's capacity. This may not be universally true but tends to be. \$\endgroup\$
    – Russell McMahon
    Commented Apr 13, 2012 at 14:14
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Consider the effective series resistance (ESR) involved and loss in power transfer.

At worst case maximum input levels:

  • Surge forward current, tp = 100 μs
  • IF = 5 A Vf = 3.5 V nominal!!
  • IF = 1 A Vf = 2.0 V nominal 2.5 V maximum
  • If = 0.2 A Vf = 1.5 V nominal

Also from the LED specifications, compute ESR [mΩ]

Vf ... If[A]... . . . delta V/delta I

  • 3.5 . . . 5
  • 2.8 . . . 3 . . . 0.7/2 => 350 mΩ
  • 2.0 . . . 1 . . . 0.8/2 => 400 mΩ
  • 1.5 . . . 0.2 . . . 0.5/0.8 => 625 mΩ
  • 1.1 . . . 0.001 . . . 0.4/.2 => 2000 mΩ

(Crude estimate of ESR)

  • ESR drops dramatically as current rises.
  • You want a power source with a capacitor and switch ESR < ~10% of 350 mΩ = 35 mΩ.

Now go find a suitable low-ESR capacitor and switch total.

Maybe decouple the battery ESR with a choke to limit current within its specifications. And use suitable fusing to prevent failure of the battery.

  • These are low-ESR $0.40 switches < 15 mΩ with a 10 V drive 35 A [FDD8778CT]
  • These are low-ESR $0.40 capacitors ~7 mΩ, CAP ALUM 68 µF 16 V 20% Thru-hole
  • Pick a bigger µF value as needed.

Assuming you can manage charging Li-ion batteries choose 4x 3 V cells for 12 V across the LED and a series awitch above to ground.

You can drive with 5 V or better 12 V so the transistor can amplify 3 V out to get 12 V out to drive the MOSFET to get 5 A from three LEDS 11.5 V, with a 0.5 V drop from the 12 V Li-ion source. You ought to design the overall current limit with a ESR of the string plus an added resistor to optimize values, that is, 0.4 V drop @5 A < 100 mΩ non-wirewound resistor.

The capacitor goes across the Li-ion battery string with perhaps a microfuse and ferrite choke inserted for good practice.

Can you run the PIC from the lowest Li-ion battery in the string @ 3 V? With 3x LEDs from 12 V with a 12 V gate drive and 5 A controlled fused pulse to the LEDs.

Got the picture?

Enter image description here

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  • \$\begingroup\$ The cap link is not working. Tony would you mind suggesting a circuit to do this. \$\endgroup\$
    – Ktc
    Commented Apr 25, 2012 at 15:51
  • \$\begingroup\$ search.digikey.com/us/en/products/EEE-FP1E471AP/PCE4440TR-ND/… Panasonic EEE-FP1E471AP \$\endgroup\$ Commented Apr 25, 2012 at 16:17
  • \$\begingroup\$ If you prefer thru hole Caps search.digikey.com/us/en/products/RR71C680MDN1/493-3715-ND/… here is one that is 7 mΩ and in stock cheap. Use the Digikey filters to sort by ESR , Stock or price and select range of ESR, voltage and uF as desired.. lots of solutions in stock \$\endgroup\$ Commented Apr 25, 2012 at 16:22
  • \$\begingroup\$ Tony, thanks for this. However I am not advanced enough to grasp some of these without a schematic or high level block diagram. That is why I kindly asked for a circuit so that I can better discuss. \$\endgroup\$
    – Ktc
    Commented Apr 25, 2012 at 16:40
  • \$\begingroup\$ What is overall comm channel rate and distance? Application? \$\endgroup\$ Commented Apr 25, 2012 at 17:27
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A Joule thief may be the answer to your problem: it's a sort of boost converter, where you open a circuit with an inductor in series to create a high voltage. Since the power is delivered by the inductor, you don't have to supply the current directly from the battery.

You have to tune the circuit to feed the LED with the proper current when the voltage goes up.

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    \$\begingroup\$ Interesting.. A solution just using a Cap would be probably much more easy to manage. \$\endgroup\$
    – Ktc
    Commented Apr 13, 2012 at 9:26
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    \$\begingroup\$ maybe, but I don't see how you would ever regulate current using just capacitors. \$\endgroup\$
    – Jason S
    Commented Apr 13, 2012 at 11:06
  • \$\begingroup\$ @Ktc the problem is that you (presumably) have a fixed load, so the way to give it a big current is to cause a big voltage; that is the work of the Joule thief. The other chance, as Steven says, is that store the charge in a capacitor and then switch it. But depending on the LED you are using, the voltage may not be sufficient \$\endgroup\$
    – clabacchio
    Commented Apr 13, 2012 at 13:01
  • \$\begingroup\$ A Joule Thief should work to charge a capacitor that you then discharge into three series LEDs with a constant current transistor setup to deliver the 5 amps constant current. \$\endgroup\$ Commented Apr 11, 2022 at 23:25

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