4
\$\begingroup\$

I need to design 40A 1.8V laser diode driver and need some word of advice as in the past I did from scratch only much smaller DCDCs. As it's a laser diode there will be no fast current changes.

My idea is: do synchronous buck DCDC, use PWM from STM32 microcontroller, then MOSFET driver then MOSFETS. 40A for mosfets does not looks like a problem. Sense current using high-side (on 1.8V rail) 0.001 Ohm shunt with opamp amplifier.

Few questions though:

1) Vin: I am looking more at 12V instead of 24/48V, to avoid too short transistor on times. Reasonable? Input current 8A vs 2A does not looks like a deal breaker.

2) Inductor. 40A inductor looks scary, and I probably won't find anything readily available. I was thinking about taking standard 13x13mm SMD inductors (which usually 3A), and parallel lots of them. But this will reduce their inductance a lot, probably below acceptable level. Any suggestion?

3) Is there a benefit of implementing multiphase DCDC for this application? I have bunch of PC motherboard inductors, they are like 1uH and too low for any paralleling, but might be ok for multiphase.

For the craziest approach I might just cut piece of PC motherboard with transistors, inductors and capacitors and redo gate drivers. Will look weird, but beating 8-layer board design is something out of reach for me at the moment. Does that look too crazy?

\$\endgroup\$
11
  • \$\begingroup\$ Out of interest what sort of laser diode are you driving? I thought most laser diodes with that sort of high current but low voltage were normally only suitable for very short period pulsed operation. \$\endgroup\$
    – PeterJ
    Commented Oct 24, 2015 at 7:46
  • 1
    \$\begingroup\$ @Will Dean: given that he's talking about reusing PC motherboard components... \$\endgroup\$ Commented Oct 24, 2015 at 8:48
  • 1
    \$\begingroup\$ RE inductors, it's not hard to find SMD inductors with 40, 50, 60, 70 A Isat. For just a few dollars each. Just visit any of the vendor websites (i.e. Coilcraft, Bourns, Eaton, ...). \$\endgroup\$
    – The Photon
    Commented Oct 24, 2015 at 10:01
  • 1
    \$\begingroup\$ @BarsMonster, is this really the kind of project you want to jerk around with eBay parts? If you are doing this for work or school (and messing around with 80 W lasers isn't something a lot of people do as a hobby), Coilcraft will probably send you free samples; then you can get exactly the parts you want and don't have to worry about whether they're really what they say they are. \$\endgroup\$
    – The Photon
    Commented Oct 24, 2015 at 16:08
  • 1
    \$\begingroup\$ Laser diodes are often fussy about over voltage spikes, take care if you plan to use a micro in a switchmode PSU feedback loop. My advice is to find a close fit laser PSU on eBay, over rated if you can find and afford it. \$\endgroup\$
    – KalleMP
    Commented Nov 2, 2015 at 8:20

4 Answers 4

9
\$\begingroup\$

This is no easy task to do by simply re-using bits of old motherboards. It's a problem that I feel requires a properly designed solution. I can't imagine that the laser diodes are cheap and because of this I would recommend a properly designed circuit board with a design solution that is tailored for this application. I'm initially thinking of maybe the LTC3882: -

enter image description here

The output parameters can be pretty much controlled from an MCU (Digitally Programmable Voltage, Current Limit, Soft-Start/Stop, Sequencing, Margining, AVP and UV/OV Thresholds) so I thinks it's a solution with high probability of working. Another chip that looks promising is this: -

enter image description here

There are plenty of options from LT - if you use their search engine it narrows down the requirements of 11 to 13 v operation with 1.2 volts and 40 amps output to this web page here.

As for using a micor to control this read what folk are saying about this here: Digital SMPS Vs. Analog SMPS

\$\endgroup\$
7
\$\begingroup\$

Multiphase is the way to go.

Ideally, build a single, low-power, multiphase controller. Then build multiple, small, power switcher sections, more than you need. This way, if one fails, you can substitute. When debugging, you will have good ones and bad ones to compare, there's nothing worse than seeing a waveform and not knowing whether it's meant to be like that or not.

If you can use a PC motherboard, do, saves so much building. But don't cut it, use it whole. There's no knowing how far vital buried tracks go a-wandering. By all means remove components, but don't take a saw to it.

\$\endgroup\$
4
+100
\$\begingroup\$

One thing to be VERY careful of is the output capacitance you typically need with buck converters of this type, it is a massive danger to your expensive pump diode.

If the output wiring becomes open circuit even for a moment the caps will charge to whatever you have set the voltage limit to be, when the diode gets reconnected the resulting current pulse will kill the diode.

When I did this for a one off (And I have, slightly different parameters but a large low voltage diode bar) I punted on the efficiency slightly and brought a couple of off the shelf isolated DC/DC module (GE Critical power as I recall), which brought my input down to a few hundred mV above the diodes operating voltage, then regulated the current with a passbank made of butch bipolar transistors on a heatsink (The isolated converters meant that I could simply parallel the current ouputs from the two halves of the passbank).

It was a while ago, and I dont have my BOM to hand, but I am thinking it was something like one of these http://uk.mouser.com/ProductDetail/GE-Critical-Power/ESTW025A0F41Z/?qs=sGAEpiMZZMvGsmoEFRKS8A9yhiuKixFL2sWuTv9sg4E%3d There are 1.8 & 2.5V versions of the same idea available which should get you something you can trim to give the required headroom for the passbank.

The advantage was that the tricky to make efficient DC/DC was a solved problem, and I had isolation so accidents around the diode bar became less destructive, and I was not trying to regulate current in the presence of hundreds of microfarads of output cap.

Modulation was brought in via a linear optocoupler and optos were used for the status, interlock loop and fault reporting output.

The isolated module was important because big laser diode bars have an annoying tendency to be positive on the mounting flange, the isolation meant that an accidental short could not bypass the current regulator.

For one off things you do NOT in general want to be designing your own heavy current switchers, apart from anything else the appropriate magnetics are a problem in small quantities.

Be careful out there, a laser diode bar like that is not a good toy, you only come with two retinas.

73 Dan.

\$\endgroup\$
1
\$\begingroup\$

Computer existing technology is valid to do your job .So is the LT chip .In other words a polyphased buck current mode hard switched running off a not too high input rail would work and would be sensible if your rail is there .If you want to go offline then you can make a 10: 1 transformer possibly using a planar E I core and prototype your winding with teflon wire .A 60 amp TO247 shottky on a 1plus 1 ct winding will do 40Amp easy .Place lots and lots of output caps across your Low volt DC .I used 680 microfarad sprauge for a 1 volt job but your job is different.Maybe you could use lots of 10microfarad ceremics .Now that the output has been dealt with your problem becomes a 4 amp problem . I ran a ZCS invertor in a half bridge on the 10 turn side .I used 2 cheap BJTs and so could you .Remember the power is less than 100watt.I fed my simple two transistor ZCS half bridge invertor with a "s trap buck convertor "This was prototyped using a mains isolation transformer for safe debug and now can go off rectified mains .You could use a buck convertor thats more familiar to you .I suggest that you control the current .My little job was to put currents through wires which wasnt a laser but showed some similarity .For current measurement I used a cheap CT on the primary AC side of the planar transformer where the currents were very managable.This meant no shunt .The error due to magnetising current was insignificant because the planar transformer had no gap and was running at low flux density because of 1 turn making such a low output voltage .

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.