5A linear current regulator?

Hello,


I want to create a linear current regulator for this IR diode:
http://www.eqphotonics.de/cms/cms/upload/datasheets/WSLD-808-005-3.pdf
It's a 5W diode, operating at 2.0V at about 5A.

I'd like to make use of a linear regulator, powered by a 5V, 8A switched mode power supply. (link)

The problem is that I can't use an LM317 because it's current rating is not sufficient.

I found this regulator from linear:
http://cds.linear.com/docs/en/datasheet/108345fg.pdf
It can handle up to 7.5A, and I think it works the same way the LM317 does.

The current set resistor (between OUT and ADJ) would consume about 1.25V*5A=6.25Watts of power.
I want to use a potentiometer here since I want to be able to adjust the output current.
This is where I no longer like this setup, but I don't know what other possibilities there are.

Is there any other way I can easily regulate currents up to 5A, with a linear regulator?


Kind regards



EDIT:
If we are going to use another IC for this, I'd prefer one which has an enable pin because the laser diode
must be controlled by Arduino to go on and off at regular intervals (about each 200ms)
It does not necessarily need to be an enable pin, but just any way of toggling the output.

Also, will toggling the output damage the diode in time?

Good regulator. Use the 1083 that you located, but integrate a current monitor so that you can use a fixed sense resistor with a low V-Drop, and then use the pot to control the current monitor gain.

If you need an enable pin, the setup above won't provide it, but you can easily use a MOSFET before the circuit above in order to give you effectively the same thing.

I was also planning on adding a MOSFET in front of the regulator. But for the current sensing, what kind of IC do you suggest?

Is the LT6105 - Precision, Rail-to-Rail Input Current Sense Amplifier - Linear Technology good for this application?
And how will I get the current monitor to create a lower output voltage when the sensed current becomes higher? (because the ADJ voltage must become lower than the OUT voltage)

It would also be great if I could control the current trough Arduino.
All I can come up with are these two possibilities:

A digitally controlled potentiometer, but I can't seem to find low resistance values (below 1ohm) or high power (above 5W) so I don't think this will be an option.

My second thought was using a transistor as variable resistor, but how will I power the base with Arduino? I'll still need an analog voltage output, which the Arduino doesn't have.
In what ways can I control a transistor in it's linear operation mode, trough Arduino?

FalloutBe said:

I was also planning on adding a MOSFET in front of the regulator. But for the current sensing, what kind of IC do you suggest?

Is the LT6105 - Precision, Rail-to-Rail Input Current Sense Amplifier - Linear Technology good for this application?
And how will I get the current monitor to create a lower output voltage when the sensed current becomes higher? (because the ADJ voltage must become lower than the OUT voltage)

Click to expand...

FalloutBe said:

It would also be great if I could control the current trough Arduino.
All I can come up with are these two possibilities:

A digitally controlled potentiometer, but I can't seem to find low resistance values (below 1ohm) or high power (above 5W) so I don't think this will be an option.

My second thought was using a transistor as variable resistor, but how will I power the base with Arduino? I'll still need an analog voltage output, which the Arduino doesn't have.
In what ways can I control a transistor in it's linear operation mode, trough Arduino?

Click to expand...

That looks like a decent IC. I think that should work.

You don't need a low ohm potentiometer. The way scaling/gain works on those ICs well typically allow you to use 1k pots (or higher). Start building the circuit with standard fixed resistors, and you'll see where the pots can go. You don't vary the sense resistor in this type of setup.

If this is for a hand held, a AMC7135 based driver like the Nanjg 105C driver should work.

You would need to stack 6 to 7 more 7135's on a 2.8 amp driver to get to 5 amps. Each chip allows 350mA of current. They "burn" off excess voltage in the form of heat. So you would probably want 1S Li-ion voltage level for input. You need at least .1 volts overhead (after voltage sag) for regulation.

There are also several AMC 7135 based circuit boards you can purchase from OSH Park if you need a different form factor than the Nanjg 105C.

ImA4Wheelr said:

If this is for a hand held, a AMC7135 based driver like the Nanjg 105C driver should work.

You would need to stack 6 to 7 more 7135's on a 2.8 amp driver to get to 5 amps. Each chip allows 350mA of current. They "burn" off excess voltage in the form of heat. So you would probably want 1S Li-ion voltage level for input. You need at least .1 volts overhead (after voltage sag) for regulation.

Click to expand...

Eeeek. This ^ won't work for you, don't do this

FalloutBe said:

I want to create a linear current regulator for this IR diode:
http://www.eqphotonics.de/cms/cms/upload/datasheets/WSLD-808-005-3.pdf
It's a 5W diode, operating at 2.0V at about 5A.

Click to expand...

The AMC7135s can't handle dropping that much voltage. You'd be talking 2.2 x 5 = 11W of heat for the ICs to burn off in total. Even if you heatsink the tops of the ICs, those IC's can't do it.

rhd said:

Eeeek. This ^ won't work for you, don't do this



The AMC7135s can't handle dropping that much voltage. You'd be talking 2.2 x 5 = 11W of heat for the ICs to burn off in total. Even if you heatsink the tops of the ICs, those IC's can't do it.

Click to expand...

11 Watts is what the diodes uses , not what the regulators burn off , assuming they are linear and in CC mode

If he feeds it 3 volts and the regulators drop 1V each at 350mA each , with enough to produce 5A , each regulator is burning 350mW of heat and a combined heat load off ,

You would need 14 350mA regulators to get 4.9A , assuming a drop of 1V and 350mA a regulator that's , 350mW * 14 = 4.9W

At 2V drop from a 4.2V source ( 2.2V over diode , 2 over regulators ) that's still 700mW of heat per IC , total of 9.8W

Total power in would be 3.2V*4.9A = 15.6W and 4.2V*4.9A = 20W respectively

ionlaser555 said:

11 Watts is what the diodes uses , not what the regulators burn off , assuming they are linear and in CC mode

Click to expand...

Nope. 10W is what the diode uses (5A x 2.0V).

11W is what the ICs burn off, assuming a lithium ion supply ((4.2 - 2.0) x 5A) = 11W.

These chips don't work well above 500mW of heat each (though the datasheet claims they can handle up to 700 without heatsinking.

Ok so the diode is droping 2 Volts at 5 amps ( read 2.2V for some reason ) so that's 10W , but the worst case as per the data sheet is 2.2V (11W ) so the maths in my last post still works , so long as the diode drops 2.2V at 5A

Well that's 780mW per IC if there are 14 of them , based on a 2V diode drop , 2.2V regulator drop , 350mA per regulator

Guys, the op said he wanted it to be powered off a 5V 8A supply.

and the fact that it will be controlled by an Arduino suggests it will be used as some sort of CNC cutter? or similar application.

FalloutBe said:

I'd like to make use of a linear regulator, powered by a 5V, 8A switched mode power supply. (link)

<snip>
EDIT:
If we are going to use another IC for this, I'd prefer one which has an enable pin because the laser diode
must be controlled by Arduino to go on and off at regular intervals (about each 200ms)
It does not necessarily need to be an enable pin, but just any way of toggling the output.

Also, will toggling the output damage the diode in time?

Click to expand...

No Li-ion cell is going to hold 4.2V under a 5 amp load. Best would probably be 3.9v. Also, the 7135 consumes .1 v. So 1.6 volts (3.9 - 2.2 - .1) to burn off.

But, I do agree that that is a lot of voltage to burn off. I miscalculated in my head when I replied. I was thinking it was an under 1 volt difference. AMC7135 might do if he is powering by 2SxP Eneloops.

EDIT: opps, djQUAN pointed out something key I missed in the OP.

I think you guys missed the part where the first post mentioned that this will be powered off a 5V 8A supply.

If you think 4.2V is bad enough already, 5V is even worse for the AMC chips so we can probably agree that it's not the best option for this.

Agreed. As I said in my second reply, I miscalculated.

So isn't he going to have this issue with any linear regulator? Really seems that he should go buck.

Yes, but a different linear regulator with a big pass transistor on a heatsink can better dissipate the heat without problems than the tiny AMC chips.

an op amp driving a darlington transistor and measuring the output current through a resistor (similar to those 358 based drivers on ePay but beefier) would be the best option for this application.

First of all, thanks for all the effort and suggestions!
But I'm going to come in between your calculation battle for a moment now, to clear out some misunderstandings.
Also, I don't think using 14 350mA regulators is a good idea. It might work, but .. .seriously? that's ridiculous..


The laser will not be handheld, it'll be used for CNC engraving, like djQUAN assumed.
It will indeed be powered by a 5V 8A supply, which I already ordered.
I also ordered the LT1083 and I still think it's a good idea to use that one.


Let's do some power calculations on the components:

2.2V*5A=11W laser diode.

The regulator uses a fixed 1.25V reference, so there we'll have 1.25V*5A=6.25W dissipating in the 'current set resistor' (located in between OUT and ADJ)

Supply was 5V, so what's left is 5V-2.2V-1.25V=1.55V
*5A=7.75W power to be dissipated in the LT1083
According to the datasheet, a maximum of 60Watts can be dissipated in the chip, so that's perfect.


After some thinking, this is what I'm planning to do for now (see attached image)

The idea is to put the laser in front of the regulator, instead of behind it. I don't think this should cause any problems; it's just components in series, so I can switch them around, right?

This, I do to make sure I can connect the source of the MOSFET to ground, so that I can easily control it from the Arduino microcontroller.

The gate voltage of the MOSFET will be set by the RC filter, which is set by a PWM output of a digital pin.
The RC filter will convert the PWM signal into a smooth DC voltage.

The MOSFET has a 0.25ohm resistor in series, to make sure the maximum current is limited to 5A, when the MOSFET is fully conducting.


What do you think about this? Will it work?