Need help driving IR emitter

[ardo]

I have a bunch of AMC7135 1050ma board leftovers from DX:
http://dx.com/p/amc7135-1050ma-regulated-circuit-board-for-diy-flashlights-10-pack-1885

I need to drive these 1W 750ma IR emitters:
http://dx.com/p/1w-850nm-ir-led-emitter-on-star-3-2v-3-5v-350ma-13961

I'm planning to drive two emitters in series from a single LiIon battery, but cannot find any instructions on modding the 7135 board, to remove one of the regulators & bring the board's current to 700ma. Could some please help me out with a link? TIA.

 

[SemiMan]

I have a bunch of AMC7135 1050ma board leftovers from DX:
http://dx.com/p/amc7135-1050ma-regulated-circuit-board-for-diy-flashlights-10-pack-1885

I need to drive these 1W 750ma IR emitters:
http://dx.com/p/1w-850nm-ir-led-emitter-on-star-3-2v-3-5v-350ma-13961

I'm planning to drive two emitters in series from a single LiIon battery, but cannot find any instructions on modding the 7135 board, to remove one of the regulators & bring the board's current to 700ma. Could some please help me out with a link? TIA.

7135 is a fixed 350mA, so if you take one off the board it becomes 700mA.

You are not going to be in regulation very long with 3.6V from the LEDs and another 0.12V from the regulator, but that may be better as the current will drop as the battery runs out.

Semiman

 

[ardo]

7135 is a fixed 350mA, so if you take one off the board it becomes 700mA.

You are not going to be in regulation very long with 3.6V from the LEDs and another 0.12V from the regulator, but that may be better as the current will drop as the battery runs out.

Semiman

Thanks, I remember seeing somewhere that the order of removed 7135 is important - not sure why, I thought they are all connected in parallel? Just wanted to make sure...

I only need the IR to run for about an hour at a time, so hopefully current will stay regulated this long.

 

[bshanahan14rulz]

What about paralleling two drivers off of the power source? Sounds like you've got plenty of drivers.

 

[ardo]

What about paralleling two drivers off of the power source? Sounds like you've got plenty of drivers.

I thought you'd parallel the drivers if you need higher currents? I actually want less - 700mA instead of the 1050mA that the board is capable of. Or am I missing something?

 

[Thr3Evo]

They are in parallel therefore removing any of them should be okay.
I'd remove either q3 or q4 to keep the board balanced for heat sinking purposes.

 

[bshanahan14rulz]

I mean, just so they'd stay in regulation longer, i.e. your batteries would be empty before the 7135 didn't have enough overhead to maintain regulation, whereas running two LEDs in series is more compact, it will fall out of regulation when the battery reaches (2*Vf+.12) which is ~3.7V. On the same hand, 3.7V unloaded is pretty close to empty for Li-Ion. Stable output for 1hr shouldn't be a prob for an 18650. What battery are you using?

 

[ardo]

They are in parallel therefore removing any of them should be okay.
I'd remove either q3 or q4 to keep the board balanced for heat sinking purposes.

Perfect, that's what I wasn't sure about. Thanks!

I mean, just so they'd stay in regulation longer, i.e. your batteries would be empty before the 7135 didn't have enough overhead to maintain regulation, whereas running two LEDs in series is more compact, it will fall out of regulation when the battery reaches (2*Vf+.12) which is ~3.7V. On the same hand, 3.7V unloaded is pretty close to empty for Li-Ion. Stable output for 1hr shouldn't be a prob for an 18650. What battery are you using?

Well, I could drive each IR from a dedicated regulator, but I thought that almost 50% of power would go into heat, hence putting 2 IRs in series. I'm using 3.2Ah unprotected cells from dead laptop batteries. An hour is about all I need: this is going to be an IR light for my chronograph, measuring bullet speed.

 

[bshanahan14rulz]

From a very basic viewpoint of a stable voltage source (i ain'ts smart enough to integrate), you should only be getting (0.12*0.7)W=0.08W of added heat by using dedicated regulators, compared to using a single regulator for both LEDs, I would think? Am I missing something? 0.08W doesn't seem like a lot of heat, would have thought a bit more than that. Anyways, so you are having a system that draws .08W more power, but will run regulated through the entire discharging life of the cell.

Then again, though, a 3.2Ah cell probably shouldn't have any trouble powering up a 3.7V load at 700mA for an hour before going out of regulation on 7135s.

 

[ardo]

From a very basic viewpoint of a stable voltage source (i ain'ts smart enough to integrate), you should only be getting (0.12*0.7)W=0.08W of added heat by using dedicated regulators, compared to using a single regulator for both LEDs, I would think? Am I missing something? 0.08W doesn't seem like a lot of heat, would have thought a bit more than that. Anyways, so you are having a system that draws .08W more power, but will run regulated through the entire discharging life of the cell.

Then again, though, a 3.2Ah cell probably shouldn't have any trouble powering up a 3.7V load at 700mA for an hour before going out of regulation on 7135s.

LOL I never thought about it this way, just saw about 50% of supplied power going up in heat. Thanks for spelling it out for me, I will drive each IR by a dedicated regulator.

 

[SemiMan]

From a very basic viewpoint of a stable voltage source (i ain'ts smart enough to integrate), you should only be getting (0.12*0.7)W=0.08W of added heat by using dedicated regulators, compared to using a single regulator for both LEDs, I would think? Am I missing something? 0.08W doesn't seem like a lot of heat, would have thought a bit more than that. Anyways, so you are having a system that draws .08W more power, but will run regulated through the entire discharging life of the cell.

Then again, though, a 3.2Ah cell probably shouldn't have any trouble powering up a 3.7V load at 700mA for an hour before going out of regulation on 7135s.

No that is not right. Power loss = (Battery voltage - LED Voltage) * current

With two LEDS in series, the power loss is much lower though it will drop out of regulation quicker.

Semiman

 

[bshanahan14rulz]

^Ooops, yep, I messed up. So, let me get this straight: 0.12V isn't the Vf of the regulator, rather it is the overhead required. i.e. the regulator will have to dissipate the power from dropping the leftover voltage, but as long as the leftover voltage is 0.12V, the regulator will be able to keep the current @ 350mA. If the leftovers is <0.12V, then regulator will not be able to hold current at 350mA, so current will drop if its up to the LED.

I'm glad you caught my mistake only a day later, I hope it was soon enough for ardo's benefit. I dunno what's been up with me lately.

 

[ardo]

No harm done: I'm still waiting for my thermal epoxy to arrive. Either way, you've convinced me to use individual regulators for each IR source. I just want to make sure the extra power will be dissipated without the need for additional regulator cooling.

 

[bshanahan14rulz]

I think I can help with calculating how much heat is dissipated by your regulators.
P=V*I, V being voltage dissipated as heat by regulator @ worst case scenario of a full battery and a single emitter:
(Vbatt-VfLED)*=P
(4.2-1.8)*0.7=P; P=~1.7W, ~0.85W per chip.

Quite honestly, I'm no longer confident in giving out advice, so I'd wait and see what Semiman suggests. Your original idea of just putting LEDs in series and letting it run that way until Vbatt gets below (2*VfLED)+0.12V may be the best route, since @ 3.72V you've had a pretty good run.

Using Sanyo 2600mAh cells as a reference (very common find in laptops), they should net you between 50 and 60 minutes of regulated runtime until their voltage falls below the (2*VfLED)+0.12V cutoff. So, two LEDs one regulator will probably be the suggested route.

- thanks to HKJ for his battery testing work :-)

 

[ardo]

Thanks! Will an unprotected battery still be OK when the circuit drops out of regulation at 3.7V?

 

[SemiMan]

In either case, the unprotected battery will not be okay if left to drain. The difference is with a single LED per regulator, the LED will be bright until the battery is far more discharged than it should be.

With the two LEDs in series, the LEDs will get quite dim which is your cue to turn off the light.

Personally, I would either use protected or have some cut-off circuit.

With one LED per regulator as pointed out you could burn 0.85Watts/regulator - 1.7 total.

With the LEDs in series, worst case power will be along the lines of (4.2-3.6)*0.75 or 0.45watts total split over two chips or 0.225Watts/regulator. Obviously that will run much cooler.

Only concern is the actual forward voltage of those LEDs, not to mention losses in wires, etc. You may not be in regulation that long, but then again, that may not be bad. Think of the regulators as protection for a full battery.

Semiman

 

[ardo]

Thank you for spelling it out for me! I will now definitely use one regulator driving two LEDs.