My intent is to build a simple driver for a laser LED with an operating voltage of 4.6 volts and forward current 1.6 amps. I have some spare LM317s that I would like to use, however they can supply 1.5 amps at most with proper heatsinking. Can I use multiple of them in parallel like this, or should I completely scrap this schematic? I suppose each of those should deliver 500 mA making it 1.5 A total.
Also, if this circuit actually works, how can I calculate the power loss on the LM317 regulators?
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\$\begingroup\$ I think you need to look into a safer laser driver. That's some serious power going into an expensive laser module. 4.6V is probably the maximum forward voltage rather than the operating voltage - if you exceed it, you no longer have a laser. At that kind of power level, you also don't want to depend on current alone - you should be monitoring the output (optical) power and regulating it to a safe level. \$\endgroup\$– JRECommented Jan 2, 2018 at 15:28
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2 Answers
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I'm not so sure about the wisdom of this circuit.
It may work ok, but I am concerned about the response of the regulators when you switch the MOSFET. That large 100uF capacitor will pull down the right side of the sense resistors to close to zero. If the regulators can respond that fast you will initially have close to 10V across each regulator at a whopping 5W each.
Further, the circuit is terribly inefficient consuming 6W to drive a 2.3W LED.
You would be far better to drive the LED from a 5V switching supply and use a low drop current limiting circuit. Perhaps something like this.
simulate this circuit – Schematic created using CircuitLab
Though this is a better enable method.
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1\$\begingroup\$ You might have to be careful with this circuit to be sure it can exit the DISABLE state without momentarily applying the whole V3 voltage across the laser. \$\endgroup\$ Commented Jan 2, 2018 at 16:46
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1\$\begingroup\$ In the end we built the latter circuit with 5 volts supplied by a LM2576-5 switching regulator and it is more effective so we don't need to cool it. Thanks! I didn't understand shunt resistors much back then. \$\endgroup\$– TefekCommented Mar 4, 2018 at 15:39
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The circuit you designed is fine.
Connecting multiple current generators in parallel can be done and is done in many applications. You need to be sure that the output voltage is within the range each of them can support.
To calculate the power loss on each regulator, you can treat them separately.
Your LED needs 4.6 V, starting from 12 V the drop is 7.4 V. Since the current is half an amp per regulator, each of them will dissipate 3.7 W. This power is dissipated on the regulator, the sense resistor and the diode. Since the resistor is 2.5 Ohm, the drop on it will be 1.25 V, while the diode will drop approximately 0.7 V. The LM317 will see only ~5.45 V of the total 7.4 V drop. Again, multiplying for the current you finally get 2.7 W on each LM317, 0.625 W on the resistors and 0.35 W on the diodes.
The total dissipated power will be 11.1 W. You will need 1W resistors, and heatsinks for the 317s.
answered Jan 2, 2018 at 13:36
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\$\begingroup\$ Your power math is off a tad. Vreg = 12 - 4.6 - 0.7 - 1.25 ~ 5.45V => 2.725W per regulator. \$\endgroup\$– Trevor_GCommented Jan 2, 2018 at 14:14
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1\$\begingroup\$ And there is no probably in "probably a small heatsink on the 317s is a good idea too" about it. Heatsink. \$\endgroup\$ Commented Jan 2, 2018 at 14:16
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\$\begingroup\$ Also. be aware when the MOSFET turns off you will have the full 12V sitting at the top. and when turning on it briefly will start up there. \$\endgroup\$– Trevor_GCommented Jan 2, 2018 at 14:24
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1\$\begingroup\$ Thanks @Trevor, I completely forgot the diodes. Corrected. \$\endgroup\$ Commented Jan 2, 2018 at 14:32
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1\$\begingroup\$ or start with 9v instead of 12v :) \$\endgroup\$– PasserbyCommented Jan 2, 2018 at 15:12