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FrozenGate by Avery
Lm317 laser diode driver
- Thread starter Petersoft
- Start date
- Joined
- May 9, 2011
- Messages
- 1,219
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- 113
Probably, but it'll be a close thing. You'll lose (at least) a bit over 1.2 volts to the regulator and a red diode will have a forward voltage of a bit less than 2 volts. Assuming a fully charged 18650 @ 4.2V you'd have a volt to spare. As the battery discharges though, the regulator will stop.....regulating.
If a '317 regulator is all that will fit in your budget you'd be better off using 2 cells. Otherwise invest a few bux in a proper driver from DTR.
If a '317 regulator is all that will fit in your budget you'd be better off using 2 cells. Otherwise invest a few bux in a proper driver from DTR.
Thanks! I think I will use 2x 16340
- Joined
- Sep 12, 2007
- Messages
- 9,399
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- 113
Pay no mind to wizardG's numbers - he's either remembering things wrong or is making things up. You only need to look at the datasheet (just google it) and it says quite plainly and I quote:
7.4.2 Operation With Low Input Voltage
The device requires up to 3-V headroom (VI – VO) to operate in regulation. The device may drop out and
OUTPUT voltage will be INPUT voltage minus drop out voltage with less headroom.
Now, in practice this is likely to be closer to 2.5V for higher current, and may be as low as 1.2V for very low output currents, but in constant current mode, you're losing another 1.25V across the current sense resistor. Therefore, you need to plan on having an extra 3.5 to 4V above the output voltage for this device to work in operation.
Not sure why so many people get this wrong since the circuit is so old and is covered so regularly here.
7.4.2 Operation With Low Input Voltage
The device requires up to 3-V headroom (VI – VO) to operate in regulation. The device may drop out and
OUTPUT voltage will be INPUT voltage minus drop out voltage with less headroom.
Now, in practice this is likely to be closer to 2.5V for higher current, and may be as low as 1.2V for very low output currents, but in constant current mode, you're losing another 1.25V across the current sense resistor. Therefore, you need to plan on having an extra 3.5 to 4V above the output voltage for this device to work in operation.
Not sure why so many people get this wrong since the circuit is so old and is covered so regularly here.
