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Old 07-09-2012, 06:41 AM #17
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Default Re: how does it add up ?

The only things you should understand about adjustable linear voltage regulators when they're in regulation (i.e. Vin > Vout + Vdo):
  • Vout = Vin - Vdo (minimum)
  • Vout - Vadj = ~1.25V
  • Output of Vout - Vadj is for the most part temperature independent (which is why your driver maintains the same current when it gets hot).

That is it.

The LM317 is a voltage regulator, period. There might, however, be some confusion about what voltage it is actually regulating: it regulates the voltage differential between Vout and Vadj to 1.25V. Everything else is how you use the IC in your circuit. So don't think in terms of "current mode" or "voltage mode". There aren't "modes" to the LM317, only how you use the 1.25V reference voltage differential. The topology of the circuit determines what function your LM317 serves, e.g. "current regulating topology/circuit".

A great set of examples for how to use the LM317 is the National Semiconductors Datasheet for the LM117/317. There you can see how the LM317 can be used for voltage regulation, or current regulation, amplification, switching, and other stuff.

For the OP's question, here is how the voltage drops work in the "DDL" current regulator (bottom of page 19 in the datasheet above):

Vin >= Vout + Vdo; Vdo ~= 3V
Vout - Vadj = 1.25V
Vadj - VfRedLaser to ground = ~3V (assuming the red drops 3V; it may be less)

VfRedLaser + (Vout - Vadj) + Vdo = ~7.25V

The ~7.25V is based on Vout = 5V (see bottom of page 6 of datasheet), which isn't the case for a red VfRedLaser = 3.0V (you could get away with Vin = 6V or less like 5.6V as above).

Remember that the DDL circuit calculations were describing a generic setup for the current regulator topology; the Vdo = 3V is just quoting the devices specifications. Your batteries aren't likely to stay at a particular value anyway, so it doesn't make too much difference. If you're using some other LDO regulator, you can also factor that into the Vdo too.


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