You have a misconception about what a current regulator does. The whole point is that it will drop ALL the voltage in the circuit that isn't taken up by the other components.

[EDIT: I wasn't thinking clearly when first adding this post... corrected now.]

That's correct. In the LM317 the regulator acts to hold its internal reference voltage (usually a bandgap reference voltage of 1.2 to 1.25V) across Vout to Vadj by adjusting the output current.

To understand the power dissipation in the LM317,
- Assume the target/desired current is fed to the load
- If the load voltage ends up being less than the max voltage possible for regulation at the LM317 output (approx. Vin - 3V?), then the LM317 will have the excess voltage across Vin to Vout. Then LM317 power is (Vin-Vout)*Iload.
- If not, then it's not regulating and you're trying to put too much current into the load.

See, for example, the applications in figures 4 and 9 in the TI LM317 datasheet: http://www.ti.com/lit/ds/symlink/lm317.pdf

Also, a few other related links:
http://users.telenet.be/davshomepage/current-source.htm
http://diyaudioprojects.com/Technical/Voltage-Regulator/

The first site there includes some power calculations, as well.


Thanks all.

 

- If not, then it's not regulating and you're trying to put too much current into the load.

I'm sorry I keep dropping in new questions, but I'm learning a ton with this discussion and whole-heartedly appreciate the feedback I'm getting from you all.

So let's assume a scenario real quick to make sure I understand 1 concept. The datasheet of the LM317 device I pointed to made it seem that the voltage drop across the regulator was anywhere from 1.3V to 1.75V. If this is true, let's take the worst-case here. On a load voltage for the higher Vf LEDs at 9.6V (3.2V each with 3 in a string) plus 1.75V regulator drop, that's 11.35V. Would this be my minimum input voltage for the circuit to operate properly, or does their have to be headroom worked in, if so, how much? Also, you're saying that if 11.35V is minimum input voltage, if the input voltage drops below this value, that the current through the load could spike?

 

If the input voltage drops below 11.35V, the regulator will "Drop Out", outputting zero current.

It is better to design for a resistor in the chain of LEDs to also drop some voltage, nothing massive, under 1 watt. That way your LM317 current source will be reliable across a wider range of voltages, and dissipate less heat.

 

If I'm already running close to being out of voltage without the resistor, how does that help? Not a facetious question. If my input voltage requirement is already at 11.35V minimum with a 12VDC source, doesn't seem to be much room to drop additional voltage. In fact, it would be preferable if the circuit would operate well at voltages as low as 10.5V.

 

If I'm already running close to being out of voltage without the resistor, how does that help? Not a facetious question. If my input voltage requirement is already at 11.35V minimum with a 12VDC source, doesn't seem to be much room to drop additional voltage. In fact, it would be preferable if the circuit would operate well at voltages as low as 10.5V.

I was thinking backwards. Apologies.

One of the LDO (Low Dropout) schematics above would work, as would a switching supply, but the DC-DC switching supply would get complicated and spendy.

 

Ok, let's talk about the LM317 for a minute then. Looking at the datasheet for this series of devices, looks like the forward voltage drop across the regulator is only 1.3 to 1.75V depending on temperature, which is good. Since all of this is pretty new to me, a lot of it is speculation, so please correct me if I'm wrong. Here's an updated circuit using the 317.

Main voltage line enters the 317, current through the LED string is set by R4(Rsc) at approximately \(\frac{V_{ref}}{R_{SC}}\). Current is stable over voltage fluctuations? Also, how is power dissipation in the 317 calculated? Would it be the forward voltage drop (approx 1.75V) times set current? In this case \(P_{diss} = 1.75V \cdot 0.085A = ~149mW\)?

Another question would be, assuming a 1.75V drop across the 317 plus the 6 to 9.6V drop across the LED string, subtracted from source voltage of approx 13.8V, \(V_{IN} - (1.75V + V_f_{string})\) can range from 2.45V to 6.05V, what happens to the extra voltage here? It has to be dissipated somewhere. Would it be in the switching MOSFET?

Thanks again guys.

Your schematic looks pretty good, and the other guys have covered most of the information you need.

The 3V includes both the 1.25V regulation of the LM317 and the voltage needed to power up the internal circuit. I am not sure of the 3V number, but I believe it is very close.

The subject of current regulators is multifacited. I have spent the last few years exploring the subject for the same application you have, LEDs.

If using a bench power supply be sure to short the output and set the current up. Leave the short in place until after you connect the LEDs up, then remove the short. This will prevent the filtering capacitors from surging the LEDs and blowing them out. Before removing the LEDs from power repeat the procedure in reverse, shorting the power supply out.