Jonathan
Enlightened
Just some thoughts. The Linear 1930 uses a 'constant current' switching control method, not a constant output current. The output is regulated with a standard resistive divider and a 1.25V reference. In the usual connection the output _voltage_ is regulated.
You could use the device to produce a regulated output current, by placing a current sense resistor between the load and ground, and connecting the load-resistor node to the feedback pin. The device will adjust the output voltage until the voltage drop across this current sense resistor is 1.25V. This is similar to the technique of using an adjustable three terminal regulator as a current source.
The only problem with the above scheme is that you will have full output current at 1.25V across your sense resistor...if you are running something like single Luxeon, then you will see about 1/3 of the power used by the Luxeon simply dissipated in the sense resistor.
I'd suggest looking at the Linear LT1618, which has a 50mV current feedback as well as a 1.25V voltage feedback. It also has a _shutdown_ pin, which means that you could do the switching with a very low current switch, eg. a reed switch.
In terms of which LEDs to use, I think that the Luxeons will have the benefit of better efficiency and a tighter beam. If efficiency _of the LED_ is your only concern, then the Luxeon will do better both for high power lighting and for low power lighting. Ideally, for lower powers, you would pulse the Luxeon rapidly so that the current level remains high enough to prevent color shifts.
But using a Luxeon will present a very serious problem: the Luxeon is a single LED with a voltage drop of about 3.2V. You can't use a boost converter with a 4.5V supply to drive a 3.2V load. You would need some sort of 'buck-boost' circuit to deal with the battery voltage falling through the load voltage.
The losses in the boost diode (about 0.5V) and in the current sense resistor will also be a very large fraction of the load voltage, meaning that efficiency would be poor. So from the point of view of actually driving the things with a boost converter, I thing that you would do better with the series strings of 5mm LEDs rather than the Luxeon.
You might be able to get away with 4AA cells, operated 2 series 2 parallel to provide a nominal 3V supply. The 1618 will operate down to 1.6V, and the combination of the Luxeon and the boost diode will be sufficiently above the battery voltage that you won't have too much leakage through the boost converter.
Regards,
Jonathan Edelson
You could use the device to produce a regulated output current, by placing a current sense resistor between the load and ground, and connecting the load-resistor node to the feedback pin. The device will adjust the output voltage until the voltage drop across this current sense resistor is 1.25V. This is similar to the technique of using an adjustable three terminal regulator as a current source.
The only problem with the above scheme is that you will have full output current at 1.25V across your sense resistor...if you are running something like single Luxeon, then you will see about 1/3 of the power used by the Luxeon simply dissipated in the sense resistor.
I'd suggest looking at the Linear LT1618, which has a 50mV current feedback as well as a 1.25V voltage feedback. It also has a _shutdown_ pin, which means that you could do the switching with a very low current switch, eg. a reed switch.
In terms of which LEDs to use, I think that the Luxeons will have the benefit of better efficiency and a tighter beam. If efficiency _of the LED_ is your only concern, then the Luxeon will do better both for high power lighting and for low power lighting. Ideally, for lower powers, you would pulse the Luxeon rapidly so that the current level remains high enough to prevent color shifts.
But using a Luxeon will present a very serious problem: the Luxeon is a single LED with a voltage drop of about 3.2V. You can't use a boost converter with a 4.5V supply to drive a 3.2V load. You would need some sort of 'buck-boost' circuit to deal with the battery voltage falling through the load voltage.
The losses in the boost diode (about 0.5V) and in the current sense resistor will also be a very large fraction of the load voltage, meaning that efficiency would be poor. So from the point of view of actually driving the things with a boost converter, I thing that you would do better with the series strings of 5mm LEDs rather than the Luxeon.
You might be able to get away with 4AA cells, operated 2 series 2 parallel to provide a nominal 3V supply. The 1618 will operate down to 1.6V, and the combination of the Luxeon and the boost diode will be sufficiently above the battery voltage that you won't have too much leakage through the boost converter.
Regards,
Jonathan Edelson