Bernhard said:
Wow! Thank you very much for your diagram and analysis, Paul.
Any idea what resistor need to be changed to other value to increase the drive current past 258mA?
I would love to get minimum drive current starting around 200mA at minimum pot and beyond that on max pot...
I did not go all the way up with the pot. It should go well above 258mA - I stopped there because I did not want to overdrive the poor LD too much. If you want to go really high, say 800mA, you will have to remove/reduce the 4.7ohm resistor - it will drop too much voltage for the circuit to stay in regulation.
The other things that must be considered is if the op amp will provide enough base current to drive the transistor and that the transistor is not over-driven.
The transistor limits:
Icmax = 500mA
Vcesat=0.5V
Max power dissipation 1000mW
hfe is around 30.
Base drive will be <20mA. The op amp should be able to drive that level.
NOTE: THIS IS A SIMPLIFIED EXAMPLE. SOME FIGURES ARE ESTIMATED, SOME CIRCUIT LOSSES ARE IGNORED AND SOME CIRCUIT DYNAMICS ARE IGNORED.
EXAMPLE 1: (258mA drive, 2x CR123)
Assumptions: Batt voltage = 5.8V(estimated sag under 250mA load), LD voltage = 3V @ 258mA.
4.7 ohm resistor (0.258 x 4.7 = 1.21V. Power dissipated = 323mW)
LD (approx 2.5V to 3.xV - depends on operating curve - let's assume 3V. Power dissipated = 774mW)
1.5 ohm resistor (0.258 x 1.5 = 0.387V. Power dissipated = 100mW)
Transistor (Battery voltage minus all the other drops. 5.8V - 1.21V - 3V - 0.387V = 1.2V. Power dissipated = 310mW)
Looking at this example, total power from the battery is approx 1496mW. Power to the LD is 774mW. Efficiency = 52%.
EXAMPLE 2: (500mA drive, 2xCR123)
Assumptions: Batt voltage = 5.6V(realistic sag under a 0.5A load), LD voltage = 3.3V @ 500mA.
4.7 ohm resistor (0.5 x 4.7 = 2.35V. Power dissipated = 1175mW)
LD (3.2V Power dissipated = 1600mW)
1.5 ohm resistor (0.5 x 1.5 = 0.75V. Power dissipated = 375mW)
Transistor (Battery voltage minus all the other drops. 5.6V - 2.35V - 3.2V - 0.75V = -0.7V!! Circuit will not stay in regulation!)
Since Vcesat=0.5V, the 4.7ohm resistor has dropped at least 1.2V (0.5+0.7) too much. Resistor would have to be reduced to 3ohms or less or removed. You would have to do these caluclations for the upper and lower of your supply voltages.
EXAMPLE 3: (500mA drive, 2xCR123, remove 4.7ohm resistor)
Assumptions: Batt voltage = 5.6V(realistic sag under a 0.5A load), LD voltage = 3.3V @ 500mA. **Remove 4.7ohm resistor.**
LD (3.2V Power dissipated = 1600mW)
1.5 ohm resistor (0.5 x 1.5 = 0.75V. Power dissipated = 375mW)
Transistor (Battery voltage minus all the other drops. 5.6V - 3.2V - 0.75V = 1.65V. Power dissipated = 825mW)
Efficiency = 1600/2800 = 57%
This circuit would stay in regulation until the battery voltage drops another 1.15V because Vcesat of the transistor is 0.5V.
Efficiency (@Vbatt = 4.5V) = 1600/2250 = 71%
HTH,
Paul
EDIT: battery voltages adjusted to indicate realistic sag under load.