justDIY said:pulsed uses less power than continuous wave, if by brightness, you mean apparent brightness, as it applies to the human vision model ... then a 50% PWM with a freq of 150 to 1000 hz would do nicely. ps, you still need the resistor!
btw, what do you mean by efficient?
if your goal is to drive a single LED from a fixed voltage source, the most efficient thing to do is match the led's forward voltage to that of the source ... in other words, you can run two red leds off a 5v rail for the same 'energy cost' as it takes to run 1 red led off the same 5v rail, you're just dropping 1v in a resistor instead of 3v.
if matching Vf to Vcc is not possible, then a switching regulator will be your most efficient means of driving the LED, for example, high power LED taillights (think 1.5 to 3a), using a resistor (or worse, a linear reg) you end up wasting tens of watts of power, using a switcher you end up wasting only a few.
linear regulators (pass transistors and the like) are the worst in terms of efficiency, much worse than a resistor, however, they're also very flexible, so that is a trade-off to be considered.
audioguru said:The LEDs are only 2.78V but the battery is 8V so only 35% of the battery voltage is used for the LEDs and the rest is wasted as heat. Additional power is wasted because the 555 isn't a low power Cmos 555, and the 1k resistor also wastes some power.
Instead if 2 LEDs are in series and 2 groups of them are in parallel (and their resistor values reduced so they have the same current as before) then the efficiency is doubled and the wasted battery power is halved. The transistor works better as a common-emitter switch instead of an emitter-follower as you described.
If you pulse it then it looks dimmer. Your vision isn't a peak and hold circuit like some people think. Brightness is sensed as the average power in a pulsed light. That is how a PWM light dimmer works.
Of course you need a current-limiting resistor. The battery voltage changes during its life so you don't want the LEDs to burn out if the battery is new at 9.5V.Overclocked said:So then is the resistor on the Collector Really Needed?
2.78 x 3= 8.34V. Then they will burn out when the battery is new at 9.5V and will turn off when the battery voltage quickly drops to 7.2V. A 9V battery is considered dead when its loaded voltage drops to 6.0V.If I just hook 3 Of them in series with each other, would it mean less Power is wasted since the resistor is no longer there?
An emitter-follower wastes about 0.7V. A common-emitter switching transistor wastes only 0.1V. The emitter-follower needs its base voltage to go much higher than the output of a 555 can go. A common-emitter switching transistor doesn't.Then if I used 2-3 LEDs in series on the collector, Couldnt I do that on the emitter?
audioguru said:You don't gain anything if you power an LED from an oscillator. The duty-cycle of the on-off time determines the brightness. The same amount of power is used if it is the same brightness as with a continuous DC current. The only thing an oscillator provides is PWM to dim the brightness efficiently.
If you use a very low power Cmos oscillator circuit then the few mA of a 555 won't be wasted, but the LED still needs current pulses of maybe 60mA. 60mA is not low power unless you use a car battery.
You need to use a battery that keeps its voltage up as it runs down. Then the voltage across the current-limiting resistor isn't high so it doesn't waste too much power.
audioguru said:You don't need to boost the voltage if you use a cell from the "lappy". It is probably Li-Ion and is about 3.6V so it is just right for powering 3.3V white LEDs through resistors.
The Li-Ion cell will be 4.2V just after charging so the current-limiting resistor for each LED is 4.2V-3.3V/30mA= 30 ohms. When the Li-Ion cell's voltage quickly drops to 3.6V then the current in each LED will be 3.6V-3.3V/30 ohms= 10mA.
Don't connect LEDs in parallel. Each one has a slightly different voltage so their currents will be different and maybe the one with the lowest voltage will hog all the current and burn out. Then the next will burn out, then the next.
Use a separate current-limiting resistor for each LED.
You will need a circuit to shut off the power from a Li-Ion cell when its voltage gets too low, or the cell will be ruined. I think the minimum voltage is about 3.0V. Even with only 3.0V, a 3.3V LED will make some light.
Be very careful when you charge Li-Ion cells. If they get overcharged then they catch on fire and make a very hot, white flame like magnesium.
audioguru said:Your flashlight doesn't need 9 LEDs since at the High Power setting they are each operating at only 66mA/9= 7.3mA. Their max rating is 30mA each. So two LEDs in series and in series with a current-limiting resistor could be doubled to 4 LEDs with two resistors and would be brighter and use half the current so the battery would last for more than twice as long.
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