6 Vdc -> 4.5 Vdc

[Magnum]

 What would I need to reduce 6 Vdc -> 4.5 Vdc.

It would power some LED lights. (3 3-cell flashlights)

I don't know what current that the LED's draw, but they only last about 5 hrs. using 3 AAA alkaline batteries.

Thanks.

[Siekmanski]

example led = 1.7 V and current draw is 0.02 A
If your source is 4.5 V you need a 140 Ω resistor

R = (Uv – Uf) / If
  = (4.5 V - 1.7 V) / 0,02 A
  = 2.8 V / 0.02 A
  = 140 Ω

[qWord]

What would I need to reduce 6 Vdc -> 4.5 Vdc.

two Si diods in series.

[dedndave]

better to use a resistor, as Siekmanski showed   :t

many LED's these days are 2.2 Vf (forward voltage)
if you can't find the spec, use a temporary battery and resistor to measure it
current rating varies - start low and go up until you get desired brightness
many LED's are rated at 20 mA

it is not a good idea to apply voltage directly across an LED without some kind of current limiter
i.e., a series resistor, or a constant current source/sink

if you have 3 LED's that are going to want 4.5 V,
then you know the Vf is 1.5 V
if you guess 20 mA

R = (6 - 4.5) / 0.02
R = 75 ohms

[qWord]

it is not a good idea to apply voltage directly across an LED without some kind of current limiter
i.e., a series resistor, or a constant current source/sink

I guess the flashlights have already a resistor (or current regulation) build in. But  maybe he has one of this flash lights that depends on the batteries internal resistance...

[dedndave]

i think you're right, qWord

but, hate to guess
you can easily burn out an LED by applying voltage source directly
however, if it's designed to be connected to a 1.5 V battery, it already has some kind of limiter built in

[Magnum]

One flashlight has 8 leds.
2nd has 12 leds.
Third (Zefal) has a 3 watt bulb using 4.5 volts V x A = Watts Amps = .65 amps

[dedndave]

i figured they were all the same
you always manage to throw a wrench in there, Andy - lol

being as they are not the same, the best thing to do is to run them in parallel, rather than in series

is it practical to use 1.5V as a source ?
that would mean connecting the batteries in parallel, as well

the reason is that one item might draw more current than the others
connecting them in series will result in different voltage drops across them

[dedndave]

i just noticed one load is designed for 4.5 V
give us complete details

what is the power source
what voltage are the other 2 loads designed for
if possible, how much current do they draw

[Magnum]

i figured they were all the same
you always manage to throw a wrench in there, Andy - lol

being as they are not the same, the best thing to do is to run them in parallel, rather than in series

is it practical to use 1.5V as a source ?
that would mean connecting the batteries in parallel, as well

the reason is that one item might draw more current than the others
connecting them in series will result in different voltage drops across them

Super Dave, :-)

There is no wrench.

I posted that ALL the lights use 4.5 volts.

Instead of burning thru 1.5 Volt batteries, I want to use a 6 Volt - 4.5 Ah - UB645 - AGM Battery
that I can recharge.

If you need more details, please let me know. :-)

Andy

[dedndave]

this isn't the ideal for efficiency, but it should work

[dedndave]

another way to go would be to use a simple shunt (zener) regulator with a pass transistor
for example, you could use a resistor through a 5.1V zener diode to get 5.1V
connect that to the base of an NPN transistor
connect the collector of the transistor to +6V
connect the emitter of the transistor to the loads
the resistor value would depend on the desired zener current
shunt regulators are not extremely efficient, though
for example, a 1N4733A zener has a test current of 33 mA
that means that up to 200 mW is wasted in the resistor/zener, alone