If a bulb says 3-9 volts ...is it not going to be very bright at 3v?
- Thread starter Norseman2
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completely wrong!
First of all, its not a BULB!
I guess it might be a led light, with a driver circuit that can run with 3-9 Volt.
Usually - better dirvers - it does not make any difference in output, just the runtime will be longer with higher input voltage.
a real BULB only has a much shorter voltage range, more a very certain one.
When powered with the correct input voltage ... it lights up
with less voltage ... it gives a glow (possibly) but is about useless
with higher voltage -->
First of all, its not a BULB!
I guess it might be a led light, with a driver circuit that can run with 3-9 Volt.
Usually - better dirvers - it does not make any difference in output, just the runtime will be longer with higher input voltage.
a real BULB only has a much shorter voltage range, more a very certain one.
When powered with the correct input voltage ... it lights up
with less voltage ... it gives a glow (possibly) but is about useless
with higher voltage -->
I'm not an expert, but my understanding is that a current regulated light will pull more current from a lower voltage battery to achieve a fixed brightness lumen level. For runtime, it is best to compare watt hours, or V x Mah of your various configs
In your case the choices are at 18650 vs 2x16340 vs 2xCR123. That's 3.7x 3100=11470 vs 7.4x750=5550 vs 6x1500=9000. So the 18650 > 2xCR123 > 2x16340 in terms of run time. Internal resistance comes into play for max outputs, but for general use across all modes, watt hours are best for comparative purposes.
Note also that single cells are always safer to run due to the potential for imbalance and reverse charging.
In your case the choices are at 18650 vs 2x16340 vs 2xCR123. That's 3.7x 3100=11470 vs 7.4x750=5550 vs 6x1500=9000. So the 18650 > 2xCR123 > 2x16340 in terms of run time. Internal resistance comes into play for max outputs, but for general use across all modes, watt hours are best for comparative purposes.
Note also that single cells are always safer to run due to the potential for imbalance and reverse charging.
fyrstormer
Banned
Pretty much correct, but one thing needs to be corrected: the light does not pull current from the battery. Nothing can pull current from a power source. Voltage pushes current through the circuit being powered; the amount of current that gets pushed is completely and helplessly dependent on the voltage of the power source and the resistance of the circuit.
Boost circuits like the kind you're thinking of operate by using induction coils with very low resistance; they temporarily short-circuit the coil, allowing power to flow in as fast as the battery can supply it, then they switch the coil to connect to the emitter, and allow the power to drain out of the coil at a higher pressure (voltage) than it initially had when it flowed in. This process is repeated several million times per second to produce what appears to us to be a smooth flow of power with a higher output voltage than input voltage.
HotWire
Flashlight Enthusiast
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3v to 9v is a range at which the LED module will work safely and correctly. Any voltage in that range should correctly illuminate that LED.
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When driving a single LED, it is usually easy to tell what kind of driver you're using. Drivers/modules labeled in the 3-9, 3-18, or similar voltages are all buck drivers, which means they step down voltage using a comparator, a switch, and an inductor. In order to work, they need to have an input a volt or two above the voltage of the output, or a few tenths of a volt for a more efficient one.
Since LEDs typically operate from 3.3-3.6V when drawing higher currents, the combined voltages of the batteries must be between 3.5-4.0V depending on the driver in order for the switching power supply to actually switch. This means that once the voltage of the battery falls below a certain point on a 3-9V driver/module, the driver is set up to bypass the "buck" part of the circuitry and will drive the LED directly from the battery. This is typically dimmer than using a current-regulated circuit, and will only work until the protection circuit in either the battery or driver kicks in (or the battery over-discharges and catches fire).
While it is possible to a construct a power supply that both steps up and steps down voltage as needed (a buck-boost converter), I've yet to see one in a hand-held flashlight since they are less efficient, more expensive, and have very limited use.
Since LEDs typically operate from 3.3-3.6V when drawing higher currents, the combined voltages of the batteries must be between 3.5-4.0V depending on the driver in order for the switching power supply to actually switch. This means that once the voltage of the battery falls below a certain point on a 3-9V driver/module, the driver is set up to bypass the "buck" part of the circuitry and will drive the LED directly from the battery. This is typically dimmer than using a current-regulated circuit, and will only work until the protection circuit in either the battery or driver kicks in (or the battery over-discharges and catches fire).
While it is possible to a construct a power supply that both steps up and steps down voltage as needed (a buck-boost converter), I've yet to see one in a hand-held flashlight since they are less efficient, more expensive, and have very limited use.
abvidledUK
Flashlight Enthusiast
LEDninja
Flashlight Enthusiast
A 3.7V 18650 is 4.2V fully charged dropping to 3.6V when it should be recharged. It will still power the light but if you let it drop to 2.7V or less it is damaged and may explode when you try to charge it.
An LED runs from 2.8V to 3.6V increasing in brightness as you get closer to 3.6V.
A 3-9V driver is usually a buck driver. Buck drivers need some voltage of its own to operate so for maximum brightness you need to supply 4+ volts.
Look at the charts here: The current out is the same from 6 to 4 volts but drops below 4 volts.
http://www.candlepowerforums.com/vb/showthread.php?192925-AMC7135-Specs-Inside-**UPDATE**
A 18650 will work properly between 4.2V and 4V and will provide steadily dimmer light as the voltage drops further.
2*18650 works from 8.4V down to 7.2V and is more suited to a 3-9V module.
An LED runs from 2.8V to 3.6V increasing in brightness as you get closer to 3.6V.
A 3-9V driver is usually a buck driver. Buck drivers need some voltage of its own to operate so for maximum brightness you need to supply 4+ volts.
Look at the charts here: The current out is the same from 6 to 4 volts but drops below 4 volts.
http://www.candlepowerforums.com/vb/showthread.php?192925-AMC7135-Specs-Inside-**UPDATE**
A 18650 will work properly between 4.2V and 4V and will provide steadily dimmer light as the voltage drops further.
2*18650 works from 8.4V down to 7.2V and is more suited to a 3-9V module.
