Maximum voltage that a bulb can take before burning out?

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Measuring the maximum voltage a bulb can withstand before burning out is challenging, as bulbs do not have a strict maximum voltage but rather a survival period that decreases with increased voltage. Rechargeable batteries can exceed nominal voltage when fully charged, potentially damaging sensitive bulbs. To protect the bulb in a 9V tactical flashlight, it's advisable to implement a circuit component that limits the voltage to the rated level. The equations provided indicate that higher applied voltage results in increased brightness but significantly reduces bulb life. Understanding these principles is crucial for safely using rechargeable batteries with high-value bulbs.
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is it possible to measure the maximum voltage that a bulb can take before burning out?

I am interested in using rechargeable batteries for a 9v tactical flashlight that I have, but rechargeable 3v batteries actually have a peak voltage of around 3.85 when fully charged. As the bulb is kind of pricey, I don't want to burn it out by putting in rechargeable batteries.

if it's not possible to measure the maximum voltage for the bulb, is it possible to put something in the flashlight circuit to limit the peak voltage to 9 volts?

thanks in advance
 
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1. The no load charge voltage on a battery is different to the load voltage. When you actually load the battery it will drop rapidly down to the nominal cell voltage. It is this nominal cell voltage that you need to use.

2. There is not really a maximum voltage for a lamp as opposed to a survival period. The more volts applied the brighter you burn the shorter you live. Something I tell new staff when working on high voltage systems.

As a rule of thumb you can derate bulbs using the following equations:

Life = Rated Life * (Rated Voltage/Applied Voltage)^12

Intensity = Rated Intensity * (Applied Voltage/Rated Voltage)^3.5

Current Draw = Rated Current * (Applied Voltage/Rated Voltage)^0.55

Note that as you get further away from the rated values these equations get less and less accurate so don't try and put 240V into your 1.5V bulb.
 
Panda said:
don't try and put 240V into your 1.5V bulb.
That would solve Aki's rocket ignitor problem... :rolleyes:
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

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