Long lived LED vs EL Wire Lamp

Question

I'm on a quest to build a small cylinder (1/2" diameter) that illuminates for extended periods of time on a single charge. I've looked at numerous small sized batteries and decided on using 2 LR44s rated at ~170 mAh @ 1.5 V (each). This would be my first true engineering project so excuse my shoddy calculations.

I'm debating between using an LED OR an EL Wire design. From what I've read, an LED DC takes ~20 mA for ~7000 mCd luminosity (and ~10 mA for ~3500 using linear relation) which would give me about 34 hours (170 * 2 / 10) of continuous use.

Now from what I've read about EL Wire,

each meter of high brightness EL draws about 10-15 mA at the high voltage

So if I have a 1" length keychain It would draw ~ 0.3 mA (0.025m * 12 mA) which would give me a runtime of ~1,133 hours (170 * 2 / 0.3). Is that at all accurate?

If it is, does an inverter of that size (1/2"x1/2") exist?

Answer 1

You have a few issues with your calculations, if you have two batteries that are rated at 170mAH at 1.5V each and you put them in series to get 3V to power the led but the capacity will still only be 170mAH (devices in series all share the same current, so any current that goes through the first battery will also go through the second). You could place the two batteries in parallel to get a capacity of 170mAH * 2 but then you would only have an output voltage of 1.5V to work with. Thus for calculating the lifetime for the LED you should be using 170mAH/10ma = 17 hours.

For the case of the EL wire you need to account for the output voltage. When you increase the voltage with the inverter the power out will be equal to the power in times the efficiency. Electrical power is voltage times current, so to have an output of .3mA at 100V (average pulled from link provided) will use .3mA*100V = 30mW of power. If you have the two batteries in series to provide 3V to the inverter it will need to supply 30mW/3V = 10mA of current. If you factor in the efficiency of the inverter it will end up needing to pull more than 10mA from the batteries so the expected life will be under that of the LED.

Answer 2

EL wire draws ~10mA per meter at 100V AC (rms). So yes your 1" piece will draw .3mA at 100V, but that's not really relevant if you're comparing it to LEDs drawing 10mA at 3V. You need to convert both to watts in order to accurately compare them.

• LED: .01A*3V = .03W =
• EL: .0003A * 100V = .03W

If you compare them like this, then you see that the EL wire takes the same amount of power as the LEDs.

Now knowing this, you can calculate how much current your inverter will draw. Assuming a perfect inverter, \$P_{in} = P_{out}\$, so \$I_{in}V_{in}=I_{out}V_{out}\$. Plugging in 100V rms for \$V_{out}\$, .0003A for \$I_{out}\$, and 3V for \$I_{in}\$, we find the inverter will draw .01A or 10mA, the same as your LEDs.

However, the inverter will not be 100% efficient, it may be 75% if you're lucky. This means the input power needs to be \$\frac{1}{.75}=1.33\$ times higher, so instead of drawing 10mA, you're now drawing 13.3mA.

Because the EL wire doesn't offer any enormous advantages over LEDs in runtime, and the fact that you will have to incorporate an inverter and deal with high voltage, you're probably better off just using the LEDs.