Playing with high voltage

Now I’m playing with high voltage.

Well,12v DC which is much higher than the 5v I’ve been dealing with. And to increase my chance of hurting myself I’ve been dealing with higher current in the 2+ amp scale, which is much higher than the 60mA I’ve been working with. So I’ve been careful not to cross the streams like Egon.

The circuit is really simple: power, MOSFET to switch on the current, and a pull-down resistor. Now that I have an idea of how to get higher-intensity lighting with Arduino, I can reevaluate how the main lamp will be constructed.

quickMOSFETsketch_bb

quickMOSFETsketch_schem

 

A note about power consumption

The Lamp is a portable object which communicates with the satellites in order to find important (or not) places in the world, and then shine a light.

Pretty simple. But the details are important.

The important part of the above sentence is the concept of portability. I can’t tether the lamp to mains power, or require a giant car battery.

Of all the limitations this project has, is the battery. Right now I’m specifying a 3.7v 2000mAh Lithium Ion Battery with an alternate being the 3.7V 6600mAh Lithium Ion Battery Pack, and connecting it to the system through an Adafruit PowerBoost 500 rechargeable 5V power shield. I guess I could design my own charging system, but lithium-ion batteries scare me. They pack a lot of power per cubic centimeter, but don’t play well with others, tending to melt or blow up if they are shorted, bent, crushed or punctured. So, I will have to create a cage for it so it doesn’t hurt me.

Let’s look at what the system draws without lighting the lamp at all:

Component unit
Arduino35.00mA
Compass board0.11mA
Compass LED20.00mA
GPS board55.00mA
Total110.11mA

So I have 2,000 mAh at worst case battery to play with, but let’s assume a 10% reduction due to recharging, storage loss, etc. This gives me 1,800 mAh to work with. I also am assuming worst-case scenario where the lamp is fully lit (for example at a center of population). In reality, most of the time all the LED’s will be lit. I’m sure there I can use calculus to find the slope and areas of illumination to derive the typical power consumption.

Item1 hr ops2hr ops3hr ops4hr ops
System power110.11220.22330.33440.44
Lamp power remaining1689.891579.781469.671359.56
WS2812 LED @ 60mA
(round down)
28 lamps26 lamps24 lamps22 lamps
Luminous Intensity @ 233 (mcd)36524 mcd6058 mcd5592 mcd5126 mcd
R20WHT-F-0160 @20mA
(round down)
84 lamps78 lamps73 lamps67 lamps
Luminous Intensity @ 600 (mcd)350400 mcd46800 mcd43800 mcd40200 mcd

Besides the total power available, the real limiting factor of the battery is the standard discharge of 500mA which can peak at 1000mA. I don’t want to go toward peaking, because see exploding, above.

I have some options: I could source alternative batteries, use multiple batteries to run different systems or parts of systems, or find a better illumination system so I can get the maximum luminance per mA.