MintyBoost - Process documentation

Minty Boost
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This page details how I went through the process of coming up with the idea, hardware, design, etc. for this project. It's not 100% correct but it's pretty close. Since this project only took 2 days (on & off) to design/test/release, it's a lot easier to keep track of than something enormous like the x0xb0x.

Original Idea

OK so where does an idea come from anyways? Its the only important question & the most difficult. I guess I'd have to say it was prompted by looking at these half-dozen projects:

• Aaron Dunlaps 9V USB charger
• Another 9V + 7805 USB charger (Instructables)
• Jason Streigel's 9V+7805 USB 'battery' (hackaday)
• Ians Firewire switching charger (Instructables)
• Chris DiClerico's 9V+AA's firewire charger

OK, there's probably even some I'm missing. So what's the overarching theme here? Almost all use 9V batteries and a 7805 (an extremely common linear 5V regulator: makes a solid 5V from 7-18V input). This design works great because, well, 7805's are awesome and 9V's provide 7-9V depending upon how 'dead' they are.

However, there's one thing about 9V's that I've learned (from lots of bad experiences). One is that they don't have a lot of amp-hours: that is, how much current (amps) they can provide and for how long (hours). A duracell 9V provides -about- 500mAh over its lifetime. That's 500 mA (or .5A) for one hour or 100mA for 5 hours. That number is somewhat idealized but its a good starting point.

Another problem is that they don't like to supply a lot of current, because they have high internal resistance (~2ohms), but basically that just means that if you want a lot of current (say to resuscitate a drained device) the 9V wont provide all 500mAh, but maybe more like 400. (Say you're drawing 250mA, then .25A*2ohm = 0.5V lost to internal resistance. For more info on 9V, read the duracell datasheet)

Another problem with the 9V+7805 scheme is that a 7805 is a linear regulator. That means if you want 100mA at 5V (basically, USB power) then you're taking 100mA at 9V and then losing the 4V*100mA = 400mW (.4W) difference as heat.
As the battery wears down to 7V the heat loss goes down to (7-5V)*100mA=.2W but you're still getting bad efficiency. At best the efficiency is 72% (5V/7V) and at worst its 55% (5V/9V) That means you're losing about a third of the battery power to heat!

I'll also throw out that the 7805 itself has a quiescent current of about 5mA so you're always losing 5% (5mA/100mA) efficiency just for regulation! (& that's at least since if you're trickle charging the battery at 50mA then the 5mA quiescent is 10%)

OK so basically the 7805+9V solution works but the efficiency is startlingly low, say 60% or so, and provides only 300mAh at 5V.

We can engineer better!

Read More... http://www.ladyada.net/make/mintyboost/process.html

Don