🚀 go-pugleaf

In article <1430013@hpvcfs1.HP.COM>, johne@hpvcfs1.HP.COM (John Eaton) writes:
> < But for Ghod's sake, if the device has the extra voltage off its battery
> < (for instance, if it's driving a voltage regulator) why not put a diode
> < in its current path to protect fuses and circuitry if you reverse the
> < battery? And diodes on the DC power lines of your devices, to protect
> < your expensive silicon from mistaken power connections? 
> ----------
> With a 6 volt battery and a typical diode drop of .6 volts you would lose
> 10% of your battery charge in the diode. Thats to much for an effective
> battery powered design. Mechanical interlocks are better.

Perhaps, if there's space for them. But I hate relying on end-users to be
sensible about how things go in. It only takes one mistake to fry very
expensive equipment and waste time and money. Also, someone working on
the circuit may need to connect another power supply to it, and what
good are the mechanical locks then?

A diode from the battery does indeed drop your *voltage*. But in some cases,
such as 9 or 12 volts driving a 5 volt regulator, it only adds to a voltage
drop you have anyway. But it draws no additional current. And the battery
charge depends on *current* drain, not power drain or voltage drop. So in
such instances you lose nothing except the price and board space of the
diode. Also, diodes from DC voltages to ground as follows:

V+ ------------GND----------- V-
	|	|	|
	|	|	|
	----<-------<----

draw extremely tiny current when the polarities are right, but protect
your circuit from negative spikes and accidental reversals. I still
think it's worth the effort.

-- 
			Nico Garcia
			Designs by Geniuses for use by Idiots
			eplunix!cirl!raoul@eddie.mit.edu