Batteries, wallwarts, voltages and currents Re: New Classic Finds and etc. (Long)

From: der Mouse <mouse_at_Rodents.Montreal.QC.CA>
Date: Mon Jan 26 13:27:24 2004

>>> Current is everything... voltage is nothing.
> Wrong! Voltage (V) determines current flow (I).

Well, not quite; voltage is one of the factors determining current.

> Ohm's Law I = V/R. The reason that your phone smoked was the it's
> design relied on the internal impedence (a fancy word for
> resistance(R) ) to limit the current.

Impedance is not just a fancy word for resistance, except possibly when
you're working with pure DC. A more accurate way of putting this is
that the design relied on the voltage sagging under load; speaking of
the power supply's internal impedance amounts to modeling it as a
perfect supply (one which sustains its output voltage under any load)
in series with a resistance, or more generally impedance, which
"explains" (more accurately, "describes") the voltage drop under load.
This is of course not a perfect model, but it's good enough for many
purposes - especially since many of the mechanisms that produce the
effect are due to internal resistances.

> When you plugged in a different supply that had a lower impedence it
> allowed more current to flow even though the voltage was the same.

This is, strictly speaking, nonsensical, though there is a grain of
truth lurking in it.

First, I should note that this discussion plays a bit fast and loose
with terminology. Strictly speaking, what I'm saying here is true only
when either the current flow is pure DC, with no AC component, or when
the load is purely resistive, with no inductive or capacitive
component. But for most purposes it's a close enough approximation to
be workable.

If the voltage *at the device*, *under load*, was the same, then there
would be no difference. When you say "allowed more current to flow
even though the voltage was the same", the truth there is more like
"allowed more current to flow *under load* even though the voltage
*without load* was the same". This is the "grain of truth" I referred
to above. The reason I say it's nonsensical is that if your statement
is taken as applying entirely to things under load, it simply won't
work that way.

> FWIW I did the same thing once when I tried to used a small 6 volt
> lead acid battery to power a camera flash that was designed to run
> off of four 1 1/2 volt AA batteries. The voltage was the same but
> the impedence was lower, that made the current much higher and the
> result was to let all the magic smoke out of the flash.

Here again - the design was expecting the voltage to sag under load,
"because of" (more accurately, "as can be described by") the internal
impedance of the batteries. That is, the voltage was _not_ the same
when it mattered - under full load - even though it may have been close
enough to the same under no, or light, load.

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Received on Mon Jan 26 2004 - 13:27:24 GMT

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