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

From: Joe R. <>
Date: Mon Jan 26 16:28:03 2004

   See my reply to Bryan. While you are technically correct, I doubt the
original poster understood a word of what you said. I tried to keep the
explanation at a level that the original poster would understand. In this
case we were talking about a battery (in other words "DC") powered device.
There was no reason to introduce reactance, inductance, impedence, phase,
frequency or any other AC related topic to this thread. You are only
somewhat correct about battery voltage. I was referring to open circuit
voltage which is how batteries are usually measured since "under load
voltage" is obviously dependent on the load.


At 02:27 PM 1/26/04 -0500, der Mouse wrote:
>>>> 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 - 16:28:03 GMT

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