Re: Resistance of ammeter caused voltage drop

From: Larry Brasfield (donotspam_larry_brasfield_at_hotmail.com)
Date: 03/16/05 

Date: Tue, 15 Mar 2005 19:42:43 -0800

"Fred Bloggs" <nospam@nospam.com> wrote in message
 news:423796E7.3090905@nospam.com...
> Larry Brasfield wrote:
>>
>> Measure the current with ammeter in at the ordinary
>> supply voltage. Call this Iao. Measure the voltage
>> at the transmitter with same lashup. Call this Vao.
>> Measure the drop across the ammeter, Vad.
>>
>> Reduce the supply voltage by an amount similar
>> to what the ammeter drops, leaving the ammeter
>> in place. Measure current and voltage, to be
>> called Iar and Var respectively.
>>
>> Calculate Rt = (Vao - Var) / (Iao - Iar)
>> This is the slope of the voltage versus current
>> characteristic for the transmitter.
>>
>> Calculate Ina = Iao + Rt * Vad
>> This is an approximation of the current the
>> transmitter draws when you have no ammeter
>> to reduce the supply voltage it sees.
>
> Yeah? So the F___ what? And since when is resistance x voltage equal a current?- and you don't even have the sign right.

(Finally, a positive contribution.)

That line should have read, of course, thusly:
   Calculate Ina = Iao + Vad / Rt
And, contrary to what the esteemed Mr. Boggs
proclaims, the sign is correct. As defined, both
Vad and Rt are positive quantities (at least if
the transmitter draws more power at higher
voltages, which is already in evidence.) Since
Ina (pronounce as: I 'n'o 'a'mmeter) represents
the current predicted when no ammeter is present,
and that is already known to be higher, adding the
positive ratio Vad/Rt to the current measure with
the ammeter present is correct for getting such a
result. I would hope that this is now obvious
even to the most critical observer.

> Here's the deal you worthless, pretentious son-of-a-bitch-with-VD, you are a worthless p.o.s.- we are wise to your dumb ass-

I perhaps should engage in some name-calling on
account of the above correction, but, as a low ranking
member of the excrement class, I am not up to it.

> go away.

I already said 'No', Fred. Do you think repetition
is going to be effective? (It would appear so.)

>> You could also put a lower shunt resistor across
>> your ammeter and calibrate the combination.
>
> Or buy/modify a p.s. with external sense compensation,

Spending money was an obvious option which
I mentioned in several of its many forms. The
OP's questions led me to believe he might be
interested in using the equipment he had. We
have seen no evidence to the contrary.

> damned worthless idiot.

Fred, I appreciate your opinion. Honestly.
I tried to tell you that earlier, but I suspect
my meaning escaped your notice. 

-- 
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

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