Re: Resistance of ammeter caused voltage drop

From: Fred Bloggs (nospam_at_nospam.com)
Date: 03/16/05 

Date: Wed, 16 Mar 2005 12:58:17 GMT

Larry Brasfield wrote:
> "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.
>

What a simple-minded little fairy and brainless *** you are- the OP
already said that he gets 1.85A at 7.0V without the ammeter- so who
needs your crap and pretentious pseudo- engineering formula? You want to
assume linearity for voltage drops less than 10%, then that there tells
you the effective power supply output resistance is 0.5V/1.85A=0.27
ohms. Then because the OP also states the load voltage is 6.14V at 1.80
with the ammeter, you have that 7.5V=1.8*0.27+1.8*Rammeter+6.14V or
Rammeter=0.486 ohms. The OP also states that without transmitting, the
drop due to the ammeter is 0.3V making I= 0.3V/(0.486+0.27)ohm=0.4 amps
at 7.2V load voltage- or a load power of 7.2*0.4=2.9 Watts. The load
power during transmit w/o ammeter is 7*1.85=13W and the load power w/
ammeter is 6.14*1.8=11W. The ammeter deprives the circuit of 2W of load
power,- so that the RF output will fall something like eta*2W where eta
is the transmit efficiency. Say for example eta is 33% more or less over
these power consumption ranges- then he must lose about 0.7W transmit
power- if he has a lot of Class A overhead stuff then eta may only be
10% and the power loss is 0.2W in transmit. Your little sissy model
conveys no usable information, and why would it- you don't what you're
doing- you are a pretentious fraud- you latch onto to same linearity
model for load Rt- you don't have a clue what's important and what's
not- and once again you can't understand what the OP is asking- like
mainly how much transmit power degradation can he expect. Get a clue and
take a hike- you are another "unwanted" NG p.o.s- just leave.

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