True RMS multimeter?

[Jasen Betts]

On 2018-02-04, jurb6006@gmail.com <jurb6006@gmail.com> wrote:

"That's all under-specified. so not even wrong... are they measuring the AC component or the total RMS voltage. "

Trying to express it seems to be the problem. Both the positive and negative sides of the waveform contribute, so it is not really half. But if you have a 1 volt P-P square wave, it is only going to read half a volt because a one volt symmetrical waveform would have the voltage going from half a volt negative to half a volt positive.Similarly if it is going from zero to one volt positive, it will read 50 % of that volt. That is where the one half comes in. Perhaps it would have been easier just to call it the peak value and forget peak to peak. But then the statement might be wrong on non-symmetrical (+ to -) waveforms. They can just treat it all as positive.

Words - dammit !

"consider a darlington pair used common collector, which node do you want to call ground? "

Not ground - COMMON. The emitter is common to both the input and the output. You put current to the B - E junction and then the transistor starts conducting from C - E. thereofre the E is always the common. How you ground it in the circuit will determine input and output impedances as well as voltage and/or current gain. But TO THE TRANSISTOR, the emitter is always common.

Why not the base, both "base" and "common" are antonyms or noble :^)

Another thing they teach wrong is that 180 out of phase is the same
as inverting the waveform. Well they should not teach it that way
because it is only true for waveforms that are symmetrical on both
axes. For example it is absolutely not true of a sawtooth wave. It IS
true of a triangle wave but that is different. The sawtooth is not
symmetrical in both axes, only one, if it even is actually.

No, it needs glide-reflection symmetry on one axis like the waveform you
get by low-passing a square-wave:

_ ____ R __ __
| | | ---/\/\/---+------ .' \ .'
|____| |__ | / `.__/
===== C
|----t----| |
--+--
/////
RC < t


In the frequency domian saying inversion is the same as 180deg phase
shift is completely accurate, in the time domain it sometimes works..

Anyway, I probably should've wrote the peak rather than half the
peak to peak. It would be a more accurate statement.

Still ambiguous.

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[Jasen Betts]

On 2018-02-04, jurb6006@gmail.com <jurb6006@gmail.com> wrote:

"yeah. but the fiction of the other topolgies makes design and anaylsis easier. "

Well put. So you DO understand it. But what about the beginner student ? I think it can cause more confusion in the future if they learn it that way.

It depends what they are learning, if they are learing transistors as a
device you can solder wires to to make useful circuits, the notion of
common base etc is useful.

If they are wanting to write dissertions on the operation of solid
state electronics perhaps not so much.

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