the finer points of electronics - I

[flippityflop1]

ok so i have questions with electronics that doesn't seem to be covered by tutorials that i've scanned throughout different websites, or i just don't want to read through a lot of material to find that small footnote that i needed to find.

i'm basically going to be posting a series of threads named "the finer points of electronics" to have people explain things to me.

this may sound like trolling (it's not), but consider this a way to get more activity in these forums.



ok so question I:

so the tutorials lead us to the idea that common ground is just the charge of the source equal to the "ambient" charge of environment, in conventional currents... but i've never been sure of it. maybe it actually exerts a "pull"... or maybe it doesn't and that's why some voltage sources provide a negative current, usually along a positive one.

can anybody tell me which is it??

View attachment 42520

 

[audioguru2]

The "ground" of a circuit is simply just a common voltage that is usually 0V and is also part of the input and output wires.
Sometimes the "ground" of a circuit is connected to earth ground.

 

[flippityflop1]

so, looking at the attached diagram, is there a "pull" on the cathode side of the capacitor or a push from the other side?? or both?

if it's a little hard to explain, then tell me how it happens in actual physics, not in conventional currents...

i need to know, because, in the future, i will be designing circuits that depend on there being a "pull". if there isn't for ground, then i guess i'll have to employ negative voltages...

 

[audioguru2]

The battery and the capacitor are simply connected in parallel. There is no ground, no push and no pull.

 

[Kevin Weddle]

Capacitor's pull current through a resistance. Then you have voltage momentarily.

 

[flippityflop1]

uhhh... nobody's explaining this satisfactorily...

c'mon... i need to know so i can ask further questions about semiconductors and VH electronics...


EDIT:
"VH speed electronics"

forgot the "S"

 

[pebe1]

As Audioguru has already explained, the 'ground' in a circuit is merely a reference point against which all other voltage nodes in the circuit are measured.  It does not push or pull anything.

It may or may not be connected to true ground (earth), or to the earthing point in an electric mains supply system, or to a metal chassis on which the circuit is constructed.

You cannot get any simpler than that.

 

[Hero999]

It's difficult to analyse circuit without a 0V node because all voltages are relative.

Ignoring that. The negative side of the capacitor gains electrons and the positive side loses electrons.

 

[audioguru2]

If all you can think about is push and pull then look at the push-pull output of an amplifier where one transistor pulls the output positive and another transistor pushes the output negative.

View attachment 42522

 

[flippityflop1]

ok, i guess i lose here... in all cases "conventional current" and the idea of 0V grounding that comes with it abstracts all cases when it comes to passive components.

in active components as long as it is only immediately in series with a passive component, then it *should be safe*. if 2 or more active components are in series, then i'd still raise questions if it could work.


ok, look at the attached diagram and apply conventional current and see if we can make it all work.


suppose the 2 identical avalanche diodes in series with a active component "Q". we also define an "ambient" voltage potential in the non affected/conducting components such as leads, and wires, any conductors. the voltage drop from each end, is of course, the whole voltage applied to the circuit. the voltage potential from the left of "D_av1", is say, 3V above ambient. the voltage potential from the right of "D_av2" is -4V below ambient. that's a voltage supply of 7V. say the avalanche diodes have a breakdown at 5.5V across.

how this would *usually* work is when we have *single active component* with at least 2 leads and have their electric potential on their immediate 2 leads interact, either by voltage potentials and/or current created... so simply as connecting a single 5.5V avalanche diode to a 7V will be enough to have avalanche breakdown. this is the same for all passive components -- resistors, inductors and capacitors.

but what will happen if we cannot continuously pass the voltage potential all the way from the other end of the circuit to make them interact in any components??

such is the case when we have Q as a hypothetical active component that DOES NOT maintain voltage potential across it the way passive components do.

the voltage across D_av1 is only 3V. and, Q is not letting it's left lead have a voltage potential negative from ambient. the same would be true for D_av2, it only has 4 volts across it and "Q" will not let it's right lead have a positive potential from ambient.


of course, this is hypothetical, as i don't know of a Q component that fits what i'm describing.

View attachment 42523

 

[flippityflop1]

also, why i am convinced that sometimes it might be helpful to consider "pulls" is that i see a lot of voltage supplies apparently provides negative voltages.

 

[Kevin Weddle]

The only addition I'll add to my post is the current of the capacitor must at least be C dv/dt.

 

[audioguru2]

Flippityflop,
I do not know what you are talking about.
You show two opdinary diodes, not zener (avalanche) diodes. Why does your circuit have two zener diodes?
There is no such thing as "an ambient voltage".

Your circuit has no input, no output, no power supply, no biasing for the active device and no load for the active device.

Kevin,
Nobody knows what you are talking about.

 

[pebe1]

If I may add to Kevin's comments:

There is a circuit that would not only supply inverse reactive current for use in unilateral phase detractors, but would also be capable of automatically synchronizing cardinal grammeters. Such a circuit is the "turbo-encabulator." Basically, the only new principle involved is that instead of power being generated by the relative motion of conductors and fluxes, it is produced by the medial interaction of magneto-reluctance and capacitive directance. 

 

[Kevin Weddle]

In the quation v = L di/dt.  The voltage goes up and the current goes up. In the equation
i = C dv/dt. The voltage goes up and the current goes up.

 

[Kevin Weddle]

The other equation is Xc - Xl. So a series LRC is only inductive or capacitive.

 

[pebe1]

........or resistive.