A question for the smarter people...

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gerhard1708

Joined Jun 17, 2016
1
I sometimes can't get my head around complex impedance. I do understand all the leading/lagging capacitive and inductive reactance and calculating all the equivalent resistances etc. I watched a video on youtube (
) where w2aew speaks about a 200 mV ac component riding on a dc component that was around 2.6V I think. Then he continues to measure the ac component in ac coupled mode and the scope actually shows the small-signal without any dc (like it is supposed to). So my head is saying there is probably some sort of filter circuit which detects the change in current and then it shows the waveform that goes with it? Now what i cant get my head around is a signal like that which have both ac and dc component. How do those electrons all sharing the same copper wire move around? I understand a little on skin effect, does it work on the skin affect theory that those ac component signals move on the outer ring of the wire while the direct current in the middle? It doesn't make sense because then there will be lots of heat generated...Don't know. Then the 2nd thing is I'm battling to get a better overview of complex impedance. Most textbooks will show lots of calculations about the subject but they never give practical explanations. Let's say I'm driving a small motor or light or whatever and my output impedance and input impedance don't match for let say 100% power transfer(not talking about impedance matching for amplifying) what happens to the rest of the power. Is the copper wire now turning the excess power into heat by heating up? Most importantly WHEN does this become a real problem. Is it only problematic in high powered applications... If I were to proble something like that what would my oscilloscope show. What will it then show in ac coupled-mode? I should probably do more homework on these subjects but unfortunately, most textbooks start by doing the maths without giving practical explanations.
 

AlbertHall

Joined Jun 4, 2014
12,346
So my head is saying there is probably some sort of filter circuit which detects the change in current and then it shows the waveform that goes with it?
All AC coupling does in the scope is to add a capacitor inseries with the input. This capacitor stops the DC component of the signal getting to the scope input but it allows the AC component in.
https://en.wikipedia.org/wiki/Capacitive_coupling
 

profbuxton

Joined Feb 21, 2014
421
AS AlbertHall says, to block the dc component one just uses a capacitor (of the correct value for the desired frequency) to allow the AC signal to "pass". The design trick is to use the right combination of resistance and capacitor to get the range of signals of interest through.
As far as electrons sharing the copper goes,they really don't care. Connect your voltmeter to the output of your power supply(adjustable). Now adjust the output to some steady value(12V). Thats DC, right! Now quickly adjust the value up and down by a volt. The meter will swing up to 13v and down to 11v right? There you have generated a AC signal on top of a steady DC voltage. Now if you connect a AC signal in series(maybe a small transformer (1volt output or whatever you have) in series with your power supply dc output , you now have a mains frequency AC signal on top of your DC output. As far as "skin" effect goes, the effect becomes greater with higher frequency AC, but I suggest you look up Wikipedia for the actual frequency calculations
As far as the complex impedance goes,a good way to look at it is to examine power distribution systems. A distribution system that has no inductance or capacitance(ideal) will dissipate(use) all the power into the load(real watts),. Even better if load matches source resistance.
Real systems generally have quite a bit of inductance(motors, transformers) which have inductance that takes a magnetising current 90 deg (lagging) out of phase with the resistive load. This is the "apparent power" since as the field increases and decreases with the AC it "takes" current from the source and then returns current to the source but doesn't do any "useful" work on the load. This is called "reactive " current and it must be supplied by the source in addition to the current used by the "real" load. So cables must be sized to allow for this total current.
So generally power systems will have a "lagging" reactive current usually of the order of 0.9 power factor.
Now if you have too much inductance in the system the power factor will drop and the source will need to generate more "apparent" power (VARS volt amps reactive) to supply the reactive current and this will affect the "real" power. So remember that capacitors take a 90 deg "leading" current so by connecting suitably sized capacitors you can balance the inductive current and increase the power factor. Now real systems do not add capacitance to bring the power factor to 1 because that is resonance and can generate very high voltages across the capacitances and inductances so it is held at a max. of about 0.98 or so.
Now the same issue applies to small signal circuits. When you tune your radio you adjust the capacitance and inductance in the circuit to be of equal impedance at the desired signal which then has the maximum signal(parallel circuit) across it and this amplified for your speakers(after more stuff happens to it). Same thing with your radio transmitter, you tune your output amplifier to be resonant(equal impedance of inductance and capacitance) at the desired frequency. This signal then goes to your aerial which is also tuned but not as sharply because it has to transmit a range of frequencies(band)
 

crutschow

Joined Mar 14, 2008
34,405
A distribution system that has no inductance or capacitance(ideal) will dissipate(use) all the power into the load(real watts),. Even better if load matches source resistance.
No, that's a common misapplication of the maximum power theorem, and is not what you want for a power distribution system, as that wastes half the energy in the source resistance.
Power distribution systems are designed to have the minimum practical source resistance (including the line, transformer, and generator) so that most of the power goes to the load (typically more than 90%).

The only time you match the load resistance to the source is when you want to deliver the maximum power, independent of efficiency.
One example of that is a solar panel, where they use MPPT to match the load to the varying solar panel impedance, so the maximum solar energy is always extracted.
 
Last edited:

MrAl

Joined Jun 17, 2014
11,453
I sometimes can't get my head around complex impedance. I do understand all the leading/lagging capacitive and inductive reactance and calculating all the equivalent resistances etc. I watched a video on youtube (
) where w2aew speaks about a 200 mV ac component riding on a dc component that was around 2.6V I think. Then he continues to measure the ac component in ac coupled mode and the scope actually shows the small-signal without any dc (like it is supposed to). So my head is saying there is probably some sort of filter circuit which detects the change in current and then it shows the waveform that goes with it? Now what i cant get my head around is a signal like that which have both ac and dc component. How do those electrons all sharing the same copper wire move around? I understand a little on skin effect, does it work on the skin affect theory that those ac component signals move on the outer ring of the wire while the direct current in the middle? It doesn't make sense because then there will be lots of heat generated...Don't know. Then the 2nd thing is I'm battling to get a better overview of complex impedance. Most textbooks will show lots of calculations about the subject but they never give practical explanations. Let's say I'm driving a small motor or light or whatever and my output impedance and input impedance don't match for let say 100% power transfer(not talking about impedance matching for amplifying) what happens to the rest of the power. Is the copper wire now turning the excess power into heat by heating up? Most importantly WHEN does this become a real problem. Is it only problematic in high powered applications... If I were to proble something like that what would my oscilloscope show. What will it then show in ac coupled-mode? I should probably do more homework on these subjects but unfortunately, most textbooks start by doing the maths without giving practical explanations.
This is understood a little better in the time domain.
A series circuit with a cap and resistor where the voltage across the resistor is the output voltage, and excited by a cosine wave, the time expression looks like this:
vR=e^(-t/(C*R))/(w^2*C^2*R^2+1)+(w^2*cos(t*w)*C^2*R^2)/(w^2*C^2*R^2+1)-(w*sin(t*w)*C*R)/(w^2*C^2*R^2+1)

If you look at that you will see three termsl. The first term is the exponential part, the other two are sinusoidal terms.
Now when we look at this after along enough time period we see that exponential part going to zero. The other two sine terms keep going though and that is all we see after some time has passed. That is the part that is represented by a complex number expression. We dont see the exponential part in a complex representation because it is assumed that all exponentials have died down.

What this would look like on a scope is that at first the voltage would shoot up, then it would turn into a pure sinusoidal wave. That is what the series capacitor causes and that is why we dont see any DC component.

Note the expression above is almost the same when we apply a DC part and an AC part:
-(C*R*e^(-t/(C*R)))/(w^2*C^2*R^2+1)+(w*sin(t*w)*C^2*R^2)/(w^2*C^2*R^2+1)+(cos(t*w)*C*R)/(w^2*C^2*R^2+1)

But the same thing happens because time 't' is negative in the exponential part so that goes to zero after some time has passsed.
 
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