Effect of not using Zero-crossing on line

[alphacat]

Thank you very much MrAl!
Please allow me to decompose your post into sections.

Hi there,
For a slightly inductive load the rise of current would be a bit slower,

I remember well our converstaion about an RL circuit.
When the voltage starts off at its peak, then in order for the current to precede the voltage, there's a DC offset given to the current.
Does it really happen in the case of inductive appliances, or is their R (ohmic resistance) high enough for to eliminate quickly the DC offset?

but geeze for a somewhat capacitive load the current would shoot way up until the capacitive part became charged.

So in that case, a rapid change in current takes place in the LIVE and NEUTRAL wires and in the appliance.
Therefore harmonics are being created in these wires.
I dont understand how do the harmonics damage the appliance or other appliances?
I know that they increase the power factor of the appliance, but dont get how they damage it.

 

[MrAl]

Thank you very much MrAl!
Please allow me to decompose your post into sections.



I remember well our converstaion about an RL circuit.
When the voltage starts off at its peak, then in order for the current to precede the voltage, there's a DC offset given to the current.
Does it really happen in the case of inductive appliances, or is their R (ohmic resistance) high enough for to eliminate quickly the DC offset?


So in that case, a rapid change in current takes place in the LIVE and NEUTRAL wires and in the appliance.
Therefore harmonics are being created in these wires.
I dont understand how do the harmonics damage the appliance or other appliances?
I know that they increase the power factor of the appliance, but dont get how they damage it.

Hi again,


[1]
As you know, when the inductance is presented with a voltage with
a step change, it takes a little while for the current to build in the inductor.
That means that the current will appear a little while after the voltage
starts out.
You can try feeding a RL circuit with a sine wave and with a cosine wave
to see the difference.

[2]
I was saying that the harmonics can interfere with another appliance
and a good example of this is probably an AM radio. The steep rise
in voltage (near the peak) would give rise to many odd harmonics that
could radiate out and interfere with an AM radio or even a stereo music
system. If you look at Fourier you'll see the number of harmonics
is infinite in theory and their amplitude depends partly on the amplitude
of the originating signal, and any one of those harmonics alone can cause
interference so groups of them have an even better chance.
On the other hand, if the voltage starts out at only 2 or 3 volts, the
originating amplitude will be much lower (perhaps 100 times lower)
so the chance of interference, although not entirely eliminated, is
much less.
Case in point: I have 5v signals in one of my devices that totally screws
up AM reception in a nearby radio. Even that low (5v) and there is still
plenty of interference at short distances. One way to get rid of it is to try to
slow down the rise of the signals wave front which isnt always easy and
just about impossible for my particular device.

 

[alphacat]

Hey,

Thank you very much for helping me out!

Hi again,


[1]
As you know, when the inductance is presented with a voltage with
a step change, it takes a little while for the current to build in the inductor.
That means that the current will appear a little while after the voltage
starts out.
You can try feeding a RL circuit with a sine wave and with a cosine wave
to see the difference.

Yes, I conducted the simulation and when the voltage sinewave starts off at zero, the current receives an exponential offset to precede the voltage wave.
However, when the voltage sinewave starts off at its peak (90 degrees' phase), the current doesnt receive an exponential offset since it already precedes the voltage wave.

Isnt it therefore disadvantagous to use zero-crossing in the case of an inductive load?
Since doesnt the exponential offset reduce the life span of the inductive load?

[2]
I was saying that the harmonics can interfere with another appliance
and a good example of this is probably an AM radio. The steep rise
in voltage (near the peak) would give rise to many odd harmonics that
could radiate out and interfere with an AM radio or even a stereo music
system. If you look at Fourier you'll see the number of harmonics
is infinite in theory and their amplitude depends partly on the amplitude
of the originating signal, and any one of those harmonics alone can cause
interference so groups of them have an even better chance.
On the other hand, if the voltage starts out at only 2 or 3 volts, the
originating amplitude will be much lower (perhaps 100 times lower)
so the chance of interference, although not entirely eliminated, is
much less.

I'm currently taking a course that deals with waves and transmission lines, and according to what i learned, in order for the power to be radiated, there must be some non-resistive load that matches the transmission line's characteristic impedance:

How does this model exist in our household system of Neutral and Live wires and appliances?
I'm not sure what causes the power to radiate outside in this system.

 

[MrAl]

Hi again,


[1]
Yes you are right that the current starts off lower when the load is turned on at
the peak of the sine voltage wave. As you say the current is at the right place
for it to be when the sine is at 90 degrees so the inductance doesnt see anything
abnormal. And yes, when the voltage is switched on at 0 degrees the current
surges up (think derivative of the voltage wave) and it goes up higher than when
the load runs normally. But this is with the *current* alone, not the voltage.
When the voltage turns on at zero degrees there is very little to be radiated
because after all it is near zero, but when the voltage is switched on at the
peak (the peak is 170v for 120vac line voltage) that's a huge steep wavefront
and so there are going to be harmonics that who knows how high in frequency
they go up to. Which brings us to [2]...

[2]
The transmission line wont put out too much at 50 or 60Hz, but at the multiple
other higher harmonic frequencies there's got to be some bad mismatch, as it cant
be matched for every single harmonic frequency. It probably becomes one big
antenna.

 

[alphacat]

Hi again,


[1]
Yes you are right that the current starts off lower when the load is turned on at
the peak of the sine voltage wave. As you say the current is at the right place
for it to be when the sine is at 90 degrees so the inductance doesnt see anything
abnormal. And yes, when the voltage is switched on at 0 degrees the current
surges up (think derivative of the voltage wave) and it goes up higher than when
the load runs normally. But this is with the *current* alone, not the voltage.
When the voltage turns on at zero degrees there is very little to be radiated
because after all it is near zero, but when the voltage is switched on at the
peak (the peak is 170v for 120vac line voltage) that's a huge steep wavefront
and so there are going to be harmonics that who knows how high in frequency
they go up to. Which brings us to [2]...

Thank you!

[2]
The transmission line wont put out too much at 50 or 60Hz,

We never talked in the course why is it that at 50Hz/60Hz you dont talk in terms of "impedance matching/mismatch".
Why isnt power radiated outside at low frequencies even if the amplitude is large?

but at the multiple
other higher harmonic frequencies there's got to be some bad mismatch, as it cant
be matched for every single harmonic frequency. It probably becomes one big
antenna.

Thats new to me, perhaps because we only learned so far about lossless transmission lines without dealing with the effects of a non-ideal environment.
The transmission lines acts like an antenna because parasitic capacitance that is created since there are wires close to each other?

 

[MrAl]

Hi again,


I guess even at 60Hz something would get radiated, but it doesnt seem to bother
anything except maybe audio amplifiers which pick up 'hum' from the line.

Wouldnt anything that carried a current act as an antenna to some degree?
I think you are right about the capacitance having an effect as that would
be somewhat low in value and therefore cause more current flow for the
steep voltage wavefronts with the high harmonics. Even lower capacitance
would cause current flow at the higher frequencies.
The electric field is probably going to cause interference too.


It's been quite a while since i looked at these issues however so i dont have
all the answers right now We havent even talked about corona yet

 

[alphacat]

hahahaha, I'm more into Vodka rather than beers

I thought of parasitic capacitance since a resistive object won't be able to radiate the transmitted power since it will dissipate all the power.
However, an inductive or capacitive load will radiate the transmitted power since it doesnt dissipate the power.

*When i'm saying transmitted power, i mean that part of the wave (power) is reflected back to the generator, and the other part goes to the load.

 

[alphacat]

The problem that switching at zero-crossing attempts to fix is L di/dt where the L is inside the appliance (motor, transformer, ballast), not in the AC line wiring.

After what me an MrAl talked about, it seems that there is no problem with inductive loads, since the current wont be changed from its natural behavior if the voltage starts off at peak.
no?

 

[MrAl]

Hi again,

There is another problem though that comes up with inductive loads, and that is
that the current through an inductor can not go to zero in zero time, so that
means that if a triac is being used to switch the inductor on/off, the inductor
current may keep the triac on for a much longer time than desired.

To see this with a simulator, you can create a triac with two transistors and
see what happens when the circuit tries to turn the triac off. It's very
interesting. Your simulator might already contain a triac model too.

If you are not using a triac but rather a transistor or two, then the problem
becomes a huge spike of voltage when the transistor(s) actually turn off
due to the inductor current through a very high resistance.

 

[alphacat]

Yes, I understand what you're talking about.
When I built a relay driver circuit, I used a diode in parallel with the relay, just for preventing a voltage spike to occur which would have damaged the transistor (which acted as a switch).

But i dont see how switching off at zero crossing would help.
On the contrary, swtiching off at the voltage's peak, is equal to switching off at the current's zero-cross.

So isnt zero-crossing just damages inductive loads?

 

[MrAl]

Hi again,


Well, if you are using a triac you can not switch off at zero cross anyway.
If the solid state relay uses a triac or two scr's that also does not turn off
at zero cross either with an inductive load.
I think you are right about switching off at the peak, but that of course would
be for a purely inductive load...if there was any resistance with that inductance
you'd have to figure out the correct angle to turn off at. With a triac the right
angle occurs naturally because the thing wont shut off until zero current and
that would happen with pure inductance or any combination of inductance and
resistance.
I actually ran simulations of this one time and keep screen shots of the result
of current and voltage (using a triac for control). I'll see if i can find those
pictures.
It's interesting that if we turn on at zero cross and try to turn off sometime
later at the zero cross, the current doesnt actually stop flowing until 90 degrees
later (pure inductance). Thus if we were using this for power control we'd have
to think about that too. If we were to turn on at zero cross and then try to turn
off at the very next zero cross we would get 1.25 cycles output, not 1 cycle.
If we were to turn on at zero cross and then try to turn off two cycles later
at zero cross we could get 2.25 cycles output, not 2 cycles.
Interesting isnt it?

 

[alphacat]

Yes, wow i learned so much by you MrAL in this thread

I sincerely thank you.

 

[MrAl]

Hi,

That's weird, i just got the thread update notice in my email today...?