AC and sinusoidal, any difference???

[samcheetah]

yes i agree with you Styx. the DC is no longer DC. nowhere did i say that it was still DC. i was just saying that you could make a DC-DC transformer by using the method that i mentioned.

but an ideal transformer has infinite permeability. so when a DC would be applied to an ideal transformer the "domains" of the core material will start to align according to the direction of the current. during this the magnetic field intensity(H) will increase and so will the magnetic flux density (B). so there is a change of flux from zero to some value. now according to the relation B=(mew)H this should go on uptil infinity. the domains will be aligning according to the direction of the current. H will increase and so will B increase. but this never happens. why?? because no material on earth is infinitely permeable. when the voltage will be applied the domains will start aligning uptil there are no unaligned domains left. there the flux will stop to change because the magnetic lines of forces per unit is not changing. this is why the practical B-H curve is not like a graph of y=x but instead of that, B goes upto a certain point and then the core saturates.

i hope you understand me. im not trying to tell you that DC will pass through a practical transformer but im telling you of a way to do that in the real world

 

[Roff]

DC can pass through a transformer!

and it doesnt need to be pulsed DC. first lets look at what happens when an AC voltage is applied to a transformer. when the voltage starts to increase in the first half cycle so does the magnetic field intensity. therefore the flux through the core increases. but at some point the flux will stop to change. but in AC that point isnt reached. before that the AC reverses direction and the magnetic field intensity goes negative and so does the flux. thats how the hysteresis loop is formed.

now lets apply a DC voltage to a transformer. the magnetic field intensity H will start to increase and so does the magnetic flux increase. but with DC the transformer gets saturated and the flux ceases to change. this time the field doesnt reverse because DC doesnt change like AC. but if we interchange the input terminals and the output terminals then we could simulate the hysteresis loop of AC with a core without even needing AC. it isnt fluctuating DC niether it is AC but still transformer action is happening.

now about the method to change the terminals. well you could mechanically change the +ive and -ive terminals by a motor or something. but you could also use a better non-mechanical alternative. for more info refer to the book "Switchmode Power Supply Handbook" by Kieth Billings

When you interchange the terminals, you create AC.

 

[Styx]

DC can pass through a transformer!

Damb Ron H you beat me to it

 

[samcheetah]

DC can pass through a transformer!

Damb Ron H you beat me to it

as i said that it can pass through an ideal transformer having infinite permeability. remember that u need a changing flux for transformer action which DC could provide if the transformer was ideal.

when i said that i also told how it could be done with a practical transformer.

 

[Styx]

DC can pass through a transformer!

Damb Ron H you beat me to it

as i said that it can pass through an ideal transformer having infinite permeability. remember that u need a changing flux for transformer action which DC could provide if the transformer was ideal.

when i said that i also told how it could be done with a practical transformer.

except the only thing that will keep teh flux changing (for the trasformer effect to happen) is for the current to keep increasing "for an ideal XFMR". that would not occur even with an ideal XFMR.

the primary winding is in effect an inductor. With DC applied to it the current will increase linearly. With as you said a core with infinite permeability this constantly increasing flux will allow the trasformer effect to continue. BUT the current on the primary will have to keep increasing.

With the current increasing through the inductor (due to the DC applied) it will eveltaully stop for a number of reasons

1) the finite resistance in the primary coil for the given primary current will then equal the DC voltage applied thus no more current change => no transformer effect

2) with the increase in current levels something will fuse either a fuse, the coil or something else => the transformer effect will stop

3) with the increasing drain of current from the utility you will hack of others at teh PCC and saturate the secondary at your local distribuition point. and you will get a nice visit from the power company.

And as you have mention "swapping the +ve and -ve terminals" I have already stated that is used every day in H-bridges to drive motors (that are in essance XFMR's especailly Induction machines) and also that is is NOT DC that is being supplied to the XFMR but AC.

as you state in theory for an ideal XFMR it "should" work BUT in theory communism works...

 

[samcheetah]

as you state in theory for an ideal XFMR it "should" work BUT in theory communism works...

:lol:

well, that solves the issue 8)

 

[Gaston]

isn't an ignition coil on an internal combustion engine an example of a transformer passing dc? i know it has to have the capacitor to make the feild collapse but i dont understand what it actually does.

 

[evandude]

*deleted*

 

[Nigel Goodwin]

isn't an ignition coil on an internal combustion engine an example of a transformer passing dc? i know it has to have the capacitor to make the feild collapse but i dont understand what it actually does.

No, it's pulsed by the points (or switching device in modern engines). It works by the back EMF from the coil, when the current passing through the coil is removed you get a large back EMF - it's the transformer action from this back EMF which generates the spark.

There's no 'ifs or buts' a transformer doesn't pass DC!.

 

[Gaston]

what is the role of the capacitor and what path does the back flow of current take?

 

[Nigel Goodwin]

what is the role of the capacitor and what path does the back flow of current take?

The capacitor (usually still called a 'condensor' in the motor trade) is to reduce sparking across the points - if you put too large a capacitor across them it prevents the fast back emf required, while too small a capacitor creates too much sparking across the points.

The back emf flows through this capacitor back to the battery.

 

[mechie]

AC or DC ?

Nobody seems to have set terms of reference yet ?

I believe that any decision on whether any waveform is AC or DC must include a clear definition of the voltage reference point used (assuming we ignore steady voltage, varying current scenarios?).

In my attached drawing is a waveform (does it matter about the shape??)
If you use the green line as your point of reference then I think everyone will agree that the waveform is AC (ignoring the pulse wave bit at the end :? - see later)

With the blue line as refernce the same waveform becomes DC? I think NOT - I see it as DC with a superimposed AC waveform - how else would you explain the operation of a simple common-emitter amplifier?

The case of the AC waveform (green ref.) is OK but for the pulse wave as this would tend to result in more energy flowing one way than the other, if a capacitor were used to remove any DC offset then the RED line would be closer to 'neutral' bias.

Maybe this is the key - if there is a neutral net energy flow (flow one way is eventually cancelled by energy flow the other way) then the waveform is AC -- if not then there MUST be some DC present.

OK rant over - my straight jacket is back from the dry-cleaners, Gotta go.

 

[samcheetah]

you are right mechie. AC is alternating current, i.e. a current that alternates between a positive and negative value. while DC is direct current which doesnt alternate between a positive and negative voltage.

the wave with the blue reference is AC superimposed on DC as you have said. the pulse wave at the end when seen from the blue reference is pulsed DC.

but i have a little doubt about the last thing you said. in an AC the power (and hence energy) is always positive. think of it, when the voltage is negative, so is the current. and power is V*I, a negative number multiplied by a negative number is a positive number. so at all times energy is positive. if your statement is true then on a 60Hz line frequency a simple lamp will be acting as a source and sink of energy 60 times in a second. i dont think that is true.

the power (and hence) would be negative when there is capacitive reactance in the circuit. but that is another case

 

[mechie]

Oops !

Point conceded :!:
My terminology is wrong - I meant to describe direction of current flow through a load.

Wow - it seemed such a simple question three pages ago ! - now I don't understand it ...

.... I'm going back to bed :!:

 

[samcheetah]

Re: Oops !

.... I'm going back to bed :!:

sweet dreams

 

[Styx]

Electronics is never an easy topic - that why i am still in a job

 

[lord loh.]

Is there any difference between AC and sinusoidal waveform ?
Sine wave is AC, AC means only sine or what ???
Is quassi square waveform AC ?

UPS/inverter genrally give Quassi-square waveform o/p .
To get Sine, we use Ott Filter.
But, can quassi square wave be called as AC.

This may have been answered, but I just could not gather the patience of going through 3 pages...

AC - It changes the direction of flow...A slow chaning AC(.5 Hz or less) may be observed on a galvenometre by viewing the deflection in the opposite direction...

sinusoidal wave is a Sine wave (or a Cos wave... )

If the sinusoidal wave changes the direction then it may be called an AC.

It need not change the direction...What people have been calling a pulsed DC may also be called an AC riding on DC(Sine wave riding on DC). It is DC!!! No doubts. It is a DC which changes it's amplitude every some time.
It may also be viewed as a sine wave with a certain DC offset...

Once this "pulsed DC" alias "Ac riding on DC" alias "AC with DC offset" passes through a coupling capacitor, It becomes a pure AC.

I believe I am 100% correct. Any clarification needed or holes in the explanation ? I shall be more than glad to answer...