Converting AC into DC -:Math:-

[Ayne]

"RMS" stands for Root Mean Square, and is a way of expressing an AC quantity of voltage or current in terms functionally equivalent to DC. For example, 10 volts AC RMS is the amount of voltage that would produce the same amount of heat dissipation across a resistor of given value as a 10 volt DC power supply. Also known as the "equivalent" or "DC equivalent" value of an AC voltage or current. For a sine wave, the RMS value is approximately 0.707 of its peak value.

Q no 1. It mean if i convert AC 30 Volts(RMS) in to Pulsating DC(Not smooth DC) then it will be 30Volts if our converter(Bridge or diodes) is ideal???

Q no 2. If i convert 30 Volts of Pulsating DC into pure(Smooth) DC What will will be the Value of Smooth DC???

Thanks.

 

[evandude]

this sounds an awful lot like a homework problem. But in the interest of trying to help you work your way through it instead of just giving you the answer...

First off, you should clarify what you mean by "pulsating DC" and "smooth DC". I think when you say "pulsating DC" you are referring to the rectified AC signal, and when you say "smooth DC" you are referring to feeding AC through a rectifier into a capacitor.

Question 1 is also ambiguous. What do you mean by "will it be 30 volts"? are you still talking RMS? and if so, are you talking about half-wave or full-wave rectification?

 

[Ayne]

First off, you should clarify what you mean by "pulsating DC" and "smooth DC". I think when you say "pulsating DC" you are referring to the rectified AC signal, and when you say "smooth DC" you are referring to feeding AC through a rectifier into a capacitor.

Yes above is right. Infact my english is not good. I have no proper words for it.

Question 1 is also ambiguous. What do you mean by "will it be 30 volts"? are you still talking RMS? and if so, are you talking about half-wave or full-wave rectification?

I am taking about RMS value on full wave.

Q no 3. I have 30 volts AC RMS i want to convert them into pure DC what will be the DC value???

 

[fingaz]

Well,

peak voltage is the actual voltage of the AC sine wave. . . RMS is the equivalent dc voltage capable of doing the same ammount of work (which is peak voltage x 0.707)

So far as i understand it, the dc output should be the same whether it is full wave rectified, or full wave rectified and filtered through a capacitor.

The only time i think there would be a difference is if it is not full wave rectified.

Maybe I'm wrong, but that's how i think it works.

I'm sure if I am wrog, one of the kind people on here will point it out for me

 

[Hero999]

You are right, the voltage across the capacitor would be the peek value which would be the same regardless as to whether it's a full of half wave rectifier. This is assuming no load conditions and the rectifier being perfect. A real diode has a volt drop of 0.6V to 1.2V (depending on the loading) and a bridge rectifier will have double this.

I could delve into calculas a lo here but I'm not going to. Root Mean Square (RMS) is the sinsoidal voltage that will do the same work as an equivalent DC voltage.

[latex]V_{pk-pk} = V_{RMS} sqrt 2[/latex]
[latex]V_{RMS} = \frac{V_{pk-pk}}{sqrt 2}[/latex]
or:
[latex]V_{RMS}= V_{pk-pk} \frac{1}{sqrt 2}[/latex]

We can't just use the average because voltage has a square relationship with power. Doubling the voltage quadrupels the power, look at the basic formula for power vs resistance and voltage:
[latex]P = \frac{V^2}{R}[/latex]

This is where we get the square root from.