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### Topic: Rectified vs Unrectified AC and power delivered to load question (ANSWERED) (Read 3270 times)previous topic - next topic

#### Voidugu

##### Apr 16, 2013, 04:30 pmLast Edit: Apr 16, 2013, 09:00 pm by Voidugu Reason: 1
Lets say i have a transformer putting out 11.4 volts AC at a max of 10 amperes. I connect it to a bridge rectifier and therefore the voltage drops down to 10 volts dc (2 x diodes of 0.7 forward voltage drop each). I have a heating element of 2 ohms whose resistance stays constant with temperature (lets just assume that) . If i connect a smoothing capacitor bank on the output of the bridge rectifier and connect the resistor in parallel will more power be delivered to the resistor when compared to:
>the same resistor connect to the AC 11.4 volts?
>the same resistor connected to the 10 volts dc without the smoothing capacitors?

Thank you very much for the help
Regards Void

#### Magician

#1
##### Apr 16, 2013, 04:39 pm
Yes, you will get higher average voltage, and consequently, more heat

#### KeithRB

#2
##### Apr 16, 2013, 05:36 pm
You will need one heck of a capacitor. According to Millman, Vdc = Vm - Idc/(4*f*C) Where Vdc is the average voltage, Vm is the peak from the rectifier, Idc is the DC load current, f is the power line frequency and C is the filter capacitor. In your case, if 11.4 is the RMS AC voltage, Vm = 16.1 V, Idc = 10 A. If we want Vdc to be about 15 V, C needs to be about 40,000 uF

#### Voidugu

#3
##### Apr 16, 2013, 06:14 pm
Keith i am aware of this equation: C = I/(2FV){where C is the capacitance in farads, I is the current in amperes, F is the frequency of the wave and V is the voltage ripple] , which is essentially the same as your equation. I used 2FV though and you used 4FV. Why is that?

Thank you both of you for your help btw.

#### KeithRB

#4
##### Apr 16, 2013, 06:17 pm
Because of full-wave rectification.

#### Voidugu

#5
##### Apr 16, 2013, 06:24 pm
I am kind of confused now with the coefficient of f. You said that f is the frequency of the power line. If the frequency of the AC waveform is 50 hertz then shouldn't the frequency of the fully rectified waveform be 100 hertz? Why is it 200 as you suggest? Please explain.

#### KeithRB

#6
##### Apr 16, 2013, 06:35 pm
First of all, these are approximations. f is set to be the power line frequency and they put in various factors depending on full-wave or halfwave and conduction angle assumptions.

#### Voidugu

#7
##### Apr 16, 2013, 07:43 pm
Keith i think you got the equation wrong. Check the link out.
http://www.electronics-tutorials.ws/diode/diode_6.html
Where: I is the DC load current in amps, ƒ is the frequency of the ripple or twice the input frequency in Hertz, and C is the capacitance in Farads.

The main advantages of a full-wave bridge rectifier is that it has a smaller AC ripple value for a given load and a smaller reservoir or smoothing capacitor than an equivalent half-wave rectifier. Therefore, the fundamental frequency of the ripple voltage is twice that of the AC supply frequency (100Hz) where for the half-wave rectifier it is exactly equal to the supply frequency (50Hz)."

#### KeithRB

#8
##### Apr 16, 2013, 07:57 pm
Don't argue with me, argue with Millman! (This is from Electronics) Again, he is not using f as *ripple frequency* it is the power line frequency. He goes through two pages of derivations to prove it out. One factor of 2 comes from the powerline frequency, the other comes from dividing the ripple by 2 to get the average - as I stated "Vdc is the average voltage.

Sheesh, I couldn't have been more clear.

#### Voidugu

#9
##### Apr 16, 2013, 09:00 pm
Thanx for the help Keith

#### MisterResistor

Let's poke the bear. How can a passive device increase "energy" in this scenario? I say at best, an imaginary lossless device.

#### KeithRB

Simple. Every time the AC voltage is higher than the capacitor voltage, A *lot* of current flows and dumps a *lot* of charge into the capacitor. This quickly drives the voltage to the AC peak voltage. When the AC voltage is less than the capacitor voltage, the capacitor loses charge to the load and causes the voltage to "slowly" drop until the next time the AC voltage is greater than the capacitors voltage.

This drop is what the simplified equation calculates.

#### MisterResistor

So you are saying that the ripple voltage increases power? How about "actual" measured power input to "actual" measured power output? Surely you're not proposing that this creates energy, the elusive perpetual motion machine of sorts.

#### afremont

#13
##### Apr 18, 2013, 05:45 amLast Edit: Apr 18, 2013, 06:27 am by afremont Reason: 1

Yes, you will get higher average voltage, and consequently, more heat

Won't you also get lower peak the same RMS voltage resulting in exactly the same power transfer to the element, minus the losses in the capacitor from resistance?  I'm really having trouble understanding where the "free" energy came from by adding a cap.
Experience, it's what you get when you were expecting something else.

#### KeithRB

There is no power increase. You are simply using the energy storage of the cpacitor and the ability for the transformer/rectifier/diode to provide high peak currents to boost the average DC voltage available.

In the example above, lets say that the diodes conduct 20% of the time, or about 2 ms. The ripple voltage is about 2 V. So the capacitor loses 2 V in the 80% period and must be charged up during the 2ms time period. Since C = q/v,  for  a 40,000 uF capacitor to increase by 2 V, you need .08 coulombs. SInce an amp is one columb per second, you need to dump .08 coulombs in 2 ms, or 40 A of peak current during the conduction time.

So, no magic involved, assuimg all the components can handle the peak currents, the capacitor is kept charged at a DC voltage higher than the RMS value of the AC voltage.

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