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whats the difference between the Vdc and Vrms ?

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Bloodninja

New Member
Hello everyone,

We know that the Vrms (root mean square) means the value of AC voltage that should deliver the same power as the DC voltage...so why is Vdc different ?

I was reading about half-wave rectifiers and found that Vrms doesn't equal Vdc...
 
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crutschow

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If you filter the output of a half-wave or full-wave rectifier you end up with a DC voltage which is near the peak voltage of the sine wave so you you will have Vdc about 1.4 times the RMS value. This is a result of filtering.

If you do not filter the rectified signal, then its value is still the RMS value of the sine wave (or ½ that value for a half-wave rectifier).
 

Sceadwian

Banned
Because if the filter capacitor is large enough and the load is light enough the Vdc output will be equivalent to the peak AC voltage, which is higher than the RMS voltage.
 

Bloodninja

New Member
so why the text i'm reading is mentioning the Vdc of a half wave rectifier regardless of the filter, it says Vdc = 0.318Vm (Electronic devices and circuit theory, Boylestad)

like, isnt it supposed to be Vm / 2(1.414)
 
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Sceadwian

Banned
You're misreading something, only a non capacitor filtered half wave rectifier will have a Vdc equivalent of .318 as stated. A capacitor filter will always cause the voltage to be higher as the capacitor will store a charge up to the peak AC voltage until it's loaded and then the effective voltage will drop with increased load relative to the stiffness of the AC source the value of the capacitor and the effective load resistance.
 
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Bloodninja

New Member
hmmm I'm sorry I need some clarification :p even wikipedia says the same :

Rectifier - Wikipedia, the free encyclopedia

i noticed that the circuit presented there does not include a filter capacitor. i remembered during class that we divided the Vm by π to find the approximate area of the halfwave...

do you agree that its more accurate to say : Vdc = Vm / 2(1.414), or maybe ∫[(Vm Sin wt)] from 0 to π

sorry again, don't mean to argue, i'm just a bit confused.

many thanks
 
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Sceadwian

Banned
You're confused because those are ideal circuits, not practical ones, you will almost never find such a circuit in the real world. The mathematical statement of what Vdc is depends on the load and on the specific rectifier circuit, this is not theory this is something you must actually measure in the real world in the given environment and circuit.
 

KeepItSimpleStupid

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Most Helpful Member
Vrms and Vdc are the same in theory. The problem is, the meters. If you look at it in terms of power delivered to a resistive load V*I and Vrms * I will be the same.

DC meters are easy. AC fluctuations are averaged out. They may use a technique called dual-slope conversion.

AC meters are a mess. Most reject the DC component, but some can be persuaded to measure Vdc+Vac or Vac.
Most are average responding, RMS reading and assume a sine wave of less than some frequency.

True RMS reading meters may be dependent on something called the "crest factor" and the frequency. Fluke once made a TRMS meter which was thermal based.

Converting AC to DC is an entirely different animal. For one, there are losses involved.
If you took 120 VAC and used a load R and you rectified that same 120 VAC and used the same R AND you did the power calculations with a TRMS meter, you would at least see the losses from the wires (possibly) and the IdRd and the IR loss across the diode or diodes. If you used the wrong meter for the job, all bets are off.
 

MrAl

Well-Known Member
Most Helpful Member
hmmm I'm sorry I need some clarification :p even wikipedia says the same :

Rectifier - Wikipedia, the free encyclopedia

i noticed that the circuit presented there does not include a filter capacitor. i remembered during class that we divided the Vm by π to find the approximate area of the halfwave...

do you agree that its more accurate to say : Vdc = Vm / 2(1.414), or maybe ∫[(Vm Sin wt)] from 0 to π

sorry again, don't mean to argue, i'm just a bit confused.

many thanks

Hi there,


You quoted Wikipedia but i could not find the place where it might say that the capacitive filter produces the 'same' output as a rectifier without filtering. Perhaps you can point out where you saw that if that is what you meant to imply.

There is a little bit of confusion that comes up when discussing half wave and full wave rectifier circuits. The main reason for this is that the complete circuit is rarely specified and the authors sometimes cross over from one kind of filter circuit to another assuming that the readers will understand what they meant from the context, or else they just make the mistake of assuming that all the rectifier filter circuits produce the same output, which they definitely do not. I attempt to outline what the differences are here.


The main source of the confusion lies in the filter section itself. In order to specify what to expect on the output we have to classify the filter circuit being used as well as the conditions that it is being used under. I know that sounds little more complicated than you probably wanted to hear, but that's the way some circuits are.

There are at least three types of filters used on the output of full wave rectifiers:
1. No filter
2. Capacitor filter
3. Inductor and capacitor filter (often referred to as a "choke input filter")

How about if we take a quick look at each one?

For the full wave rectifier with no output filter, the output is a sine wave that has it's negative half cycles flipped so they are positive too. The average DC voltage is calculated by integrating over time (summing small contributions over one cycle) and the value comes out to:
Vavg=2*Vpeak/pi
where
Vpeak is the peak value of the original siine wave.

Note here that Vavg is the average voltage, also called the average DC voltage or the average DC value. This is not really a DC voltage, it's just a number we would get if we had taken the average of the full wave signal over time. This value is lower than the peak because the wave fluctuates between the peak and zero.

Now we look at the simple capacitor filter.

The cap filter takes the input and filters it into a smoother value by storing some of the energy so that some energy is still available between the peaks of the sine, unlike the rectifier with no filter which does not do this. This means it can supply energy for more of the total cycle time so the average output goes up. What it actually goes up to depends largely on the value of the cap and the value of the load current. It's not the simplest thing to calculate because the cap keeps the lower part of the output above some level while the upper part still approaches the peak value.
An approximation would be some DC level with a triangle wave riding on top. The average of the triangle plus the DC level would give us the new average DC level, and it would be higher than with no filter at all. Since the loading affects the valley part of the wave, more load would mean a lower triangle dip and so the average would come out lower too. Less load is just the opposite. The overall result is that with a cap filter the average output changes according to how much load we put on it and how much capacitance we use. This is unlike the non filtered full wave which does not change with load but stays at the set value of 2*Vpeak/pi.

Now lets look at the choke input filter.

The inductor and capacitor filter (choke input filter) is entirely different. This is where the expression, "The output DC of the full wave rectifier is the same with or without a filter." comes from. That expression isnt entirely true, but it is true within a certain range of loads when it is designed right. The average DC output of the filter is approximately equal to 2*Vpeak/pi just like the full wave rectifier without a filter, except of course with an LC filter (LC means inductor and capacitor) we get a much smoother output DC level.
There is a limit to this too though, because if we load the filter too much we wont see this relationship anymore, but it does hold approximately over a wide range of loads so this is where that expression comes from. We do have to keep in mind however that it wont always hold with extreme values of load (too light or too heavy).



Summary:

1. The average DC output of a full wave rectifier with no filter is 2*Vpeak/pi and does not change (much) with load.
2. The average DC output of a full wave rectifier with only capacitor filtering varies with load and value of the capacitor.
3. The average DC output of a full wave rectifier with a choke input filter (inductor plus capacitor) varies with load and value of inductor and capacitor, but remains approximately equal to the same as with no filter (2*Vpeak/pi) within a certain relatively wide range of load values. The DC output does vary, but not as dramatically as with a capacitor only filter.
 
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