How does the RMS voltage capture the time varying part in the conversion ratio of the half wave rectifier?

Question

I am studying conversion ratio of a halfwave rectifier from here.

It says that

Conversion ratio (also called "rectification ratio", and confusingly, "efficiency") η is defined as the ratio of DC output power to the input power from the AC supply. Then the definition of conversion ratio is given as: $$\eta=\frac{P_\text{DC}}{P_\text{AC}}$$ where \$P_\text{DC}=I_\text{DC}^2R_L\$ calculated using average dc current and \$P_\text{AC}=I_\text{RMS}^2(R_L\$+Resistance of secondary coil+ diode forward resistance).

I am unable to understand its intuitive meaning and have the following questions:

1. If the output of halfwave rectifier is given by DC voltage that has ripples, which we consider as the time varying part, then how does RMS voltage contribute to that time varying part so that it can be considered as an AC component.

2 How exactly does RMS voltage capture the time varying part when its definition says that here.

For alternating electric current, RMS is equal to the value of the constant direct current that would produce the same power dissipation in a resistive load.

3. Does RMS voltage some how relate to the concept of variance and DC voltage to mean, so we have a mean part and RMS is the variation around it? No doubt RMS voltage is calculated like square root of second moment of a PDF.

Most of the answers seems to say that efficiency as the percentage of power delivered to the load, relative to the input power.

This is not the case here. Please note that for input I_rms is used while for output I_dc.

Interestingly, I_rms is also supplied to the load resistor for the halfwave rectifier (\$I_{RMS}=I_m/2\$ and not \$I_\text{RMS}=I_m/\sqrt(2)\$)definitely it is the \$I_\text{RMS}\$ after passing through diode. Why not take efficiency in terms of only I_rms? Definitely there is some time varying part of DC that is playing a role here. Therefore it is not efficiency but conversion ratio.

An interesting observation:

For AC current with zero DC:

I_dc=0, I_rms=I_m*0.707

For half wave rectifier output:

I_dc=0.318 I_m, I_rms=I_m*0.5

For pure DC:

I_dc=I_rms

As the amount of AC component/ripples decreases the I_rms is getting closer to I_dc

Please check here to not get confused conversion ratio with efficiency

Answer 1

While I have not encountered this term before, after some reflection, I think it makes sense. The "conversion ratio" is a measure of how completely the AC power has been converted to DC.

The conversion ratio approaches 1 as the output power approaches pure DC with no ripple. The intention behind using RMS in the denominator is to make the denominator = to total power. It is not intended that the denominator be only AC power. RMS current * resistance = total input power. This relationship (Ptotal = Irms * Rload) holds true generally for AC or DC or any mix of the two.

The reason the numerator uses average current is because average current * resistance has the property of approaching total power when the current approaches pure DC with no ripple.

So the numerator approaches the denominator when the output current is pure DC. Any ripple in the output will reduce the numerator compared to the denominator.

Answer 2

Your definition of conversion ratio with the bracketed part removed for clarity and red text used in the appropriate area is this: -

Conversion ratio is defined as the ratio of DC output \$\color{red}{\text{power}}\$ to the input \$\color{red}{\text{power}}\$ from the AC supply.

Conversion ratio is average watts in to average watts out.

It has nothing to do with time varying voltages in this definition.

• So, for an ideal rectifier (half or full), power out equals power in under all circumstances.
• For a non-ideal rectifier, there will be power losses in the diode (s).

Answer 3

No. RMS is simply a kind of average, it's not a higher moment. It's used because RMS voltage times RMS current equals mean power (when they're in phase). You can't use mean voltage and mean current for AC, since those are zero. If you tried multiplying mean absolute voltage and mean absolute current, you would get a value that was lower than the true power, and if you used peak voltage and peak current, you would get a value that was higher. RMS just works right.

Also, I don't see anything confusing about calling the ratio of useful output power to input power "efficiency".