The output power is evaluated as V_DC*I_DC since the rectifier is intended as an AC-to-DC converter, so the (useful) output power is just that delivered at DC. As a matter of fact, some power is also delivered to the load at harmonics of the AC input frequency because of the ripple in the output voltage, so the overal power delivered to the load will be slightly larger than V_DC*I_DC.

A further remark: the definition of the (active) input power P_in=I_rms*V_rms is valid for a sine wave input where V_rms and I_rms are the rms values of the components of the input voltage and current at the fundamental frequency (the only at which both V and I are different from zero) and in phase. In particualr, I_rms is not the overal rms value of the (non-sinusoidal) input current. If a non-sine-wave input is considered, the input power should be calculated considering the contributions of all harmonic frequencies, and not as I_rms,overal*V_rms,overal.

Final remark: efficiency = output power / input power

Adding to what has been said by the Research gate colleague Paoloa, i would like to revert to the active power consumed in a load or delivered by a source. If the voltage is v(t) and the current is i(t), then the

average power p= integral v(t)  i(t) dt / T.........equation1

where t is the integration time. In case of periodic wave forms the integration time is the periodic time.

In case of DC,v(t) =VDC, and i(t) = IDC. Substituting in the definition expression of the active average power we get P= IDC VDC T/T =VVDC IDC.

We do not need to repeat the integration every time for the DC power, it is always given by the above formula.

In case of sinusoidal wave forms with v= Vm sinwt and i= Im sinwt, substituting in equation 1 and integrating we get P= Vm Im /2= (Vm/sqqare root of 2)( Im/square root of 2)= Vrns Irms.

This is a quick proof of the given efficiency expression. The power expressions are

derived from the definition of the average power in its integral form. 

my best wishes

Thank you very much Paolo and Abdelhalim Zekry. A quick question though.

Assume a sine wave AC signal fully rectified using four ideal diodes, What would be the efficiency ? Because a quick calculation based on the above formula shows around 80% Then what happens to the remaining 20 % ? Or do I miss something ?

dear Gayan,

I suppose you are considering a full wave rectifier with a resistive load (no output capacitor). In this case, the remaining 20% of the input power is delivered to the load at harmonic frequencies, since the rectified sine wave load voltage includes a DC component but it is not a pure DC voltage. If you consider the overal power delivered to the load, including the AC components at harmonic frequencies, it can be easily shown that the overall output/input power efficiency would be 100%, since |sin(omega t)|^2 = sin^2(omega t).

Nonetheless, the definition of efficiency for the rectifier is given considering that it is an AC-DC converter, so the "good" output power is only the one delivered at DC. As such, the efficiency is just 80%.

Gayan,

The efficiency as defined above is the HARMONIC EFFICIENCY where the load selects the DC component of the the applied waveform on its terminals and rejects the harmonics. May also expressed in an other way, functionally, only the DC component of the waveform will be the useful one while the ripples will be not useful for the load and lost in form of heat. The quality of the DC power supply voltage depends on the ripple contents in the output waveform. Therefore, normally filter are used to reduce the ripples. The simplest filter is a large capacitor on the output terminals of the rectifier. LC or CLC filters can be used. Also electronic voltage regulators can reduce also the riplles and render thigh quality DC supply. The presence of such ripples may be annoying in electronic circuits such as amplifiers,

best wishes

Dear Paolo and Abdelhalim,

I think I got it now. I should have consider the harmonics. Thanks for the quick reply!

I confused about power conversion efficiency of AC DC converter

PCE= Pload/(Pload + Ploss + Pactive) 100%

@OR PCE= Pout/Pin= IDc*VDC/Irms*Vrms 100%