Consider we have the instantaneous voltage and current of a full wave (first image).
The active (average) power is obtained as shown in the second image (continuous and discrete values respectively).
Now consider we have the rectified voltage and current of a full wave (third image). How can the average (active) power of the nonrectified signal be obtained?
Note: I have the instantaneous rectified values of voltage and current and need to obtain the real power of the original nonrectified signal.
Well, if you have the rectified waveforms for both current and voltage you can deduce what are the non-rectified waveforms (basically the full sine wave with the same peak value of the rectified waveform). By the plots you can figure out the peak values (I_peak, V_peak), the frequency (w) and the phase shift between the two sine waves (phi), then you can write something like:
i(t) = I_peak * sin(w*t)
v(t) = V_peak * sin(w*t+phi)
Then, knowing that p(t) = v(t) * i(t) and T = 2*pi/w, you can calculate the integral value. That is what comes to my mind at the moment. By the way, what are the units used in those plots?
The average power is computed identically for rectified wave also but the expression for p(t)- product of voltage expression v(t) and current expression i(t) change. Both current and voltage can be expressed to include the window in which they occur for one period T.
hi
. Are you dealing with switched half wave bridge fired at approximately 90deg with inductive load L(high) r ( small) load then the voltage wave should have negative portion unless there is a freewheeling diode .The best approach is to use integration of instantaneous power= v*i on load side + losses in switching device .You can then use energy balance to find supply side energy and if it is repetitive average it .
Cheers
Thank you for your answers.
Abner Andrade, the graphs in my post are simply something i made up in excel to represent the situation. I know that in case of ideal sines i could simply estimate the negative part however i am dealing with nonideal ones and the negative parts are not necessarily ideal zeroes (there might be some noise) so it's hard to decide when the sine starts and ends.
Sudha Balagopalan, could you elaborate further? That's the problem, the expression changes but how? I cannot find anything for such situation on the net.
Narasim Ramesh, the load is indeed inductive with some resistance - it is a motor. i tried to estimate Fi (the phase difference) by obtaining the time of V_max and time of I_max and then obtain Fi in seconds. However for some reason Fi obtained using such approach varies greatly so i am trying to figure out alternative approach to estimate the power factor or active power.
hi
Perhaps one approach is to Fourier analyze the input Voltage and Current.
1.Make the reasonable assumption that input voltage source is a pure sinusoid.
2 Power is transmitted through only the fundamental of the current input waveform.
3.The phase angle of fundamental current wrt pure voltage source gives desired angle.
4. Check with input V*I cos(phi) = integral vdc*idc / T on dc side.
Cheers
With a single diode rectification and a resistor load, the power is reduced by a factor two, as it is only present during 50% of time. The integral tells you that also. A lot of distribution companies do not allow this kind of waveform as it saturates their distribution transformer. Electricity meters without current transformer do still bill the correct power, so it is not a solution to pay less... . Don't put the waveform in a lamp as it results in 50Hz Flicker. Don't put it into a transformer as it will saturate and overheat. However I did try to put a diode in the 230V side of a electronic transformer for a halogen lamp, and it resists as it is not a transformer but a half bridge with a full bridge rectifier at the grid side and a ferrite transformer at 25kHz. It was not for a lamp but for a small heater of 20W resistor with two positions: 10W and 20W. The manufacturer will decline any warranty on the device, but it can work in practice.