# How do I calculate the average voltage of this wave?

I need to calculate the average voltage of the wave below, the DC voltage is 120 V , the curve starts at 21.65 degrees and the peak voltage is 325.269 V. The solution is 221.3 V. This is from a uncontrolled full-wave bridge rectifier with RE load, the source is 230V rms at 50Hz, R=2 ohm and E=120V. I tried calculating the average voltage of the rectified sine wave and subtracting the average voltage of the portions before the curve starts.

• Take the positive period of a sine-wave and shift it down so within that period the same portion is above 0. Then integrate it over just the interval that is positive in a single period, and divide that result by half the period. Don't forget to adjust the amplitude so after shifting down it still matches up with your measured peak voltage. – DKNguyen Jan 17 at 3:33
• It depends on the Diode current rating Vf 100A, but I get Vavg=233V with 323Vp and 230Vrms in but with ~ 50% of 200Vpp + 120Vdc you get 220V Avg – Tony Stewart Sunnyskyguy EE75 Jan 17 at 3:35
• "I need to calculate the average voltage of the wave below" is that really the actual wave you need to average, or just something similar? – Bruce Abbott Jan 17 at 6:42
• Why not measure it AND why is the top clipped? – Andy aka Jan 17 at 8:22
• @Andyaka that is just the simulation program problem – Pedro Jan 17 at 9:44

Here's a more mathematical approach to your problem. I understand your graph as the flat parts are at 120V. We can divide the area of the half period into a DC part before and after the sine wave (green) when the sine voltage is higher than 120V (blue).

We start by finding the angles, $$\\theta_1\$$ by setting the sine voltage equal to the dc voltage and $$\\theta_2\$$ by symmetry:

\begin{align} V_{dc} = \sqrt{2} V_{RMS} * sin(\theta_1) \rightarrow \theta_1 &= sin^{-1} \left( \frac{V_{dc}}{\sqrt{2} V_{RMS}} \right) \\\\ \theta_2 &= \pi - \theta_1 \end{align}

Using $$\V_{dc}=120V\$$ and $$\V_{ac}=230V\$$, we get the values $$\\theta_1 = 21.649^o \$$ and $$\\theta_2 = 158.35^o\$$. We then calculate the average voltage from the basic formula:

\begin{align} V_{avg} &= \frac{1}{T_p} \int^{T_p} v(t) d \theta \\\\ &= \frac{1}{\pi} \left[ \int_0^{\theta_1} V_{dc} d \theta + \int_{\theta_1}^{\theta_2} \sqrt{2}V_{ac}sin(\theta) d \theta + \int_{\theta_2}^\pi V_{dc} d \theta \right] \\\\ &= \frac{1}{\pi} \left[ V_{dc} \theta \bigg\rvert_0^{\theta_1} - \sqrt{2}V_{ac}cos(\theta) \bigg\rvert_{\theta_1}^{\theta_2} + V_{dc} \theta \bigg\rvert_{\theta_2}^\pi \right] \\\\ &= \frac{1}{\pi} \bigg[ V_{dc} (\theta_1 + \pi - \theta_2) - \sqrt{2}V_{ac}\Bigr(cos(\theta_2) -cos(\theta_1)\Bigr) \bigg] \\\\ &= \underline{\underline{221.33V}} \end{align}

• Thank you, I realize my error now. I used your approach but I divided by the wrong period. For the green part I divided by theta1. – Pedro Jan 17 at 12:37

Vmax = 323 = 325Vp -2V for Vf=1V@ 150A rated diodes.
Vmin = 120V dc

Therefore Vpp = 323-120 = 203Vpp or 101.5Vp

Vacg = Vdc (min) + Vp = 120V + 101.5 = 221.5 ( close enuf for gov't work.)