# 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. Jan 17, 2020 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 Jan 17, 2020 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? Jan 17, 2020 at 6:42
• Why not measure it AND why is the top clipped? Jan 17, 2020 at 8:22
• @Andyaka that is just the simulation program problem Jan 17, 2020 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. Jan 17, 2020 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.)