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I have a question that reads:

A full-wave controlled bridge rectifier is considered, shown below. Vs = 230V. Load is represented by \$i_o = 10\$ A. Frequency is 50 Hz and the delay angle is 45\$^{\circ}\$.

(a) Sketch \$V_{o(avg)}\$, \$V_{SCR1}\$ and \$i_s\$

(b) Derive the formula for calculating the average value and RMS value of the output voltage

enter image description here

Do I assume that this FWB rectifier has an RL load or an R load?

I am assuming that there is a purely resistive load:

So, they look roughly like this

(a)

enter image description here

(apologies for paint)

(b)

$$V_{o(avg)} = \frac{1}{\pi} \int_\alpha^\pi V_m \sin{\omega t} \text{ d} \omega t = \frac{V_m}{\pi}\left[-\cos{\omega t}\right]_\alpha ^\pi = \boxed{(1+\sin{\alpha}) \frac{V_m}{\pi}}$$

and RMS:

$$V_{rms} = \sqrt{\frac{2}{\pi}\int_\alpha^\pi V_m^2 \sin^2{\omega t}\text{ d}\omega t }$$

Then, I think I just use \$\sin^2{\omega t}=\frac{1-\cos{2\omega t}}{2}\$ to contiune the integration. Is this right, or am I making things too complicated?

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  • \$\begingroup\$ Think about it a bit. The average voltage is going to be zero because you are modifying positive and negative half cycles equally. \$\endgroup\$ – Andy aka May 1 '17 at 18:42
  • \$\begingroup\$ @Andyaka Are you sure? My answer for the average voltage agrees with the vast literature surrounding this topic... It would only be 0 if \$\alpha = 180^{\circ}\$... The average value of a rectified sinusoid is not 0... \$\endgroup\$ – lmsavk May 1 '17 at 18:46
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    \$\begingroup\$ Yeah I had a brain reset!!! \$\endgroup\$ – Andy aka May 1 '17 at 18:57
  • \$\begingroup\$ @Imsavk you can provide an answer to your own question \$\endgroup\$ – Voltage Spike May 1 '17 at 20:37
  • \$\begingroup\$ And you can mark it as answered \$\endgroup\$ – Voltage Spike May 1 '17 at 21:44
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Following the assumption that there is a purely resistive load: by the definition of the root mean square, you can easily get to:

$$V_{rms} = \sqrt{\frac{1}{\pi} \int_\alpha^\pi V_m^2 \sin^2{\omega t} \text{ d}(\omega t)}$$

without any trouble, for any signal. The rest is trig identities and calculus.

$$V_{rms} = V_m\sqrt{\frac{1}{\pi}\left( \int_\alpha^\pi \frac{1}{2} -\frac{\cos{\omega t}}{2} \text{ d}(\omega t)\right)}$$

$$V_{rms} = V_m\sqrt{\frac{1}{\pi} \left(\left[\frac{\omega t}{2}\right]_\alpha^\pi - \left[\frac{\sin{2\omega t}}{4}\right]_\alpha^\pi\right)}$$

$$V_{rms} = V_m\sqrt{\frac{1}{\pi} \left(\frac{\pi}{2} - \frac{\alpha}{2} +\frac{\sin{\alpha}}{4}\right)}$$

$$\boxed{V_{rms} = V_m \sqrt{\frac{1}{2}-\frac{\alpha}{2\pi}+\frac{\sin{2\alpha}}{4\pi}}}$$

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    \$\begingroup\$ Mark your own answer as accepted instead of changing the title as "solved". It fits better the site policies. \$\endgroup\$ – Enric Blanco May 1 '17 at 21:39

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