What is the average voltage for a fully controlled bridge rectifier with firing angle \$\alpha\$ ?
In a lot of web resources it is said to be:
$$\frac{V_{max}}{\pi} (1+ cos(\alpha))$$
But in my lecture notes it says it is:
$$\frac{2\sqrt{2}V_{rms}}{\pi}cos(\alpha)$$
which is equivalent to:
$$\frac{2V_{max}}{\pi}cos(\alpha)$$
Why have I been taught a different formula to what is online? Which is correct, and if they are both correct when should I use each formula?
Extra Detail
In the first link, the formula is derived from:
$$\frac{1}{\pi}\int_\alpha^\pi V_{max} sin(\omega t)\;d(\omega t)$$
which is \$\frac{1}{\pi}\$ multiplied by the area under the input voltage between alpha (the angle at which the output voltage will appear), and the zero crossing point. In the diagram below this is the area under the first "bump" in the \$V_{out}\$ curve.
In my lecture notes the formula for average output voltage is derived from:
$$\frac{1}{\pi}\int_\alpha^{\pi+\alpha}\sqrt2V_{rms}sin(\omega t)\;d(\omega t)$$
The explanation for this is that two of the thyristors conduct until \$\pi+\alpha\$, as shown in the picture below:
So the output waveform looks something like this:
Which is correct?
Thanks!