Electronic – Average output voltage of a fully controlled full wave rectifier

converterpower electronicsrectifierthyristor

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:

$$\frac{V_{max}}{\pi} (1+ cos(\alpha))$$

For example here and here.

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.
enter image description here

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:

enter image description here

enter image description here

So the output waveform looks something like this:

enter image description here

Which is correct?

Thanks!

Best Answer

Consider the situation when alpha is the full 180 degrees. Clearly, the output is zero i.e. no conduction and the formula \$\frac{V_{max}}{\pi} (1+ cos(\alpha))\$ resolves to: -

\$\frac{V_{max}}{\pi} (1+ cos(180))\$ = \$\frac{V_{max}}{\pi} (1-1)\$ = zero

So, if the options for guessing which formula is correct are : -

  1. \$\frac{V_{max}}{\pi} (1+ cos(\alpha))\$ or
  2. \$\frac{2V_{max}}{\pi}cos(\alpha)\$

I'd have to choose option 1. I'm not planning on deriving the formula from 1st principles because it's too early and I had some vodka last night which doesn't help.

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