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:


which is equivalent to:


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?



3 Answers 3


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.


Figure 4 in your first link agrees with your lecture notes. It seems the difference is that the first equation is for a resistive load, while the second is for a resistive inductive load. Note that the first equation is non negative in the domain \$[0, \pi]\$, as it is impossible for a resistive load to return energy to the source.

  • \$\begingroup\$ Thanks for your answer. It says in my notes that for an inductive load the equation becomes: \$\frac{2V_{max}}{\pi}cos(\alpha)-\frac{2}{\pi}\omega L I\$, which is different from the second equation, so something still isn't right... \$\endgroup\$
    – Blue7
    Apr 22, 2014 at 5:36
  • 1
    \$\begingroup\$ @Blue7 The added term 2/pi * omega * L * I is the "fall of voltage" (inductive load) and is the "internal resistance " of converter ( 2/pi * omega *L). So formula is not false. In fact, in your last picture, there is also a phenomena not shown, "commutation" between thyristors ... which lead to a short between the "bridge" and there is a point where the voltage is equal to zero, the time of commutation which can be not "short". \$\endgroup\$
    – Antonio51
    Nov 7, 2021 at 20:30

The first diagram is the waveforms for the circuit with a freewheeling diode on the load side, which blocks the negative part of the waveform and the formula is (Vmax/π)(1+cos(α))

The second waveform has a negative part (doesnt have free wheeling diode and releases energy from the inductor) so in the waveform there is a negative part and the formula is: (2Vmax/π)cos(α).

Hope this helps.


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