# Find the DC component of a Full wave rectifier (DC Power supply)

I was asked: Find the DC component of

---|Full Wave rect| -- Vr(t) ---|Filter| -- Vo(t)

$$V_r(t), V_{dc} = ?$$

Where $$V_r(t) = |V_m sin(\omega_o t)|$$

I know that the DC component of #$V_r(t)#$ is the Avg value of the wave which I believe is an integral but I don't know for sure and am having trouble finding a direct answer to the question.

• Homework? What have you tried? Dec 19, 2012 at 18:41
• I'm studying for a final, its an old test question. There is not a whole lot to try its just a straight forward question. I put V_m (t) / T where V_m is the wave amplitude and T is the period
– Nick
Dec 19, 2012 at 18:42
• Actually, it's a trick question, as anyone who has built a power supply can tell you. The output voltage has nothing to do with the average value of the absolute value of a sinewave. The problem is that the bridge rectifier is a very nonlinear device, and its output impedance is far from constant. The actual waveform seen at Vr(t) depends very strongly on the nature of the filter. Consider a simple capacitor vs. a more complex L-C or R-C filter... Dec 19, 2012 at 19:55

What you need to do is compute the average of a function. The definition of the average of a function over the interval $[a,b]$ is:

$$\overline{f} = \frac{1}{b-a}\int_a^bf(x)\,dx$$

The sine is a periodic function, so the average must be computed over a period. The absolute value of a sine is still a periodic function, and its period is T=$\frac{\pi}{\omega_0}$, so $a=0$ and $b=T$. Let's throw in the formula what you've got:

$$V_{dc}=\overline{V_r(t)} = \frac{1}{T-0}\int_0^T|V_msin(\omega_0t)|\,dx=...$$

But in that interval the sine is always positive so we can take away the absolute value:

$$...= \frac{1}{T-0}\int_0^T V_msin(\omega_0t)\,dx=\frac{V_m}{T}\cdot\Bigg[-\frac{cos(\omega_0t)}{\omega_0}\Bigg]_0^T = \frac{V_m}{T\omega_0}\cdot\Bigg(1-cos(\omega_0T)\Bigg) = \frac{V_m\omega_0}{\pi\omega_0}\cdot\Bigg(1-cos(\omega_0\cdot\frac{\pi}{\omega_0})\Bigg) = 2\cdot\frac{V_m}{\pi}$$

And there you go.

Sorry for bringing up such an old post but I feel it is beneficial to provide another approach for solving this problem.

The output of the full-wave rectifier circuit is basically the absolute value of the sine-wave. In other words, it is a new periodic signal.

The average value of a continuous signal (aka dc component) is computed as follows:

$$V_{ave}=\frac{1}{b-a}\int^b_a f(x) dx$$

Now it is straighforward to compute the dc component of the output $v_{o}$ of the full-wave rectifier circuit, hence \begin{align} v_{o,dc} &= \frac{1}{\pi} \int^\pi_0 V_m \sin{t} dt \\ &= \frac{1}{\pi} ( -V_m \cos t )\Big|_0^\pi \\ &= \frac{1}{\pi} (V_m + V_m) \\ &= \frac{2}{\pi}V_m \end{align}