Problem 62-7    Fonte: Problem elaborated by the author of the site.

A half-wave rectifier has a rectified voltage V_Cmax = 28 V. The ripple factor counts r = 1% . The load draws a current of 200 mA. Find:

a) the value of the filter capacitor.

b) the peak current value in the diode.

c) compare the item b) result by an alternative method.

Solution of the Problem 62-7   

Item a   

With the values provided we can calculate the value of V_r (rms) using eq. 62-11, reproduced below:

To calculate the capacitor value, we can use eq. 62-17 reproduced below, as we know that I_DC = 0.2 A.

Item b   

To calculate the peak current value of the diode, let's calculate the value of r_pp, as this will enable us to calculate the values of θ₁ and θ₂. So, matching the eq. 62-06 and eq. 62-09, we found that

With this value in hand, let's use the eq. 62-24, reproduced below.

So, the value of θ₁ is:

For the calculation of θ₂ we will use eq. 62-26, or

So, the value of θ₂ is:

Now we can calculate what the conduction angle of the diode is, that is, φ_c. Then

To find the diode's peak current value, let's use eq. 62-28, that is

So, we have

Item c   

Since we know the frequency of the input voltage, we have that the period is T = 1 / f = 16.67 ms. With this value we can calculate T_c ( diode conduction time in ms ), according to eq. 62-28, or

On the other hand, we can calculate the charge value (in coulomb) that the capacitor needs to return to V_Cmax, that is, ΔQ = C x ΔV_r = 3.5 x 10^-3 x 0.034 x 28 = 3.33 x 10^-3 C. So using the known fact that I = ΔQ / Δt, we get

As we can see, a perfect agreement of values, which attests to the veracity of the equations used.