Question

Diode conduction angle:
prof this formula

Answer 1

Peak Rectifiers

Lets consider the Half Wave Rectifier with Capacitor for smoothing the voltage profile (with less ripples)

(Assuming Diode to be Ideal)

Add capacitor in parallel with the Load as show an below

İD Ty AC mains or other Rest of the circuit Halfcycle peak rectifier

Here Capacitor charges during positive cycle and discharge during the Negative cycle

To make the voltage profile smooth we need to increase the discharging time (so that it discharges slowly)

For this we need to use large value of capacitor C and R is the given load in the circuit

Now drawing the output voltage waveform v_o and output current waveform i_Land Diode current waveform i_D

ti Conduction interval Ar

We’ll require that $\tau = RC>>T$ i.e. time constant of the RC circuit must be much greater that the period of input sinusoidal signal. so the response of the waveform will be like

Now we have to determine the Ripple voltage V_r in the output; Assuming $\tau = RC>>T$

Essentially, a straight line for τ >> T Vo V, At Vi Diode on

When diode is OFF;   $V_{0}(t) = V_{p}e^{-t/\tau}$..........................(1)

At the end of the discharge time,t_d, the output voltage equals

...........................(2)

Substituting the V_o from eq(1) at this time t_d

.  .........................(3)

V Ys-a/r =V,-V, -(1-e-a/r) or

This equation has the two unknowns V, and td, assuming τ is known. If we can determine ta, then we can find V,. Finding ta can be done numerically by equating (1) to the expression for the input voltage v, (t)-V, cos(or) and solving for the time ta when the two are equal as V, cos(V or cos(e This needs to be done numerically since (5) is a "transcendental equation." Alternatively, if Δι is small compared to T (true when τ»T, as assumed), then from (3)

Again, because rT then we can truncate the series expansion of the exponential function to two terms (see Lecture 4) giving V T This simple equation gives the ratio of the ripple voltage to the peak voltage of the input sinusoidal signal for the half-cycle rectifier. It's worth memorizing, or knowing how to derive Often R and Tare fixed guantities. So from (7) to obtain a small ripple voltage we need a large C in this case.

So, We have to proceed to determine the Conduction Interval of diode $\Delta t$, it is the time for which the diode is conducting the current, as we have seen in the above waveforms

The diode conduct from time t_d to T , using eq (4) at time t_d will give following Eq

$V_{p}\cos [\omega (T-t_{d}))] = V_{p}-V_{r}$ or   $V_{p}\cos (\omega\Delta t) = V_{p}-V_{r}$(9)

The conduction interval is assumed to very small therefore Truncating the series expansion of cosine to 2 terms Result in

2V

Hence Conduction angle of the diode is proved above

Thankyou