Filtering Scheme

Filtering

We have already seen that the output waveforms of the rectifier aren't pure dc. It has harmonics. The filter can be used to smooth out the dc output voltage.Two types of filter are used and they are:

1)DC Filter: L Type, C Type, and LC Type

2)AC Filter : LC Type

Capacitor Filter For Half Wave Rectifier

During half cycle, Diode conducts and Vs appears across the load. At the same time, capacitor gets charged, when wt = π/2, the capacitor gets charged to Vpeak at wt = π/2. When wt > π/2, Vs decreases to zero at wt = π. But the capacitor discharges slowly through the load according to the time constant = RC. Here, the output voltage (V₀) = Voltage across the capacitor,therefore V₀ is smooth. At t = t₂. V_S is positive and increasing. Therefore, the capacitor charges again and discharges as before.

Here, t1 = Charging period and t2 = Discharging Period

At t = t1, Capacitor Voltage V_C(t₁) = V_P

During Discharging period,

V_c + Ri₀ = 0 --------------------------(i)

Where V_c is the voltage across the capacitor at any instant at any time during discharging.

We can write i₀ = (dq)/dt and q = C * V_c

I₀ = (d( C * V_c))/dt = C * dV_c/dt

From eqⁿ(1), we gt

Vc + R C * dV_c/dt = 0

Or, dV_c/V_C = (-dt/(RC))

Now, integrating this equation, we get:

$$\eqalign{
& \int {{{dVc} \over {Vc}}} = - {1 \over {RC}}\int {dt\; + K} \cr
& where\;K = const,\;whose\;value\;depends\;upon\;initial\;condition \cr
& Loge(Vc) = {{ - t} \over {RC}} + K \cr
& when\;t = 0,\;Vc\; = \;Vm \cr
& k = Loge(Vm) \cr
& Subsitute\;the\;value\;of\;k\;in\;above\;equation, \cr
& Loge(Vc) = {{ - t} \over {RC}} + Loge(Vm) \cr
& Loge(Vc) - Loge(Vm) = {{ - t} \over {RC}} \cr
& or,Loge({{Vc} \over {Vm}}) = {{ - t} \over {RC}} \cr
& or,Vc = Vm\;{e^{{{ - t} \over {RC}}}}\; \cr
& \cr
& Equation\;of\;Load\;Current\;is\;given\;by: \cr
& I0 = {{Vm} \over R}\;{e^{{{ - t} \over {RC}}}} \cr
& Now,\;for\;the\;calculation\;of\;ripple\;current \cr
& Vripple = \;Vm - \;Vm\;{e^{{{ - t2} \over {RC}}}} \cr
& Note: \cr
& {e^{ - x}}\; = (1 - x) \cr
& Therefore, \cr
& we\;can\;write \cr
& Vripple = Vm(1 - 1 + {{t2} \over {RC}}) \cr
& Vripple = {{Vm\;t2} \over {RC}} \cr
& Assu\min g\;t2\; \approx \;T, \cr
& Vripple = {{Vm\;T} \over {RC}} \cr
& Average\;value\;of\;Vo\;can\;be\;written\;as \cr
& Vdc = \;Vm\, - {{Vripple} \over 2} \cr
& \;\;\;\;\;\; = \;Vm({{2fRC - 1} \over {2fRC}}) \cr
& Ripple\;Factor: \cr
& R.f = {{Vripple/(2*\sqrt 2 )} \over {Vdc}} \cr
& \;\;\;\;\;\; = {{{{Vm\;T} \over {2*\sqrt 2 fRC}}} \over {\;Vm({{2fRC - 1} \over {2fRC}})}}\; = \;{1 \over {\sqrt {2(2fRC - 1)} }} \cr} $$

Hence, the value of ripple factor can be calculated by the use of the capacitor as the d.c filter and can be compared with the half wave rectifier without the use of any filter.