Center Tap Transformer

The transformer and rectifier

For a transformer that has two identical windings, the voltage from its center to one side is 22.0 V. The rectifier is made with two identical diodes and the filter is built with a capacitor C=400 mF. Draw the schematics and find the DC voltage and the ripple when its output resistor value is 1400 ohms. Please do the same for 130 ohm resistor.

(A center tap transformer has two identical windings in series. At the center of the transfomer half of the voltage applied is available. They are used with two rectifying diodes situated antiparallel and with the center grounded. This is a netter construction than the usual one winding transformer together with a four diodes bridge. If gives the same rectified wave , with less materials. This is named a full wave rectifier.)

Answer

The figure that contains the schematic and the voltage on the capacitor is at the bottom of this post. (It is taken from wiki) . The DC voltage at the output of the device is:

$V(DC) = Vpp(secondary) -V(diode forward) = 22-0.7 =21.3 V$

On the capacitor the voltage is

$V(t) = V0*exp(-t/\tau)$  and has a time constant for charging and discharging of:

$\tau = R*C = (V0/I)*C$

$V(t) =V0*\exp( -(t*I) / (V0*C))$

$T =1/F =1/60 Hz$

The capacitor is discharging a time interval of about $T/2$.

$V(T/2) = V0*exp(-(I/(2F*V0*C))$

$t=T/2$ because there are 2 diodes (full wave rectifier)

The ripple

Ripple is:

$Vpp(ripple) =V(T/2) -V0 =V0*(1-\exp( -I/(2*F*V0*C)) =$

$=V0*(1 – 1+ I/(2*F*V0*C)) = I/(2FC)$

(check Time domain ripple, Wiki)

For $R=1400 ohm$ the current is:  $I = V(DC)/R = 21.3/1400 =15.21 mA$ and therefore the ripple voltage is:

$Vpp(riplle) = 0.01521/(2*60*400*10^-6) =0.317 V$

For $R =130 ohm$ the current is $I = V(DC)/R = 21.3/130 =163.84 mA$ and the ripple voltage:

$Vpp(ripple) = 3.413 V$

(CC Licence for pictures, free to  reuse and modify)