Full wave rectifier without center tapped transformer

Figure 9.13: Full wave rectification without the center tap

We can do away with the center tap and therefore use a single a.c. supply to the circuit by the use of a slightly more complicated circuit invloving four dioded as shown above.

The diode D2 and D3 conduct in the positive half cycle while D1 and D4 in the negative half cycle. when the load is purely resistive. The output is the same as center tapped transformer based full wave rectifier.

Figure 9.14: Output

We started with the objective of getting a ripple free waveform at the output. We notice that in the circuits that we have discussed so far, this aim is not achieved and the primary reason for that is the change in output voltage as the input voltage changes. We overcome this shortcoming by the use of Energy Storage Elements whose function is to absorb energy at higher voltages and release it at lower voltages, thereby reducing the ripple factor..

Examples of such elements are inductors and capacitors. The circuit below shows the configuration of a full wave rectifier with a capacitor.

Figure 9.15: Full Wave Rectifier with a capacitor

Without any load, the capacitor once charged remains charges and therefore the ripple factor become zero.

Figure 9.16: Output without any load

Now,if a load $R_L$ is connected, some amount of discharging of the capacitor takes place in each half cycle. The magnitude of the ripple factor therefore depends on the $C$ and $R_L$.

Figure 9.17: The effect of load on output

Neglecting the voltage discharge that takes place after the completion during the next cycle, i.e. from the point $T/2$ onwards, we get, at $t=T/2$, $V_R=V_Pe6{-\frac{1}{2fRC}}$, where $f=1/T$. Therefore,

$$\Delta = V_P-V_R$$

$$= V_P(1-e^{-t/RC})$$

$$= V_P(1-(1-\frac{1}{2fRC}+(\frac{1}{2fRC})^2)\frac{1}{2!}-\ldots)$$

Neglecting the square and higher terms for $\frac{1}{2fRC}\ll 1$, we get,

$$\Delta = \frac{V_P}{2fRC}$$

From the figure above, the average value of the voltage can be approximated to be $V_P-\frac{\Delta}{2}$. Therefore,

$$\mathrm{R.F.} = \frac{\Delta}{V_P-\frac{Delta}{2}}$$

$$= \frac{\frac{1}{2fRC}}{1-\frac{1}{2fRC}}$$

$$= \frac{1}{2fRC-1}$$

When $2fRC\gg 1$,

$$\mathrm{R.F.} \approx \frac{1}{2fRC}$$

Next, we will consider the diode current calculations.

In the period P, the diode current, $i_d$ is given bym

$$i_d = \frac{v_m \sin{\omega t}}{R_L} + C\frac{dv_m \sin{\omega t}}{dt}$$

$$= \frac{v_m \sin{\omega t}}{R_L} + C\omega V_m\cos{\omega t}$$

$$= I_m\sin{\omega t + \psi}$$

$$\mathrm{where,}$$

$$I_m = \sqrt{(\frac{V_m}{R_L})^2+(\omega CV_m)^2}$$

$$\psi = \tan^{-1}{\omega C R_L}$$

Note that the peak diode current, $I_m$, increases with reduction in $R_L$ or increase in $C$ or $\omega$.

We see that as $R=R_L$ reduces, R.F. increases.But it cannot become zero for finite resistances. A way to achieve a better (smaller) ripple factor is through the use of a Zener Diode

Figure 9.18: The circuit with a zener diode

The zener diode when connected in parallel with the load makes the load voltage approximately equal to the zener voltage ( $V_0\approx V_Z$)as long as it is reverse biased. This happens if and only if the voltage across the capacitor is greater than the zener voltage.

Figure 9.19: The effect of the zener diode on the output

For this condition to be met, we must have

$$V_Z+(\frac{V_Z}{R_L}+I_Z)R = V_C$$

$$V_Z(1+\frac{R}{R_L})+I_ZR = V_C$$

$$\therefore , I_Z=\frac{V_C}{R}-\frac{V_Z}{R}(1+\frac{R}{R_L})$$

When $V_Z=0$, $I_Z=V_C/R$ and when $I_Z=0$, $V_Z=\frac{V_CR_L}{R_L+R}$.

For the figure above, slope = $(\frac{1}{R}+\frac{1}{R_L})$. When discharge rate is higher so that at some point, $V_C$ becomes less than $V_Z$, then, $V_)$ goes below $V_Z$ as shown below

Figure 9.20: The effect of capacitor discharge on output

When the capacitor voltage is less than the input voltage, the diodes are input dioded are in cutoff and the capacitor sees the following discharge circuit for the case when $v_Z/R_l<(v_C-v_Z)/R$.i.e. $v_C>v_Z(1+\frac{R}{R_L})$. In this case,

$$C\frac{dv}{dt}=\frac{v_c-v_Z}{R}$$

Note that when $v_C$ is lesser, output is not constant.

In the above case,

$$v_C(t) = v_Z-(v_Z-v_P)e^{-t/RC}$$

$$v_Z(1+\frac{R}{R_L}) \leq v_Z-(v_Z-v_P)e^{-t/RC}$$

$$\therefore t \leq RC\log(\frac{R_L}{R}(\frac{V_P}{V_Z}))$$

Now, if this t is greater than the half time period, then no ripples will be seen ever. Therefore, we must have,

$$RC\log(\frac{R_L}{R}(\frac{V_P}{V_Z})) \geq \frac{1}{2f}$$

$$\therefore v_P \geq v_z(\frac{R}{R_L}e^{\frac{1}{2fRC}}+1)$$

Example: If $V_z\;=7.5$ volts, $R_L = 5\Omega$, $R=5\Omega$ and $C=1$mF, we must have $V_P\geq 62.9$ V. Conversely, for a given $V_p$, one can find a range of $R_L$ for no ripples.