Circuit

You are to design a power supply based on the circuit of Figure 59 and meeting the specifications below. The power supply consists of a transformer supplied by the 240 V mains, a bridge rectifier, a capacitor smoothing filter and a load.

**Figure 59:** Power supply circuit.
$\begin{figure} \begin{center} \epsfig{file=images/psup.eps}\end{center}\end{figure}$

AC voltages.

$\begin{displaymath}v(t) = V_{peak} \sin ( \omega t) \end{displaymath}$

$\begin{displaymath}\omega = 2 \pi f, \ \ \ f=\frac{1}{T} \end{displaymath}$

$\begin{displaymath}V_{rms} = \frac{V_{peak}}{\sqrt{2}} \end{displaymath}$

Fuse.

The fuse is included to protect the transformer from possible damage due to high or short circuit currents.

Transformer.

$\begin{displaymath}\frac{V_p}{V_s} = \frac{N_p}{N_s} \end{displaymath}$

$\begin{displaymath}\frac{I_p}{I_s} = \frac{N_s}{N_p} \end{displaymath}$

(The resistance of the transformer secondary is several ohms, obviating the need for a surge resistor here.)

Rectifier.

$\begin{displaymath}v_s(t) = V_m \sin (\omega t) \end{displaymath}$

$\begin{displaymath}v_{s(out)}(t) = V_{m(out)} \sin (\omega t) \end{displaymath}$

$\begin{displaymath}V_{m(out)} = V_m -1.4 \ {\rm V} \end{displaymath}$

$\begin{displaymath}{\rm PIV} = V_{m(out)} + 0.7 \ {\rm V} \end{displaymath}$

Filter.

Ripple voltage (bridge full-wave rectifier, peak-peak):

$\begin{displaymath}V_{rpp} = \frac{V_{m(out)}T}{2R_L C} \end{displaymath}$

where T is the period of the input waveform.

Ripple (peak-peak):

$\begin{displaymath}r = \frac{V_{rpp}}{V_{DC}} \end{displaymath}$

DC output voltage

$\begin{displaymath}V_{DC} = V_{m(out)} - \frac{V_{rpp}}{2} \end{displaymath}$

Load.

The load supplied by the power supply is modelled by a load resistor R_L. The value of R_L used in the design corresponds to full load current.

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