This power supply delivers plus and minus 9 V
to replace two 9-V batteries. The rectifier circuit is
actually two separate full-wave rectifiers fed from
the secondary of the transformer. One full-wave
rectifier is composed of diodes Dl and D2, which
develop +9 V, and the other is composed of D3 and
4, which develop —9 V.
Each diode from every pair rectifies 6.3 Vac, half the secondary voltage, and charges the associated
filter capacitor to the peak value of the ac
waveform, 6.3 x 1 Á14 = 8.9 V. Each diode should
have a PIV, Peak Inverse Voltage, rating that is at
least twice the peak voltage from the transformer,
2x8.9=18 V. The 1N4001 has a PIV of 50V.
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