Posted on Sep 20, 2012

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 D1 and D2, which develop +9 V, and the other is composed of D3 and D4, 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 1.414 ~ 8.9 V.

Click here to download the full size of the above Circuit.

Each diode should have a PIV, Peak Inverse "Voltage, rating that is at least twice the peak voltage from the transformer, 2 x 8.9 ~ 18 V. The 1N4001 has a PIV of 50 V.

Leave Comment

characters left:

New Circuits



Popular Circuits

Power Supply for TTL
Low cost microwave field strength meter
Few Switching regulator circuits
Pulse Generator using Op-Amp
200 solar charged remote control lawn mower
Automatic Switch Project
Simple 100W inverter schematic design