Final answer:
To design the specified DC power supply, include a step-down transformer, full-bridge rectifier, and a filter capacitor in the schematic to achieve an average DC output voltage of 12 V with a ±1 V ripple and accommodate for a load impedance of 200Ω, considering the diode voltage drop of 0.7 V.
Step-by-step explanation:
To design a DC power supply with an average DC output voltage of 12 V and a ripple voltage of ±1 V for a load impedance of 200Ω, using a full-bridge rectifier supplied by a transformer with a line voltage of 120V (rms) which translates to an approximate peak voltage of 170V, we need to consider the specifications given. The design will include a transformer to step down the voltage, followed by a full-bridge rectifier to convert the AC voltage to pulsating DC, and a filter capacitor to reduce the ripple voltage. Since the diode drop is 0.7 V, the peak voltage after the rectifier would be 170 V - 2(0.7 V) = 168.6 V. This peak voltage will result in a slightly higher average DC output before filtering. The filter capacitor must be chosen to achieve the desired ripple voltage across the specified load impedance.
A sketch of the overall schematic should include:
- A transformer connected to the 120V (rms) lines stepping down to approximately 24V (rms)
- A full-bridge rectifier connected to the transformer's secondary side
- A filter capacitor connected to the output of the rectifier
- The load impedance connected across the capacitor