Introduction

AC-DC converter and RC low-pass filter is constructed and simulated using OrCAD/PSpice.

Procedures

The AC-DC converter presented is a half wave rectifier. This circuit consists of an AC sinusoidal

voltage source in series with a diode and a resistor.

The RC low-pass filter presented filters out high frequency signals. This circuit consists of an AC

voltage source in series with a resistor and a capacitor.

Solutions

The following schematic models a half wave rectifier.

Figure 1. half-wave rectifier schematic

The following graph displays the input voltage (green) vs. the output voltage (red). The output

voltage is 0V when the sinusoidal voltage enters its negative phase. The output voltage is also

close to 9.2 volts at its peak.

Figure 2. half wave rectifier graph

The following schematic models an RC low-pass filter.

Figure 3. RC filter schematic

The following graph displays the voltage input (green) vs. the voltage output (red). The voltage

output is 70% of its voltage input when the frequency reaches 1 kHz.

Figure 4. RC filter graph

Figure 5. dB loss as frequency increases

Performance Analysis

AC-DC converter:

The value for the load is 60 ohms. The DC output is very close to 9.2V.

RC filter:

The value for the resistor is 1.59 ohms. The 3dB point is at 1 kHz.

Conclusion

The bridge rectifier is known as a full wave rectifier. This rectifier produces a DC output voltage

for an AC sinusoidal signal at the positive and negative amplitude. This is formed by DC pulses.

First, the positive terminal of the voltage source is wired to the middle of a set of diodes. Next,

the negative terminal of the voltage source is wired to the middle of a different set of diodes.

Finally, each set of diodes are wired in parallel to the load resistance and the ground.

The diodes are aligned in the same direction so that when the voltage enters the area between

diodes, it is allowed to pass through one terminal only, and subsequently into the load

resistance. The current is then transferred through the ground terminals back into the "bridge"

of diodes. The power lost through the resistive load give it a lower potential than the point it

entered the bridge. Therefore, it is returned to the opposite terminal through the other set of

diodes. This is done for the opposite side of the voltage source when the sinusoidal wave enters

its negative phase. This creates DC pulses since the current is allowed to pass through the load

at all times.

The filter was designed with the formula 1/2πRC = frequency3dB. The frequency3dB was identified

as 1 kHz. This is attained when the value of the original voltage is at 70% when it reaches a 1kHz

frequency. Solve the formula above for the resistance and you have the value to apply to the

resistor in the circuit.