11.4.1: Simple Rectifier
1. Consider the circuit of Figure 11.3.1. For an input voltage significantly larger than 0.7 volts, the diode will be forward biased for the positive half of the input sine wave. Therefore, all of the input signal (less 0.7 volts) will appear across the load resistor, R. Conversely, during the negative polarity of the input, the diode will be open, thus blocking any current and producing no voltage across the load.
2. Build the circuit of Figure 11.3.1 using Vin = 10 volts peak at 1 kHz and R = 10 k\(\Omega\). Set the oscilloscope inputs to DC coupled. Place one oscilloscope probe across the input generator and a second probe across the load resistor. Record the peak amplitude of the output load waveform in Table 11.5.1. Also, save an image of the scope trace showing both the input and output waveforms.
3. Measure the load voltage with the DMM (DC volts) and record this in Table 11.5.1.
4. Reverse the diode and repeat steps 2 and 3.
5. Reverse the diode so that it is back to the original orientation. Reduce the input to 800 mV peak and repeat step 2.
11.4.2: Filter Capacitor
6. The circuit of Figure 11.3.2 adds a filtering capacitor across the load. This should help to “fill the gaps” created by the missing portions of the waveform. Build the circuit of Figure 11.3.2 using Vin = 10 volts peak at 1 kHz, R = 10 k\(\Omega\) and C = 22 nF. Making sure that the scope inputs are DC coupled, place scope probes across the input and load, and capture the resulting image. Record the peak value in Table 11.5.2.
7. Measure the load voltage with the DMM (DC volts) and record this in Table 11.5.2.
8. Replace C with the 470 nF capacitor and repeat steps 5 and 6.
11.4.3: Computer Simulation
9. Perform a Transient Analysis simulation of the circuits shown in Figures 11.3.1 and 11.3.2, and compare the resulting waveforms to those captured from the oscilloscope.