1. Half-Wave and Full-Wave Rectifier Materials and equipment Needed: Materials: One 240/24 Vrms center-tapped transformer Two diodes 1N4001 Two 2.2 kΩ resistors One 100 μF, 50 V electrolytic capacitor (any voltage rating is fine since is simulation only) One fuse (any rating is fine since is simulation only) Equipment: Oscilloscope Function generator Procedure: In this experiment, use Multisim to connect a low-voltage (24 V ac) transformer to a 240V 50Hz ac line (use the function generator for simulation and consider whether the typical specification is peak or rms). Connect the half-wave rectifier shown in Figure 1. Notice the polarity of the diode. Be sure to set the tolerance of the resistor to 5%. Connect the oscilloscope so that channel 1 is across the transformer and channel 2 is across the load resistor. View the secondary voltage, V SEC, and the load voltage, V LOAD , for this circuit and observe their waveforms.
8. Assume the input signal to a rectifer circuit has a peak value oF V m = 12 V and is at a Frequency oF 60 Hz. Assume the output load resistance is R = 2kΩ and the ripple voltage is to be limited to V r = 0.4 V. Determine the capacitance required to yield this specifcation For a (a) Full-wave rectifer and (b) halF-wave rectifer. Show all work. Vpp=12 V F= 60 Hz R=2Kohm Vr=0.4 V Vpp=I/2FC ±or Full wave rectifer, V dc =0.707*6 =4.242 V 12=4.242/4*1000*60*C C=1 u± 9. A Full-wave rectifer is to be designed to produce a peak output voltage oF 12 V, deliver 120 mA to the load, and produce an output with a ripple oF not more than 5 percent. An input line voltage oF 120 V (rms), 60 Hz is available. Consider a bridge type rectifer. SpeciFy the transFormer ratio and the size oF the required flter capacitor. Show all work. Vp(dc)=12V IR =120mA Vrms =120V ±=60 H TransFormer ratio =V2/V1 =120/12=10:1 V(ripple)=0.05*6=0.3V V(ripple)=120/60*C C=120*10^-3/60*0.3 =0.0066± =6.6m±