ECEN 325

LAB REPORT 6

**Part A:**

Figure 1: I-V characteristics of Si Diode

Figure 2: I-V characteristics of Ge Diode

**Discussion:**

The graph we obtained is not very clear but we can see that there is a significant difference in the cutoff voltage of both diodes. For our graph, each division is 500mV. Therefore, we can determine that the cutoff voltage for Si diode is about 700mV, which is 0.7V. For Ge diode, the cutoff voltage is about 300mV, which is 0.3V.

Figure 3: Plot of Vout and Vin

**Discussion:**

There is a constant voltage drop across the diode that is about 0.7V. This is identical to the cutoff voltage of the Si diode we used. This constant voltage drop reduced the maximum output voltage in the positive half circle from 9.8V to 9.1V. As for the negative half circle, the voltage swing is significantly small compare to the input. Our input is a 20V peak to peak sine wave and the voltage swing is in uV and nV range. Therefore, is almost not visible in the plot.

**Discussion of Diode Current vs Time:**

According to ohm’s law, current is the ration of voltage and resistance. Therefore, based on Figure 3, when the input is in its positive half circle, the current should also be a sine wave but divided by 1000 which is the resistance value we choose in our circuit. When the input is in its negative half circle, since the voltage is significantly small, the current is also very small. We can assume there is no current go through. Therefore, the plot of current vs time should be similar to the plot of Vout vs time.

**Part B:**

Figure 4: Input voltage for Part B

Figure 5: Output voltage for Part B

Figure 6: Design for Part B

When the input goes through its positive half circle, diode 2 and 3 are forward biased. When the input goes through its negative half circle, diode 1 and 4 are forward biased.

According to the oscilloscope, the mean of output voltage is 4.07V

**Discussion of current flowing in the load resistor in Fig. 6:**

Figure 7: Fig. 6

Since our diode is not ideal, even it is reversed biased, there will still be a very small voltage drop through the diode(This can be found in Discussion of Diode Current vs Time). Therefore the current should be considered as bidirectional even the negative half is significantly small.

**Part C:**

Figure 8: Result for 10uF

Figure 9: Result for 100uF

A) The ripple voltage of 100uF power supply is 1.4V

B) The ripple voltage of 10uF power supply is 6.6V

Our motto is deliver assignment on Time. Our Expert writers deliver quality assignments to the students.

Get reliable and unique assignments by using our 100% plagiarism-free.

Get connected 24*7 with our Live Chat support executives to receive instant solutions for your assignment.

Get Help with all the subjects like: Programming, Accounting, Finance, Engineering, Law and Marketing.

Get premium service at a pocket-friendly rate at AssignmentHippo

Tap to ChatGet instant assignment help