Diode circuits and DC Power

Introduction 

The information on ripple voltage and p-n junctions, as well as semiconductors and metals, are all contained in unit 2. Metals are free to move within the lattice of the metals due to the presence of free electrons that make them highly conductive. Chemicals doping, on the other hand, happens when a positive charge is formed to have holes because of the electrons making the semiconductors unlikely. The hole doped (d-type), as well as the electron dope (n-type), are the two examples of the semi conductor's regions. The movement of the electrons is restricted and occurs only through the hole because of the band. There is a negative reverse bias version (reduced current) as well as the positive case that is usually forward when explaining the p-n junctions in the semiconductor. Various colors, which could be LED, could get produced when an individual decides to use the band gaps. 

The electric field could get created in the depletion region that is usually between the p-n junctions’ area. The process of determining the performance behavior of semiconductors is through the use of a diode as stated in the third unit. The process is essential in the creation of the rectifier. A full wave rectifier is made possible in unit 3 limited to a specific direction flow of the currents. A voltage ripple is seen from the current flow showing the fluctuations which are said to be the change of voltage when the battery is discharging. Thus, there will be three questions in this unit 3, first measuring the diode, and make I-V curve, which shows the break down voltage –drop as 0.65 V and second is for create diode clipping circuit, to see the voltage drop and last will be the full-wave rectifier, which makes 0.5V ripple and we can compare the theory of it. 

Useful Formula for 

1. Measure the diode I-V from –15 V to about +0.6 V (apply forward voltage with caution). 

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Data 

Table 1. 

I (current)  A	V(voltage)
-13	0.07
-12	0.07
-11	0.099
-10	0.029
-9	0.024
-8	0.043
-7	0.148
-6	0.093
-5	0.143
-4	0.08
-3	0.08
-2	0.105
-1	0.071
0.118	0.20
0.025	0
0.38	0
0.79	0
0.184	0.57
0.268	0.2
0.409	6.12
0.601	13.23

The plot of Diode I-V curve 

The above I-V curve also referred to as the exponential wave can get defined from this plot showing the diode’s current-voltage characteristics even though there is a cutoff line. When the voltage goes over 0.6 V it becomes hazardous hence the measurements were done ranging from -15 V to 0.6 V. The depletion region where the p-n junction is present has an inbuilt potential of 0.6 V. The diode will have a positive bias due to the N-doped region which is negative with respect to the P-doped region which is positive will be across the voltage of 0.6 as per the knowledge from the curve. From the graph, the forward bias is clearly shown even though the value of the reverse bias was too small to get recognized; hence the anemometer’s reading ranges needs to get changed. The value might change when a different diode is introduced even though 0.8 V is the value of the voltage at the cut off line. 

2.Design and test a diode-based asymmetrical clamp (also called “clipping circuit”) tolimit the upper value of the input voltage to + 5.6 V. Use the output from the signalgenerator (SG). [Typically, one clamps all non-zero voltages in a circuit for protection.] How would you add to this circuit to also clamp the lower value of theinput at -5 V? Draw the circuit in your lab report (assume you have an additional power supply of any voltage you choose). 

Table 2. 

Resistor	2
Voltage	5.0V
Voltage drop across diode	0.6v
Vout	5.6V

Circuit drawing picture 1. 

 The total expected voltage is 5.6. The primary voltage is 5.0 V, and 0.6 V is the value of the voltage drop across the diode. The gap between the flat surface and the waveform is 0.6 V with the function wave having a value of 5.6 V as per the oscilloscope. 

3 . Design and build a full-wave rectifier to deliver 20 V DC with less than 0.5 V ripple using a 165 V AC transformer output. Choose the capacitor and the “bleeder” resistor. Test this rectifier on the scope. 

For this question, we built a full-wave rectifier circuit drawing 

In order to make 0.5 ripple voltage, we need to set up based on the calculation 

We use 12V as input, and  = 19V, because of RMS value 0.707*13.43V, 

the input value has some difference. 

The ripple is out to be 0.48V±0.02. 

Fig 1.0 shows the voltage ripple 0.5V. 

Fig 1.1 shows the full-wave of the rectifier. 

A Villard cascade voltage multiplier 

Analysis 

The experiment was successful since the three questions in unit 2 were addressed and their result was known. All the knowledge learned in class such as measuring diode, making I-V curve, and break down voltage –drop as 0.65 V was applied. The experiment agrees with our theory since we were able to determine that the non-linearity nature of diodes are very useful in building circuits especially rectifier circuits. Statistical errors and systematic errors were present throughout the experiment. However, before the experiment had commenced all measuring apparatus was evaluated to ensure it does not have zero error, this help in eliminating the systematic error. To eliminate statistical error reading were recorded three times and its average is obtained to get the best suitable value. Most statistical errors in our experiment emanated from the reading of measurement and it is eliminated by averaging three figures obtained from a particular reading. 

Conclusion. 

The reverse bias, as well as the forward bias and its implications, the various distinctive characteristics of the diode and its general concepts, are some of the things that we have learned from the three questions asked. Also, another critical detail noted is how the different cut off value gets arrived at after using two diodes to build a circuit value. Lastly, the voltage ripper can get observed from a full wave rectifier that was created. The values arrived at are 0.48 V ± 0.02 with a margin error of four percent whereas the value that was expected is 0.5 V. the value is acceptable since it is within the error range.