Lab 8

Prelab 8

Richard Lang

The Sketch for question one can be located on the next page before this one, the results of the PSPice simulation can be found on the following pages.

C = .1u

Voltage, Red is place before the diode and Green is after the diode.

Current through the diode and resistor.

C = 1u

Voltage, Red is place before the diode and Green is after the diode.

Current through the diode and resistor.

C = 10u

Voltage, Red is place before the diode and Green is after the diode.

Current through the diode and resistor.

Richard LangLab 8: Diodes and Their Applications

Professor SchugPartners: Kyle Crandall, Christian Burton

11/27/14

I - Introduction

In this lab, our objective is to analyze the voltage and current characteristics of diodes and use the diodes in applications of rectifiers circuit and wave-shaping circuits. Much like capacitors, inductors and resistors, this lab introduces us to a new circuit component, diodes. Diodes are circuit devices that only allow current through one way while blocking off all of the current from the other direction. This means that with ac sources where the direction of current switches, only one direction of the circuit will actually work.

The materials we need for this lab includes basic breadboard parts and instruments to analyze the circuit.

Equipment SummaryPower Supply This tool provides power to the circuit, the supply on the breadboard doesn't meet the

requirements to complete the lab correctly.Function Generator

This creates a function of a variable frequency which when hooked up with the power supply will vary the power given by the function generator.

Digital Multimeter

This piece of equipment is used to measure the voltage and current across parts of the circuit.

Oscilloscope This piece of equipment is used to view the voltage as a function of time and is useful in order to see how the circuit behaves.

Solderless Breadboard

This piece of equipment is used to actually put together the circuit.

Resistors The values of resistors used in this lab are 1 KΩ and 10 KΩ.Capacitors The values of capacitors used in this lab are 1 uF and 10 uFDiode The diode used is a IN4007 diode.

II - Models

We will use one equation in this lab, most the analysis in this lab will revolve around viewing the properties of the voltage and current through the diode. Only equation is Vs=Vd+V

III - Algorithm

The first exercise of this lab involves setting up the simple circuit shown in figure 8.3; then varying the power voltage and recording the current through the diode and the voltage through the diode. This show the voltage and current relationship that is seen in a diode. The next exercise is setting up the circuit as seen in Figure 8.4 with a sinusoidal source with 5 V and a frequency of 1000 Hz. Using the oscilloscope to view both the voltage across the resistors and the voltage coming from the source. Next exercise involves setting up the figure 8.5, using the oscilloscope to view the output of the function generator and the voltage across the resistor. The last exercise has us create the circuit seen in figure 8.6 and viewing the output of the function generator and the output voltage on the same graph.

IV - Implementation/Experiment/Analysis

Exercise 8.1

The resulting data from this exercise is as follows, the source voltage is given by Vs, current through the diode is given be id, voltage through the diode is given by Vd.

Vs (Volts) id (mA) Vd (Volts)-3 -.0003 -3-2.5 -.0003 -2.5-2 -.0002 -2-1.5 .0000 -1.5-1 .0000 -1-.5 -.0001 -.50 .0000 0.5 .012 .3821 .0628 .45051.5 .0973 .46982 .1535 .49042.5 .1914 .49963 .2423 .51113.5 .2912 .52004 .3356 .52704.5 .3805 .53315 .4329 .5388

-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1

-0.1

0

0.1

0.2

0.3

0.4

0.5

id V.S. Vd

id V.S. Vd

Vd (Volts)

id (A

mps

)

Looking at this plot of id and Vd, we can confirm that the voltage and current characteristic of diodes is correct. The current peaks when the voltage is positive while when the voltage is negative, there is no current going through the diode. The positive side of the diode reaches it's threshold voltage immediately at around .5 volts. The slope of Δid/ΔVd increases as the current through the diode, this is because the relationship between current and voltage is exponential. The voltage of the diode is equal to the power source when on the bottom half of the graph.

Exercise 8.2

After setting up the circuit, here is the resulting voltage through the diode and resistor. The diode voltage is given by the blue line and the resistor voltage is given by the yellow line. As we can see, the voltage source is sinusoidal at 1000 Hz with a peak voltage of 5 V. When the circuit changes the direction of current, depending on the direction, it will allow current through the circuit or not. The plot of the diode voltage follows the resistor voltage expect when the voltage is negative. This is exactly what we expect and the voltage through the resistor peaks at 5 V and the diode voltage peaks slightly lower due to thermal causes.

Exercise 8.3

C = 1 μF

Now that we have a capacitor hooked up in parallel in this circuit, the voltage across the resistor is affected by the capacitor always trying to hold in charge. When the circuit goes through a loop, the voltage of the resistor goes to a max on the ripple when not opposing the diode. Then the capacitor begins to charge up slowly as well. The capacitor charging up means that the voltage across the resistor slowly drops. When the voltage switches to opposing the diode, the voltage across resistor goes to a max again because the capacitor will discharge. The average voltage is around 4 V and there is a noticeable ripple effect in the graph.

C = 10 μF

In this instance, the time constant has increased therefore the circuit never gets enough time to charge the capacitor completely. As seen, the voltage across the resistor is almost straight in this case because of this effect. Comparing these to PSPice, we can see that they act as expected.

Exercise 8.4

The figure 8.6 shows the circuit we have built, in this case we have two voltage sources. One of the voltage sources is a normal voltage source which only drives the circuit in one direction. The other voltage source is an AC voltage source which changes direction based on the frequency.

Vdc = 2 V

Vdc = 3 V

As seen in the oscilloscope, Vout is related to Vin. The voltage at which the V out begins to straighten out is due to the single voltage source Vdc. For the most part, Vout is directly equal to Vin and only stops when the voltage becomes positive enough. Due to the additive voltage of Vdc and Vin, we can see that Vin and Vout are directly related to the sum and Vin and Vdc.

V - Conclusion

So overall, we explored the many properties of diodes and performed a multitude of different exercises. We proved that the current and voltage relationship of the diode are indeed true. Using the oscilloscope, we were able to see the voltage of a sinusoidal voltage source and the voltage through the diode directly. This checks off another essential circuit element that has been studied in the lab. Understanding how diodes work is an important thing to understand another way to control the current of the circuit.