# Experiment 4

### What is Ohmic? I-V Characteristics of a Diode vs. a Resistor

Objective: The object of this experiment is to compare the I-V characteristics of a diode with those of a resistor. By measuring the voltage drop across the diode or resistor as the current is varied, the student will discover the relationship between the current and the voltage.

Time: 40-50 minutes

Review of Scientific Principles:

Requirements: To make a current flow through a material, three requirements must be met.

1) An electric field must exist; 2) charge carriers must be present in the material; and 3) the charge carriers must be mobile. To establish an electric field, a voltage is applied to the circuit. The charge carriers are the valence electrons in a conductor, or the electrons in the conduction band and the holes in the valence band of a semiconductor or insulator. The mobility is dependent on the crystal structure and the temperature.

Conductor: For a conductor, such as a metal, the valence electrons occupy partially filled energy levels to form a valence band. The crystal structure of a metal allows the valence electrons in the valence band to move freely through the crystal. However, as the temperature increases, the atoms vibrate with greater amplitude, and move far enough from their equilibrium positions to interfere with the travel of the electrons. Only near absolute zero is the mobility at its maximum value.

Semiconductor: For a semiconductor or insulator, the valence electrons occupy a filled valence band. Electrons must move from the valence band to the conduction band (leaving holes, vacancies, in the valence band). Both the electrons in the conduction band and the holes in the valence band are considered charge carriers. The number of these charge carriers is dependent on the temperature and the material. As the temperature increases, more electrons have the energy needed to "jump" to the conduction band. (Important: The electrons do not move from a place in the crystal, called the valence band, to another place, called the conduction band. The electrons have the energy associated with the valence band, and acquire enough energy to have the energy associated with the conduction band. An energy change occurs, not a position change.)

Doping: Doping of a semiconductor material, by adding atoms with one more or one less valence electron than the base material, is one method of increasing the number of charge carriers (such as adding Ga, with three valence electrons, or As, with five valence electrons, to Ge or Si which has four valence electrons). Addition of a Group V element, such as As, forms an n-type material, which provides new "donor" energy levels. Addition of a Group III element, such as Ga, forms a p-type material, which provides new "acceptor" energy levels. The energy needed for an electron to move from the valence band to the acceptor level as with Ga (forming a hole), or from the donor level to the conduction band as with As (yielding a conducting electron) is less than the energy needed to make the original "jump" from the valence band to the conduction band of the pure semiconductor material. Thus, for a doped semiconductor material as compared to a pure semiconductor material (at the same temperature), the doped semiconductor would have more electrons in the conduction band (n-type), or more holes in the valence band (p-type). For and n-type material, the carrier of electricity is a negative electron. For a p-type material, the carrier is a positive hole. As the temperature increases, the atoms do vibrate with greater amplitude. However, the increase in number of charge carriers has a greater effect on increasing the material's conductivity than the reduction caused by the vibrating atoms.

Resistor: When a voltage is applied across a resistor, an electric field is established. This electric field "pushes" the charge carriers through the resistor. This "push" gives the charge carriers a "drift velocity" in the direction from high potential energy to low potential energy. As the voltage increases, the drift velocity increases. Since the amount of current flowing through a resistor is directly proportional to the drift velocity, the current is directly proportional to the voltage, which produces the electric field, which produces the drift velocity. This is the origin of Ohm's Law.

Diode: However, in a diode, the number of charge carriers is dependent on the number of electrons that have enough energy to move up an energy hill and across the p-n junction, producing current flow through the diode. The size of this hill, or energy barrier, is dependent on the amount and type of dopants in the semiconductor material of which the diode is made. As a voltage is applied (in the forward bias), the size of the hill is decreased, so more electrons have the energy needed to cross the p-n junction producing current flow. The number of electrons with the energy needed to move up the hill and across the junction increases exponentially as the voltage increases. Thus, the current increases exponentially as the voltage increases.

Applications:

The behavior of components in a circuit is a very important aspect of circuit design. Diodes are found in many semiconductor circuits. Their non-linear I-V behavior makes them quite useful for a variety of applications. Resistors are often used in series with another circuit component to reduce the voltage across that component or in parallel to reduce the current through a component.

Materials and Supplies:

DC Power Supply

Germanium or Zener Diode

2-1K Ohm Resistors

6 lead wires (including those on the power supply)

Milliammeter or Galvanometer

Voltmeter

General Safety Guidelines:

* Always reset the power supply dial to zero, before building or changing the circuit.

* Keep your hands and the work area dry to avoid shock.

* Follow safe and correct procedures for operating the power supply.

Experimental Setups:  Procedure:

Circuit set-up:

1. Build a circuit as shown in Figure 1. Do not turn on the power supply.

2. Check to make sure the lead wires on the power supply are connected to the DC

terminals.

3. Rotate the voltage and current (if applicable) dial to zero. Rotate the current dial one

quarter turn clockwise.

4. Now turn on the power supply.

5. Slowly rotate the voltage dial clockwise, and watch the milliammeter and voltmeter dials.

If the needle moves to the right, the meters are correctly connected. If the needle moves

to the left, reverse the lead wires on that meter.

Resistor (forward):

6. Rotate the voltage dial clockwise slowly until the milliammeter needle shows full

deflection. Record this milliameter and voltmeter reading as the maximum, Imax and Vmax.

7. Divide the value of Imax by 5, This the increment, I, by which you will increase the current. (You will collect 5 sets of data.)

8. Rotate the voltage dial to zero.

9. Rotate the voltage dial clockwise slowly, until the milliammeter reads I.

10. Record the values of I and V in the resistor data table.

11. Increase the voltage until the milliammeter reads 2 I.

12. Record the values of I and V in the data table in rows 1-5.

13. Continue to increase the voltage and record I and V, until you reach Imax.

14. Rotate the voltage dial to zero.

Resistor (reverse):

15. Reverse the resistor, so the current will flow through it in the opposite direction.

16. Repeat steps 10-16, recording the values of I and V as negative numbers in the resistor

data table in rows 6-10.

Diode (forward):

17. Remove the resistor from the circuit and replace it with the diode as shown in Figure 2.

18. Check to make sure the positive end of the diode is connected to the positive terminal of

the power supply.

19. Repeat steps 6-14, however, this time divide the Imax by 10, and record the data in the

diode data table.

Diode ( reverse):

20. Reverse the diode and repeat steps 9-14, using the same values of [[Delta]]I as in step 19. Record these values of I and V as negative numbers.

21. Turn off the power supply.

22. Disconnect the lead wires and replace the equipment in their appropriate places.

Data and Analysis:

Resistor Data Table

V (volts)I ( mA )V/I ` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `

Divide the value of V (volts) by the value of I (amps) to find values of V/I ( ), and complete the data table.

Diode Data Table

V (volts)I (10-6A )V/I ` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `
` `

Divide V (volt) by I (amp) to find V/I ( ), and complete the data table.

Questions:

1. Plot the voltage (horizontal axis) vs. the current (vertical axis) from the resistor and

diode data table on graph paper.

2. What is the shape of the graph of the data for the resistor?

3. What is the shape of the graph of the data for the diode?

4. If the shape is linear, determine the slope and the equation of the line.

5. Compare the slope with the V/I values.

6. According to Ohm's Law, V/I represents what measurable quantity?

7. The slope of the line, I / V, represents what measurable quantity?

8. For the graph that is nonlinear, how did the values of V/I vary as the values of V

increased?

9. Which device conducts electricity both directions?

10. Which device conducts electricity only in one direction?

11. Name the 2 types of charge carriers.

12. In a metal, what conducts electricity (carries charge)?

13. In a semiconductor, what carries charge?

14. In a resistor, increased voltage has what effect on the charge carriers?

15. In a diode, what changes to allow more current to flow as the voltage is increased?

Extension:

Plot V (volt) on the x-axis and ln I (natural log of the values of I in amps, not milliamps) on the y-axis, for the data from the diode section for the experiment (in forward bias). The equation for this relationship is:

I = Io (expeV/kT-1)

where the values of the variables and constants are:

e = 1 electron volt/volt

V = volts

k = 8.62 x 10-5 electron volts / K

T = Temperature in K

Io = current value when V = 0

I = amps

Solve this equation by taking the natural log of both sides.

ln I = ln Io + ln ( expeV/kT -1)

substituting the values of e, k, and T in the equation will prove that the value of eV/kT will be about 100V. Therefore, expeV/kT will be exp100Vwhich is much greater than 1, so we can disregard the 1. Now the equation is:

ln I = ln Io +eV/kT

The slope of the graph (line) is e/kT. From the graph, find the slope. Set the value of the slope equal to e/kT. Slope = e/kT, using the values of k and e, solve for the value of T. Compare this value of T to the room temperature in K.

Teacher Notes:

*Teacher preparation time is approximately 30 minutes.

*This experiment is designed to be used in the electricity unit of a physics class with students

who have already learned how to set up circuits and use test meters.

*For the procedure steps:

1. The teacher should demonstrate the proper procedure for connecting an ammeter and

voltmeter in a circuit.

2. The teacher should demonstrate the proper procedure for operating a power supply.

3. If the power supply does have a current dial, the student may have to adjust this dial to

allow sufficient current to flow through the circuit, as the voltage is increased.

4. If digital multimeters are used, use voltages from 0-2V as shown in the sample.

1. Use separate graph papers, because the scale of each will be different.

2. This should be a straight line. Make sure the students draw the best line through the data

points; they should not "connect the dots".

3. This should be exponential. Have the students use a ruler (straight edge) to approximate

the slope of the graph as V increases, by drawing a tangent line at various points on the

curve.

4. Have the students draw a large right triangle, representing the [[Delta]]I and [[Delta]]V, the sides of

the triangle. The units should be part of the description of the slope. The units may help

the students relate the slope to the measurable quantity it represents.

5. Since the slope is I/ V, its value should be reciprocal of V/ I.

6. Resistance.

7. Conductance is the reciprocal of resistance.

8. The values of V/I decreased as V increased.

9. Resistor

10. Diode

11. Electrons and holes.

12. Electrons in the valence band.

13. Electrons that have jumped to the conduction band and the corresponding holes in the

valence band.

14. Increased voltage causes a stronger electric field, which pushes the electrons harder in

the direction opposite to the field, which increases the drift velocity; so more current

flows.

15. As the voltage increases, the size of the hill (energy gap) is decreased, so more

electrons (at this temperature) can move up the hill and across the p-n junction.

Data and Analysis:

Sample Resistor Data Table
(forward same as reverse)
V (volts)I ( mA )V/I 0.210.211000
0.410.411000
0.610.611000
0.810.811000
1.091.091000
1.201.201000
Sample Diode Data Table

V (volts)I (10-6A )V/I .1450 2800
.181001800
.201501300
.222001000
.24250960
.26300870
.27350770
.28400700
.29450760
.14 0 .18 0 .20 0 .22 0 .24 0 .26 0 .27 0 .28 0 .29 0 *No current in the reverse bias