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

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**
Review of Scientific Principles:**

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**

**
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.

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**
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, I_{max} and V_{max.}

7. Divide the value of I_{max} 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 2I.

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 I_{max}.

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 I_{max} 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.

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**
Data and Analysis:**

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.

V (volts) | I (10^{-6}A ) | 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 = I _{o} (exp^{eV/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

**
I _{o}** = current value when V = 0

**
I** = amps

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

**
ln I = ln I _{o} + ln ( exp^{eV/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, exp^{eV/kT} will be exp^{100V}which is much greater than 1,
so we can
disregard the 1. Now the equation is:

**
ln I = ln I _{o} +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.

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**
Teacher Notes:**

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*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.

**
Answers to Questions:**

**
**

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:**

(forward same as reverse)

V (volts) | I ( mA ) | V/I |
---|---|---|

0.21 | 0.21 | 1000 |

0.41 | 0.41 | 1000 |

0.61 | 0.61 | 1000 |

0.81 | 0.81 | 1000 |

1.09 | 1.09 | 1000 |

1.20 | 1.20 | 1000 |

V (volts) | I (10^{-6}A ) | V/I |
---|---|---|

.14 | 50 | 2800 |

.18 | 100 | 1800 |

.20 | 150 | 1300 |

.22 | 200 | 1000 |

.24 | 250 | 960 |

.26 | 300 | 870 |

.27 | 350 | 770 |

.28 | 400 | 700 |

.29 | 450 | 760 |

.14 | 0 | |

.18 | 0 | |

.20 | 0 | |

.22 | 0 | |

.24 | 0 | |

.26 | 0 | |

.27 | 0 | |

.28 | 0 | |

.29 | 0 |