[ECEN 1400]
Introduction to Digital and Analog Electronics
R. McLeod
HW #3: LEDs and Transistors (100 pts)
1
Light Emitting Diode IV Response (20 Points)
Create a multisim circuit and include a schematic including a voltage source (Source → POWER SOURCES →
DC POWER), an ammeter (Indicators → AMMETER → AMMETER H) and a generic red light emitting
diode (Diodes → LED → LED red) and ground (Sources → POWER SOURCES → GROUND). Your
circuit should look like this:
You will now create a I/V response graph of a LED. To do this, increment the voltage of the DC POWER
appropriately and record the current through LED .Take enough data points to give you a smooth curve of the
I/V response, but dont waste your time by taking more data than required. Concentrate points where something
interesting is happening. Current limits on LEDs vary, but a typical upper limit is 20 mA, so use this to find an
appropriate voltage range. To gather each data point you will:
1. Start the Simulation
2. Record I and V
3. Stop the simulation
4. Increment the voltage appropriately
5. Repeat
Note what voltage causes the LED to turn on to an accuracy of at least 0.05 V. (Turn on is indicated by the
arrows to the right becoming red.)
Make a plot of this IV curve in a graphing program such as excel including both negative and positive voltage.
Show the turn-on voltage where the LED lit up on the graph.
What do you notice about the I/V characteristics at this particular voltage?
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[ECEN 1400]
Introduction to Digital and Analog Electronics
R. McLeod
Solution:
(a) Schematic
(b) Sample Data
Figure 1
Turn on voltage is between 1.75 and 1.8 V, which is approximately the knee in the IV curve.
Grading:
• (5 pts) Schematic including four components.
• (10 pts) Data table (sufficient data taken correctly.)
• (5 pts) Data graphed appropriately (labels, units, sufficient points, marked turn on point on graph)
2
Limit the Current Through an LED (15 Points)
The resistance of the LED is sufficiently low that it would draw too much current and be permanently damaged
when hooked across a typical voltage source (e.g. 5 V). A solution is to include a current limiting resistor in series
with the LED. Since the LED response is nonlinear, its not completely trivial to solve for the required resistance.
A simple method is to use the LED I/V response that you found in problem one. Assume you want to limit the
current through the LED to be 10 mA when the supply voltage is 5V. Use your graph to find the voltage drop
across the LED at this current. From that, solve for the required resistance using Ohms law for the resistor. Check
your work with multisim.
Solution:
Figure 2
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[ECEN 1400]
Introduction to Digital and Analog Electronics
R. McLeod
1. Find the voltage across the LED at the specified current from the graph.
2. Subtract from the source voltage to find the required voltage across the resistor.
3. Solve for the resistance from Ohms law.
Vr = 5V − VLED = 5V − 1.79V = 3.21V
R=
VR
3.21V
=
= 321Ω
I
10mA
Grading:
• (4 pts) Find the numerical value of voltage across the LED at the specified current from the graph.
• (1 pt) Correct answer of voltage.
• (5 pts) Write correct equation for the required voltage across the resistor.
• (3 pts) Write Ohms law for the resistance.
• (2 pts) Correct numerical answer and right units.
3
Regular Diode IV Response (20 Points)
Delete the LED and replace it with the diode in your circuit kit which is a 1N4148 (Diodes → DIODE
→1N4148). Repeat the experiment, but now that youve done it the tedious way, use multisim to do the hard
work for you. In the top menu, select Simulate → Analyses → DC Sweep. You should see a dialog box like
this:
Set the Source in the first tab (shown above) to be V1 (the default) and let it sweep from -0.5
V to 2 V in increments of 0.01 V. This will sweep the voltage of your source over the specified range. Then
select the Output tab. On the left is a list of variables you can calculate as the voltage is swept. Select I(D1[ID])
which is the current through the diode. Click Add to move it to the right-hand list. Then click Simulate
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[ECEN 1400]
Introduction to Digital and Analog Electronics
R. McLeod
at the bottom of the dialog box. This should plot a graph for you like you did by hand in the previous
problem.
No light is emitted from this device, but it will have a similar IV shape including a turn-on voltage. Which
device has a larger turn-on voltage?
Solution:
Figure 3
The LED has the higher turn on voltage. The regular diode turn-on voltage of .6 to .7 volts is a characteristic
of the silicon material out of which it is made and is thus common to a large number of devices.
Grading:
• (3 pts) Correct voltage range from -0.5 V to 2 V. (Beyond this range is OK, i.e. -1 V 4 V.)
• (3 pts) Correct current range corresponding to the voltage range.
• (4 pts) Correct tendency (the knee of the curve is around 0.6-0.7 volts; current increases with voltage beyond
turn on point. )
• (5 pts) Read the turn on voltage from the graph, around 0.6
0.7 volts.
• (5 pts) Make comparison of turn on voltage with LED.
4
Current Limit Your Diode Circuit (15 Points)
Add a 1.69K resistor in series with your 1N4148 diode to create this circuit (the actual value of the voltage source
wont matter since well use a DC sweep analysis).
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[ECEN 1400]
Introduction to Digital and Analog Electronics
R. McLeod
Use the same DC sweep analysis as you did on the previous problem, but now sweep from 0 to 5 V. You can
plot either the current through R1 or through the diode since they are the same. What is the current through the
diode when the supply is at 2V?
Solution:
Figure 4
The current at 2V is about 833 mA. This is important for the next problem.
Grading:
• (3 pts) Correct voltage range from 0 to 5 V. (Beyond this range is OK.)
• (3 pts) Correct current range corresponding to the voltage range. . (0-3 mA for 0-5 V)
• (4 pts) Correct tendency.
• (4 pts) Correct numerical value of the current at 2V.
• (1 pt) Correct unit.
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[ECEN 1400]
5
Introduction to Digital and Analog Electronics
R. McLeod
LED Controlled by a Transistor (30 Points)
Create the following circuit in multisim. The LED is the same red LED used in problem 1. The transistor (labeled
Q1, part number 2N2222A) is the one in your kit, found under (Transistors → BJT NPN). It is the top entry
on the component list. For R2, use the value you found before that limits the current through the LED to 10 mA
when using a 5V supply.
Part a: Use the DC Sweep Analysis to sweep V1 from 0 to 5V and plot the current through R1. Compare this
to your result with the 1N4148. What do you conclude about the base/emitter terminals of this transistor?
Part b: Change your DC sweep analysis to plot the current through R2 as a function of sweeping V1 (not V2
leave this at 5V). This is a very interesting result. Imagine you had a voltage source, V1, which could deliver only
1 mA of current, was noisy, and the value of the voltage varied randomly with an amplitude of 100 mV. What
advantages would this circuit have as an LED driver over that of problem 3? These advantages are the foundations
of digital logic.
Solution:
Figure 5: Schematic
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[ECEN 1400]
Introduction to Digital and Analog Electronics
R. McLeod
(a) Sample Data
(b) Sample Data
Figure 6
1. The I/V relationship of the 1N4148 diode is very similar (if not identical) to the 2N2222A transistor.
2. V1 is supplying only small current (< 1mA) but the LED can draw 10 mA or more from V2, controlled by
the transistor.
3. The LED could be switched from off at 0-0.5 volts to on at 1-2 volts and the noise would have almost no
effect on the LED current because the LED current is flat everywhere except the switching region near 0.75
V.
Solution:
• (3pts) Correct voltage range for I(R1)-V1 curve.
• (3 pts) Correct current range for I(R1)-V1 curve.
• (4 pts) Correct tendency for I(R1)-V1 curve.
• (3 pts) Correct voltage range for I(R2)-V1 curve.
• (3 pts) Correct current range for I(R2)-V1 curve.
• (4 pts) Correct tendency for I(R2)-V1 curve.
• (2 pts) Comparison with 1N4148. (the same)
• (3 pts) Point out the base/ emitter terminals act as a diode.
• (5 pts) Point out the noise has almost no effect on LED current beyond certain voltage value.
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