ELEC 350
Electronics I
Fall 2012
Homework Assignment #3 – due in Dana 301 at 3:30 pm Friday, Sept. 28, 2012
(revised 9/27/12)
Instructions, notes, and hints:
You may make reasonable assumptions and approximations in order to compensate for missing
information, if any. Provide the details of all solutions, including important intermediate steps.
You will not receive credit if you do not show your work.
Prob. 4.18: Assume that the emission coefficient n has a value of one and that the diode operates
at a temperature of 20°C.
Probs. 4.39 and 4.41: In the constant-voltage-drop model, rd = 0 is assumed.
Prob. 4.48: Apply the principle of superposition, and use the constant-voltage-drop model for the
DC case(s). You may assume that the signal (AC) currents are negligibly small compared to the
DC currents flowing along the same circuit branches. The capacitors are large enough to have
negligibly small reactances at the signal frequency. Assume that n = 1 and that the load
impedance connected between the vo node and ground is infinitely large. In the diagram, the
arrows above and below the ideal current sources imply that they are connected to other circuits
(not shown) that allow them to operate normally. The values for vo/vi obtained for the various I
values should be presented in tabular form.
Assignment:
Probs. 4.18, 4.39, 4.41, and 4.48 in the textbook, plus the following additional problems:
1. [revised 9/27/12] As explained in the textbook [Sedra & Smith, Microelectronic Circuits, 6th
ed., Oxford University Press, 2010, p. 175], the value of the saturation current IS doubles for
every 5°C (or 5 K) rise in temperature. This implies that the value of IS can be found at other
temperatures using the formula IS(T) = IS30020.2(T – 300), where IS300 is the value of IS at exactly
300 K. As shown below, two diodes are used to drive the two inputs of an instrumentation
amplifier with a differential-mode gain of 5. Assuming that the emission coefficient n = 1 for
both diodes and that IS300 = 5 × 1015 A for D1 and 10 × 1015 A for D2, find the output
voltage of the instrumentation amplifier at temperatures of 300 K and 301 K.
VPOS = 15 V
R
5 k
R
5 k
v1
v2
D1
Instrumentation
Amplifier
Ad = 5
D2
(continued on next page)
vo
2. The diode attenuator circuit shown below has been designed for use in the VHF (roughly 30300 MHz) spectrum. The capacitors act like shorts and the inductors act like opens at AC.
The signal source is represented by the Thévenin equivalent circuit made up of vg and Rg, and
the load is represented by RL. (In RF/wireless circuits, it is common for sources and loads to
have 50- input and output impedances.) Assuming that the diodes have DC turn-on (or cutin) voltages of 0.7 V, find the approximate numerical value of the attenuation vo/vin (not
vo/vg). The diodes are identical and have saturation currents of IS = 20 nA and emission
coefficients of n = 2. The circuit operates at room temperature so that VT is 25 mV. You may
ignore the loading effect of resistor RB at AC because it is so much larger than RL.
C3
VS = 12 V
RB
1 k
Rg = 50 
vg
+
−
C1
+
id
vin
L1
ID
ID
D1
D2
L2
C2
+
vo
−
RL
50 
−
3. The two back-to-back diodes used in the previous problem might seem to be redundant. After
all, it should be possible to remove D2 and L2 and still have an effective attenuator. That
would be true if the signal current id were very small relative to the DC quiescent current ID.
To see why the back-to-back arrangement is superior for large signal currents,
a. [revised 9/27/12] Compute the diode resistance for both diodes for at least ten values
of id ranging from –1 mA to +1 mA (and ID at its constant value), and plot the sum as
a function of the instantaneous value of id (i.e., plot rd1 + rd2 vs. id, where rd1 and rd2
are the two diode resistances). To calculate the diode resistance values accurately,
you will have to consider the total current flowing through them. Note that id flows
from left to right in both diodes.
b. [revised 9/27/12] Imagine that D2 and L2 have been removed from the circuit and that
RB has been increased to 4 k. This should result in the same diode resistance in line
between the source and the load as in the two-diode case (if the resistances are
computed based on the value of ID alone). Compute the diode resistance rd1 for values
of id ranging from –1 mA to +1 mA (and ID at its new constant value), and plot it as a
function of the instantaneous value of id on the same graph as the plot in part a. The
total in-line diode resistance in both cases should be exactly the same when id = 0.
c. Based on your results, explain why the two-diode attenuator is superior to the onediode attenuator in the presence of relatively large-amplitude signals. Hint: Imagine
applying a purely sinusoidal signal to both attenuators. How might the output signal
(vo across RL) be affected in each case?
(continued on next page)
4. In the circuit below, the voltage Vin varies from 0 to 12 V. It is supplied by a circuit with a
negligibly small output resistance. The two diodes are identical LEDs (light-emitting diodes),
and their turn-on voltages are each VF = 1.5 V. The circuit is designed so that both LEDs are
off when the voltage Vin is low; D1 turns on and D2 remains off when Vin rises above a
relatively low threshold; and then both LEDs turn on when Vin rises above a higher threshold.
The circuit is a very simple (and coarse) bar graph display of the applied voltage. Using the
constant-voltage-drop model for the diodes when they are in the on state, answer the
following questions:
a. Above what value of Vin is LED D1 on while D2 is off?
b. Above what value of Vin are both D1 and D2 on?
c. What values are the diode currents ID1 and ID2 when Vin = 12 V?
Vin = 0 to 12 V
R3
300 
I3
R2
91 
ID2
D2
R1
220 
ID1
D1