Lecture #14 Example circuits, Zener
diodes, dependent sources, basic
amplifiers
Reading: 4.10, 5.1, 5.8
Next: transistors (chapter 6 and 14)
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Topics
Today:
Examples, circuit applications:
• Diode circuits, Zener diode
• Use of dependent sources
• Basic Amplifier Models
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Notes on Use of Models
• Most of the diode models are piecewise defined:
–
–
One function for reverse bias
Another for forward bias
• You will need to:
–
“Guess” which diode or diodes are reverse (or forward)
biased
– Solve for V, I according to your guess
– If result is impossible, guess again
• Rarely, both guesses may lead to impossibility.
–
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Then you must use a more detailed model
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Example 1: Ideal Diode Model
Find ID and VD using the
ideal diode model.
1 kW
ID
2V
• Is the diode reverse biased
or forward biased?
• Make a guess, substitute
corresponding circuit
for diode.
+
+
-
V
_D
I
I
Forward bias
+
Reverse bias
• “Reality check”
answer to see if we need to re-guess.
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V
V
_
4
Guessing the Diode Mode: Graphing
• Look at the diode circuit as a Thevenin equivalent linear circuit
attached to a diode.
IL
VL = VD
ID
+
+
Linear circuit
VL
IL = -ID
V
D
_
• Graph the diode I-V curve and the linear circuit I-V curve on the
same graph, both in terms of ID and VD.
• This means draw the diode I-V curve normally, and draw the
linear I-V curve flipped vertically (IL = -ID).
• See where the two intersect—this gives you ID and VD.
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Example 1: Ideal Diode Model
ID
• Forward biased
• VD = 0 V
• ID = 2 mA
2 mA
2V
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VD
6
Guessing the Diode Mode:
“Common Sense”
• Notice the polarity of the 2 V
falling over the resistor and diode
• The 2 V is in same direction as VD
• Diode is probably forward biased 2 V
1 kW
ID
+
-
+
V
_D
• It’s generally easier to guess reverse bias first since it is easy to
check.
• No matter what piecewise model we use, reverse bias is always
open circuit.
• So when you don’t know what to do, put in open circuit for the diode,
and see if it violates reverse bias conditions (zero current, negative
voltage).
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Example 1: Ideal Diode Model
1 kW
• Guess reverse bias:
ID
• Since no current
+
+
2V
V
is flowing,
_D
VD = 2 V (by KVL)
• This is impossible for reverse bias (must have negative VD)
 So the diode must be forward biased
1 kW
• VD = 0 V
2V
• ID = 2 V / 1 kW = 2mA
• Same as what we got graphically.
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ID
+
+
-
V
_D
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Example 2: Large-Signal Diode Model
• The large-signal diode model takes into account voltage
needed to forward bias, (VF = 0.7 for silicon)
1 kW
to find ID and VD.
• To be in forward bias
+
mode, the diode needs
0.5 V
0.7 V.
ID
+
V_ D
• The source only provides 0.5 V.
• The resistor cannot add to the voltage since the diode could
only allow current to flow clockwise.
• Reverse bias => open circuit => ID = 0 A, VD = 0.5 V
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Zener diodes
• A Zener diode is the name commonly used
for a diode which is designed for use in
reverse breakdown
• Since the diode breaks down sharply, at
accurate voltage, it can be used as a
voltage reference
• The symbol for a Zener diode:
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Zener diode as a simple regulator
• The Zener diode shown here will keep the
regulated voltage equal to its reverse
breakdown voltage.
R
~
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Constant voltage
power supply
to load
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Resistor sizing
• How big should R be in the regulator shown?
• If the load draws a current I, then the resistor
must carry that current when the unregulated
voltage is at the lowest point, without letting the
regulated voltage drop.
• Lets say the load draws 10 milliamps, the
regulated voltage is 2 volts, and the minimum
unregulated voltage (low point of ripple) is 2.5
volts)
• The resistor must be
• R=(2.5v-2v)/10 milliamps=500 ohms.
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Ripple calculation
• How much ripple will be observed on the
unregulated supply?
• The maximum current will be:
I peak 
(Vunregulated ) peak  Vregulated
R
• And we can estimate the amount of charge lost:
Q  I  t
• So the ripple will be
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Q
V 
C
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Dependent Voltage and Current Sources
• A linear dependent source is a voltage or current source that
depends linearly on some other circuit current or voltage.
• We can have voltage or current sources depending on voltages or
currents elsewhere in the circuit.
Here the voltage V is proportional to the voltage across the element c-d .
c
+
V = A v x Vcd
Vcd
+
-
+
-
OR
-
d
Sometimes a diamond-shaped symbol is used for dependent
sources, just as a reminder that it’s a dependent source.
Circuit analysis is performed just as with independent sources.
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WHY DEPENDENT SOURCES?
EXAMPLE: MODEL FOR AN AMPLIFIER
AMPLIFIER SYMBOL
Differential Amplifier
V+
V
+
A

input
output
V0  A( V  V )
V0
V0 depends only on input (V+  V-)
EXAMPLE: A =20 Then if input (V+-V-) = 10mV; Vo = 200mV.
An actual amplifier has dozens (to hundreds) of
devices (transistors) in it. But the dependent source
allows us to model it with a very simple element.
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EXAMPLE OF THE USE OF DEPENDENT SOURCE
IN THE MODEL FOR AN AMPLIFIER
AMPLIFIER SYMBOL
Differential Amplifier
V+
V
+
A

V0  A( V  V )
AMPLIFIER MODEL
Circuit Model in linear region
V0
Ri
+

V1
+

AV1
+

V0 depends only on input (V+  V-)
See the utility of this: this model, when used correctly
mimics the behavior of an amplifier but omits the
complication of the many many transistors and other
components.
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V0
NODAL ANALYSIS WITH DEPENDENT SOURCES
Example circuit: Voltage controlled voltage source in a branch
R5
R1
R3
Va
Vb
Vc
R2
+

VAA
+

AvVc
R4
R6
ISS
Write down node equations for nodes a, b, and c.
(Note that the voltage at the bottom of R2 is “known” so current
flowing down from node a is (Va  AvVc)/R2.)
Va  VAA Va  A v Vc Va  Vb


0
R1
R2
R3
Vc  Vb Vc
Vb  Va Vb Vb  Vc

 ISS


0
R5
R6
R3
R4
R5
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CONCLUSION:
Standard nodal
analysis works
17
NODAL ANALYSIS WITH DEPENDENT SOURCES
Finding Thévenin Equivalent Circuits with Dependent Sources Present
Method 1: Use Voc and Isc as usual to find VT and RT
(and IN as well)
Method 2: To find RT by the “ohmmeter method” turn
off only the independent sources; let the dependent
sources just do their thing.
See examples in text (such as Example 4.3) and in
discussion sections.
Pay most attention to voltage-dependent voltage
sources and voltage-dependent current sources. We
will use these only.
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NODAL ANALYSIS WITH DEPENDENT SOURCES
Example : Find Thévenin equivalent of stuff in red box.
R3
Va
Vc
R2
+

A v Vcs
R6
ISS
With method 2 we first find open circuit voltage (VT) and then we
“measure” input resistance with source ISS turned off.
R 2 (R 6  R 3 )
ISS R 6 (R 2  AR 3 ) R 
Verify the solution:
VTH 
TH
R 2  R 3  R 6 (1 - A)
R 2  R 3  R 6 (1 - A)
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EXAMPLE: AMPLIFIER ANALYSIS
USING THE AMPLIFIER MODEL WITH Ri = infinity:
A: Find Thévenin equivalent resistance of the input.
B: Find the Ratio of the output voltage to the input voltage (“Voltage
Gain”)
R
AMPLIFIER MODEL
F
VIN
RS
V+
V-
Circuit Model in linear region

A
+
V0
Ri
+

V1
+

AV1
+

V0
Method: We substitute the amplifier model for the amplifier, and
perform standard nodal analysis
You find : RIN and VO/VIN
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EXAMPLE: AMPLIFIER ANALYSIS
USING THE AMPLIFIER MODEL WITH Ri = infinity:
How to begin: Just redraw carefully!
RF
VIN RS
V+
V-

A
+
RF
VIN
RS
V-
-
V0
V+
+
V1 +

V0
AV1
Method: We substitute the amplifier model for the amplifier, and
perform standard nodal analysis
- AR F
R F  (1  A)R S
Verify the solution: RIN =
VO/VIN =
R F  (1  A)R S
1 A
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