ECE 3410 – Homework 1
Problem 1. Complete the following review problems on Ohm’s law, voltage dividers, current dividers and Thévenin/Norton equivalent circuits.
(A) Measurements taken on various resistors are shown below. For each case, calculate the power dissipated in the resistor and the power rating necessary for safe
operation using standard components with power ratings of 1/8 W. 1/4 W, 1/2 W,
1 W or 2 W.
i.
ii.
iii.
iv.
v.
vi.
1 kΩ conducting 30 mA
1 kΩ conducting 40 mA
10 kΩ conducting 3 mA
10 kΩ conducting 4 mA
1 kΩ dropping 20 V
1 kΩ dropping 11 V
Solution
Power (W)
Safe Rating
0.9 W
1.6 W
0.09 W
0.16 W
0.4 W
0.121 W
1W
2W
1/8 W
1/4 W
1/2 W
1/8 W
(B) You are given three resistors whose values are 10 kΩ, 20 kΩ, and 40 kΩ. How many
different resistances can you create using series and parallel combinations of these
three? List them in value order, lowest to highest. (Hint: In your search, first
consider all parallel combinations, then all series combinations, then combined
series/parallel combinations of which there are two kinds).
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ECE 3410 – Homework 1
Solution
10kΩ
10kΩ
40kΩ
20kΩ
40kΩ
40kΩ
20kΩ
20kΩ
10kΩ
10kΩ
20kΩ
Total 5.7kΩ
Total 23.33kΩ
40kΩ
Total 46.67kΩ
Total 70kΩ
20kΩ
10kΩ
10kΩ
10kΩ
20kΩ
40kΩ
Total 8.57kΩ
20kΩ
40kΩ
40kΩ
20kΩ
Total 14.28kΩ
40kΩ
10kΩ
Total 17.14kΩ
Total 28kΩ
(C) Consider the voltage divider shown below alongside the Thévenin equivalent circuit
seen looking into node X. Find expressions for V0 and R0 .
VDD
R1
R0
V0
≡
R2
X
V0
−
+
X
Solution
To find the equivalent resistance seen at node X, we first zero out any DC
independent sources (i.e. VDD = 0) and then solve for the equivalent resistance
between X and ground. By visual inspection, we see that R0 = R1 k R2 . The
equivalent voltage is obtained from the voltage divider relation:
V0 = VDD
R2
R1 + R2
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ECE 3410 – Homework 1
(D) For the circuit below, choose a value of R to achieve IL = 0.2I0 .
IL
I0
R
1 kΩ
Solution
Here we have a current divider, so
R
R + 1kΩ
R
⇒ 0.2I0 = I0
R + 1kΩ
R
⇒ 0.2 =
R + 1kΩ
0.2 × 1kΩ
⇒R=
1 − 0.2
= 250Ω.
IL = I0
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ECE 3410 – Homework 1
Problem 2. For the two circuits shown below, complete the following analyses:
• Derive an expression the Laplace-domain transfer function H (s).
• Solve an expression for the magnitude response |H (ω)|2 .
• If the component values are R = 1kΩ and C = 10µF, calculate the time constant
and the pole freqency.
• Draw a Bode magnitude plot for the signal VB using the calculated pole frequency,
given a constant input magnitude of VA = 60dB.
Complete these tasks for the high-pass and low-pass circuits shown below.
(A) Low-pass circuit
R
va
vb
C
Solution
H(s) =
1
1+sRC
|H (jω)|2 =
1
1+ω 2 R2 C 2
|H (ω)| (dB)
τ = RC = 0.01 s
1
ωc = RC
= 100 rad/s
80
60
40
20
0
10−1 100
101 102 103
ω (rad/sec)
104
105
(B) High-pass circuit
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ECE 3410 – Homework 1
C
vb
va
R
Solution
H(s) =
s
1+sRC
|H (jω)|2 =
ω2
1+ω 2 R2 C 2
|H (ω)| (dB)
τ = RC = 0.01 s
1
= 100 rad/s
ωc = RC
80
60
40
20
0
10−1
100
101
102
ω (rad/sec)
103
104
Problem 3. A non-linear one-port device has a voltage v applied across its terminals, and the
corresponding current i is some some function of v. The device is said to have a
differential resistance defined by
∆v
req =
,
∆i
where both ∆v and ∆i are assumed to be very small so that the i, v relationship is
approximately linear. Consider the following scenarios and determine the differential
resistance in each case.
(A) Suppose a device has an exponential response i = (10 µA) exp (v/0.25 V). If the
device’s voltage is kept very close to 0.5 V, what is the differential resistance?
(Hint: apply a first-order Taylor approximation centered at 0.5 V).
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ECE 3410 – Homework 1
Solution
v d
(10 µA) exp
dv
0.25
V
v=0.5 V
10 µA
= ∆v
exp(2)
0.25 V
∆i ≈ ∆v
= ∆v × 2.956 × 10−4 Ω−1
1
⇒ rd =
2.956 × 10−4 Ω−1
= 3383 Ω
(B) A temperature sensor is specified to provide 2 mV/◦C. When the sensor is connected to a load resistance of 10 kΩ, and the temperature is changed by 10 ◦C, the
output voltage is observed to change by 10 mV. What is the sensor’s differential
resistance?
Solution
RS
+
−
+
∆vs
∆vL = 10 mV
10kΩ
−
10 kΩ
10 kΩ + RS
∆vs × 10 kΩ
− 10 kΩ
⇒ RS =
∆vL
∆vL = ∆vs
Now since ∆T = 10 ◦C, ∆vs = ∆T × 2 mV/ ◦C = 20 mV, and ∆vL = 10 mV,
the result is
RS = 10 kΩ.
Problem 4. The differential voltage amplifier shown below has a non-linear transfer characteristic
described by a non-linear equation:
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ECE 3410 – Homework 1
+
vIN
−
+
vOUT vOUT = VR
−
exp(kvIN )−1
exp(kvIN )+1
In this equation, VR is interpreted as the rail voltage, i.e. the power supplies are at +VR
and −VR . The constant k is related to the amplifier’s gain.
When dealing with a non-linear amplifier such as this, we usually approximate the
device using a first-order Taylor expansion, so that it looks like a linear amplifier:
vout ≈ Avin ,
dvOUT where A = dvIN vIN =0
In this problem, we will take a closer look at this approximation.
Please perform the following tasks:
(A) Using Matlab, Octave or other software, plot the transfer characteristic (vout on
the vertical axis and vin on the horizontal axis) over the domain vin ∈ (−2V, +2V)
in steps of 0.01V, using the parameter values VR = 10V and k = 10V−1 .
vOUT (V)
Solution
10
5
0
−5
−10
−2
−1
0
1
2
vIN (V)
(B) Notice that the slope is steepest near vIN = 0. In your plotting software, zoom
into the domain vin ∈ (−0.1V, +0.1V). Ideally, an amplifier’s transfer characteristic should be a straight line. Comment on the straightness of the zoomed
characteristic, and how it might affect amplifier function.
Solution
The dashed red line is a straight-line approximation centered at zero. The
approximation is very close near (0,0), but is not as good toward the extremes.
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ECE 3410 – Homework 1
vOUT (V)
Solution (cont.)
4
2
0
−2
−4
−0.1
−5 · 10−2
0
5 · 10−2
0.1
vIN (V)
(C) In your zoomed plot, measure the slope ∆vOUT /∆vIN for the transfer characteristic
near the point where vIN = 0.
Solution
The slope is measured as 50 V/ V.
(D) Now find the amplifier’s gain by solving an expression for the derivative dvOUT /dvIN ,
and evaluate the derivative at the point vIN = 0. Give your answer as an equation
in terms of VR and k.
Solution
The gain is
d
exp (kvIN ) − 1 VR
dvIN
exp (kvIN ) + 1 vIN =0
k exp (kvIN )
exp (kvIN ) − 1 = VR
− k exp (kvIN )
exp (kvIN ) + 1
(exp (kvIN ) + 1)2 v
IN =0
k
= VR
2
(E) Finally, evaluate your gain equation using the values specified in part (a). How
closely does it match to the numerical measure you made in part (c)?
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ECE 3410 – Homework 1
Solution
It evaluates to 50 V/V, an exact match.
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ECE 3410 – Homework 1
Problem 5. An amplifier has the following small-signal characteristics:
Av (dB) = 40dB
Rin = 1MΩ
Rout = 10Ω
This amplifier is used to drive a load of RL = 100Ω. The signal source is an ideal
voltage source with no series resistance (i.e. Rsig = 0).
(A) What is the open-circuit voltage gain with the signal-source connected, but with
no load connected? (AV O , vvout
, excluding the coupling divider ratios).
sig
Solution
Since Rsig = 0, there is no coupling effect at the amplifier’s input, and since
RL is not connected, there is no coupling effect at the output. Therefore the
gain is exactly equal to the specified gain of 40 dB. Expressed in volts-per-volt,
this is AV O = 100 V/V.
(B) What is the loaded voltage gain with both the signal source connected and the
load connected? (AV L , vvout
, including the coupling divider ratios).
sig
Solution
Since Rsig = 0, there is no coupling effect at the input. Now that the load is
attached a voltage-divider will appear at the amplifier’s output, so the loaded
gain is
RL
Rout + RL
100 Ω
= (100 V/V)
100 Ω + 10 Ω
= 90.9 V/V.
AV L = AV O
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ECE 3410 – Homework 1
Problem 6. A current amplifier has the following characteristics:
Rin = 1kΩ
Rout = 10kΩ
Ai = 100A/ A
The amplifier’s signal source is a sinusoid with amplitude Isig = 1mA and parallel
resistance of 100kΩ. The amplifier also drives a load resistance of RL = 100Ω.
What are the values of the current gain, the voltage gain and the power gain for this
circuit?
Solution
The described circuit:
iin
iSIG
100kΩ
iout
Rin
Ai iin
Rout
100Ω
The unloaded (short-circuit) current gain is Ai = 100 A/A. When the signal source
and load are connected, current-divider relationship appear due to the resistive
coupling at the input and output terminals. Therefore:
iout
isig
= Ai
AIL =
100kΩ
100kΩ + Rin
Rout
Rout + 100Ω
= 98 A/A
To obtain the voltage gain, we first solve for vin and vout :
vin = iin Rin
vout = iout × 100Ω
vout
Av =
v
in 100Ω
iout
=
iin
Rin
Rout
100Ω
= Ai
= 9.9 V/V
Rout + 100Ω
Rin
Lastly, the power gain is the product of voltage and current gains:
AP = AIL Av
= 970 W/W
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ECE 3410 – Homework 1
Problem 7. The circuit shown below is a transresistance amplifier used to convert a small current
signal into a voltage signal.
Amplifier Circuit
Signal Source
iSIG
RF
−
Rsig
vOUT
iIN
+
By solving for Rin , Rout and Rm , you obtain the equivalent circuit model:
Signal Source
Equivalent Model
Rout
vout
iin
Rsig
Rin
−
+
iSIG
Rm iin
If the op amp is ideal, obtain expressions for the amplifier circuit’s input resistance,
the output resistance and the transconductance gain Rm = vOUT /iIN .
Solution
Since the op amp is ideal, it has infinite input resistance and zero output resistance.
Therefore:
iin = isig
⇒ Rm
vout = −isig RF
vout
,
= −RF
isig
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ECE 3410 – Homework 1
Problem 8. Two amplifiers are to cascaded between a signal source and load, as shown below.
Amplifier 1
Amplifier 2
Rout 1
Rsig
iin 1
−
+
iSIG
Rin 1
Rm1 iin 1
Load
Rout 2
vin 2
Rin 2
−
+
Signal Source
Av2 vin 2
vout
RL
isig = (10µA) sin (2πf t)
Rsig = 10kΩ
Rin 1 = 100kΩ
Rm1 = 10V/ µA
Rout 1 = 10kΩ
Rin 2 = 1kΩ
Av2 = 10V/ V
Rout 2 = 100Ω
RL = 10Ω
(A) Accounting for all coupling effects, give an expression for the circuit’s overall
transresistance gain, Rtot = vout /isig . Calculate the value for this gain.
Solution
For the current signals, the resistive coupling obeys a current-divider relation.
For the voltage signals, there is a voltage-divider relation. Chaining them
together:
Rsig
Rsig + Rin 1
Rin 2
= Rm1 iin 1
Rin 2 + Rout 1
RL
= Av2 vin 2
RL + Rout 2
iin 1 = isig
vin 2
vout
Then the overall gain is
Rtot = Rm1 Av2
Rsig
Rin 2
RL
Rsig + Rin 1 Rin 2 + Rout 1 RL + Rout 2
(B) Calculate the amplitudes of iin 1 , vin 2 and vout .
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ECE 3410 – Homework 1
(C) If the amplifiers’ input and output resistances could be modified, describe the
changes that would be required to maximize the total transresistance gain. State
the value of the maximum gain that can be achieved under these modifications.
Solution
To achieve ideal behavior, we want to eliminate the coupling effects, i.e. we
want all the divider ratios to approach 1. To achieve this:
Rin 1 → 0
Rout 1 → 0
Rin 2 → ∞
Rout 2 → 0
With these changes, the total transresistance gain is
Rtot → Rm1 Av2 .
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