HW Solution#4 - ShanghaiTechsist.shanghaitech.edu.cn/faculty/zhoupq/Teaching/Fall15/HW4...Β Β·...
Transcript of HW Solution#4 - ShanghaiTechsist.shanghaitech.edu.cn/faculty/zhoupq/Teaching/Fall15/HW4...Β Β·...
Electric Circuits Fall 2015 Homework #4
1
HW Solution#4
4.1 [8%]
The op amp in Fig. 1 has π π = 100kΞ©, π π = 100Ξ©, π΄ = 100,000. Find the differential
voltage π£π and the output voltage π£π.
Fig. 1
Solution
[2]
At node 1, VsβV1
10k=
V1
100k+
V1βV0
100k (1) [2]
At node 2, V1βV0
100k=
V0β(βAVd)
100 (2) [2]
Since Vd = V1, A = 100,000,Vs = 1mV, From (1) and (2),
Vd β 100nV [1]
V0 β β10mV [1]
Electric Circuits Fall 2015 Homework #4
2
4.2 [6%]
Determine the output voltage π£π in the circuit of Fig. 2.
Fig. 2
Solution
Transform the current source as shown below. At node 1,
10βV1
5k=
V1βV2
20k+
V1βV0
10k (1) [2]
At node 2,
V1βV2
20k=
V2βV0
10k (2) [2]
Since V2 = 0, From (1) and (2), V0 = β2.5V [2]
Electric Circuits Fall 2015 Homework #4
3
4.3 [9%]
In the circuit shown in Fig. 3, find π in the voltage transfer function π£π = ππ£π .
Fig. 3
Solution
We notice that v1 = v2. Applying KCL at node 1 gives
v1
R1+
v1βvs
R2+
v1βv0
Rf= 0 (1) [2]
Applying KCL at node 2 gives
v1
R3+
v1βvs
R4= 0 (2) [2]
From (1) and (2),
v0 = Rf [(1
R1+
1
R2+
1
Rf) (
R3
R3+R4) β
1
R2] vs [4]
i.e.
k = Rf [(1
R1+
1
R2+
1
Rf) (
R3
R3+R4) β
1
R2] [1]
Electric Circuits Fall 2015 Homework #4
4
4.4 [11%]
It is common in electronics to convert a current signal into a voltage signal. One case where
this is necessary, is when converting the photocurrent generated by a photodiode into a
voltage signal for downstream processing electronics. A photodiode acts similar to a normal
diode except it converts input optical power into a photocurrent. A photodiode must also be
reverse biased for optimal operation.
As shown below, we can model a photodiode as an ideal current source with photocurrent
given by πΌπβ = π πππππ‘ where ππππ‘ is the aborbed optical power [W] and π π is the current
responsivity assumed to be 0.9 [Amps/Watt]. Note that πΌπβ flows from the cathode to the
anode since the diode is reverse biased. For this problem, we can ignore the diode leakage
current.
Fig. 4(a)
(a) One way to both reverse bias the photodiode and convert the current into a voltage is with
the circuit below in which the photocurrent is sent into a resistor. For this configuration
calculate:
i) The voltage responsivity, π π£ =πππ’π‘
ππππ‘
ii) A drawback of this configuration is that the diode bias is not constant. Find the input
optical power at which the diode is no longer reverse biased.
Fig. 4(b)
(b) A good way to convert the diode photocurrent into a voltage is with the following
transimpedance amplifier (TIA). For this problem calculate:
i) The voltage across the diode. Does it depend on ππππ‘?
Electric Circuits Fall 2015 Homework #4
5
ii) The transimpedance gain, πΊ =πππ’π‘
πΌπβ and,
iii) The voltage responsivity, π π£ =πππ’π‘
ππππ‘
Fig. 4(c)
Solution
a) Employing the model for the photodiode as a current source gives the figure below:
i) From ohms law:
πππ’π‘ = πΌπβπ πΏ = π πππππ‘π πΏ [1]
π π£ =πππ’π‘
ππππ‘= π ππ πΏ = 0.9 Γ 500 = 450[π/π] [1]
ii) From KVL:
ππ = πππ’π‘ β 11V [1]
The diode will become forward biased when ππ > 0. Which occurs when πππ’π‘ > 11π. Setting
πππ’π‘ = 11π gives the point at which the diode is no longer reverse biased. [1]
So,
πππ’π‘ = π πππππ‘π πΏ = 11V [1]
ππππ‘ =11
0.9Γ500β 24ππ [1]
(b)We have the following for the TIA circuit with the current source model for the photodiode
employed:
Electric Circuits Fall 2015 Homework #4
6
i) We have ππ = 0 = ππ. Now the anode is at a virtual ground. The diode voltage is still
defined as the voltage from the anode to the cathode. KVL gives:
ππ = β11 [1]
ii) Note that πΌ2 = 0 because the voltage on both sides of the 100Ξ© is the same. KCL at Vn with
ππ = 0 gives:
πΌπβ = βπππ’π‘/500 [1]
πΊ =πππ’π‘
πΌπβ= β500Ξ© [1]
iii)
β500 =πππ’π‘
πΌπβ=
πππ’π‘
π πππππ‘ [1]
π π£ =πππ’π‘
ππππ‘= βπ π β 500 = β450[V/W] [1]
4.5 [10%]
For the circuit shown in Fig. 5, find the Thevenin equivalent at terminals a-b. (Hint: To find
π πβ, apply a current source ππ and calculate π£π.)
Fig. 5
Solution
ππβ = πππ (1) [1]
Apply KCL we find
Electric Circuits Fall 2015 Homework #4
7
ππ =π 1
π 1+π 2πππ (2) [2]
From (1) and (2),
ππβ = (1 +π 2
π 1)ππ [2]
To get RTh, apply a current source Io at terminals a-b as shown below.
Since the noninverting terminal is connected to ground, v1 = v2 = 0, i.e. no current passes
through R1 and consequently R2 . Thus, vo = 0 and
π πβ =π£0
π0= 0 [5]
4.6 [10%]
Design an op amp circuit such that
π£π = 4π£1 + 6π£2 β 3π£3 β 5π£4
Let all the resistors be in the range of 100Ξ© to 1kΞ©.
Solution
A summing amplifier shown below will achieve the objective. An inverter is inserted to invert
v2. Since the smallest resistance must be at least 100 Ξ©, then let R/6 = 100Ξ© therefore let R =
600 Ξ©. (Other reasonable answers will be ok.)
Electric Circuits Fall 2015 Homework #4
8
4.7 [11%]
Determine π£π in the op amp circuit of Fig. 7.
Fig. 7
Solution
The output of amplifier A is
vA = β30k
10kΓ 1 β
30k
10kΓ 2 = β9V [3]
The output of amplifier B is
vA = β20k
10kΓ 3 β
20k
10kΓ 4 = β14V [3]
va = vb =10k
60k+10kΓ (β14) = β2V [2]
At node a, vAβva
20k=
vaβv0
40k [2]
Therefore, v0 = 12V [1]
Electric Circuits Fall 2015 Homework #4
9
4.8 [12%]
(a) For the following schematic, determine a constraint on π 1, π 2, π 3 and π 4 such
that the circuit behaves as a difference amplifier, i.e. π£π = K(π£2 β π£1).
(b) Suppose that the resistors values chosen for part (a) are real resistors with 1% precision.
What is the common-mode gain? In other words, if the inputs contain a common-mode signal
π£1 = ππ +π£π
2
π£2 = ππ βπ£π
2
how much of ππ appears at the output?
Fig. 8
Solution
(a)
Now we write two nodal equations at v1 and v2.
V1
R1=
VxβV1
R2 [1]
V2βVx
R3=
V0βV2
R4 [1]
Solving for these equations gives us the following relationship:
Electric Circuits Fall 2015 Homework #4
10
v0 =R3+R4
R3v2 β
R4
R1R3(R1 + R2)v1[1]
In order for this expression to be in the form vo = K(v2 β v1), we must have the following
constraint:
R3+R4
R3=
R4
R1R3(R1 + R2) [1]
This simplifies to
R1R3 = R2R4 [1]
(b) If the resistors have 1% precision, then any resistor Ri can have a resistance as high as
1.01Ri or as low as 0.99Ri. Letβs assume that each resistor takes on a value R = Rnom(1 + β)
where Rnom is the nominal value, and |β| < 0.01 is the tolerance (positive or negative). Now
re-write the expression for the gain as
v0 = π2v2 β π1v1[1]
The a2 coefficient is expanded and then simplified assuming β are small
π2 = 1 +Rβ²4
Rβ²3= 1 +
R4(1+β4)
R3(1+β3)β 1 +
R4
R3(1 + β4)(1 β β3) β 1 +
R4
R3(1 + β4 β β3) [1]
Notice that a2 can be written in terms of the ideal gain G = 1 +R4
R3
π2 = G + (G β 1)(β4 β β3) = G [1 +Gβ1
G(β4 β β3)][2]
Likewise, since
R1
R3=
R4
R3
we can expand the coefficient a1
π1 β G + (G β 1)(β4 β β3) = G(1 + β4 β β3 +β2ββ1
G) [2]
For a common-mode input Vc we have
v0 = π2Vc β π1Vc = VcG (Gβ1
G(β4 β β3) + β3 β β4 β
β2ββ1
G) = Vc(β1 β β2 + β3 β β4)[1]
|β| < 0.01
or
Ac = β1 β β2 + β3 β β4
|β| < 0.01
Electric Circuits Fall 2015 Homework #4
11
4.9 [11%]
In the schematic below, the voltage source π£π is separated from the load resistor π πΏ
by three amplifier stages. We have three different amplifier configurations as shown
below.
Fig. 9
(a) Suppose we go with the sequence BAC. If π 1 = 2kΞ©, π 2 = 6kΞ©, π 3 = 4kΞ©, and π 4 =
8kΞ©, calculate the overall voltage gain for the above circuit. What is the load resistance as
seen by the voltage source?
(b) Repeat the above calculations for the sequence ABC. Which of the two is better?
(c) What is the voltage gain in both cases if we omit the third stage (C)? What is its
purpose of having it in the circuit?
Solution
(a) For the sequence BAC, we first have a noninverting amplifier, followed by an inverting
amplifier, and finally a voltage follower. The gain is thus
G = (1 +R4
R3) (β
R2
R1) = (1 +
8k
4k) (β
6k
2k) = β9 [3]
The load resistance seen by the voltage input is infinite, since no current flows into the
noninverting input of op amp B.
(b) The gain here is the same, since it is order-independent (multiplication is commutative).
[2]
However, the input resistance is 2 kΞ© higher in this case, since the source sees the inverting
amplifier first. So in terms of overall gain, both orderings are equally good.
[1]
However, the ABC former is better for voltage amplification, as its input resistance is much
higher. [1]
(c) The voltage gain remains the same in both cases, since the third stage does not affect it.[2]
C is a voltage follower which has high input resistance. It is usually useful to prevent loading
effects at either the input or output of an op amp, like A and B, especially if it is non-ideal
(although it doesnβt play any role here since all the op amps are ideal).[2]
Electric Circuits Fall 2015 Homework #4
12
4.10 [12%]
a) For the circuit shown in Fig. 10, show that if ΞR βͺ R, the output voltage of the op amp is
approximately
π£π βπ π
π 2
(π + π π)
(π + 2π π)(ββπ )π£ππ
b) Find π£π if π π = 470kΞ©, π = 10kΞ©, Ξπ = 95Ξ©, and π£ππ= 15 V.
c) Find the actual value of π£π in (b).
Fig. 10
d)If percent error is defined as
% error = [approximate value
true valueβ 1] Γ 100
show that the percent error in the approximation of π£π above is
% error = [βπ
π
(π + π π)
(π + 2π π)] Γ 100
e) Calculate the percent error in π£π.
Solution
Let R1 = R + βR
Apply KCL,
Vp
Rf+
Vp
R+
VpβVin
R1= 0 (1) [2]
Vn
R+
VnβVin
R+
VnβV0
Rf= 0 (2) [2]
Electric Circuits Fall 2015 Homework #4
13
From (1) and (2),
π£π =(π +π π)π π
π [(R+βR)(π +π π)+π π π](ββπ )π£ππ [1]
if R β« βR
π£π βπ π
π 2
(π +π π)
(π +2π π)(ββπ )π£ππ [2]
(b) π£π β β3.384π [1]
(c) π£π β β3.368π [1]
(d) % error = π π
π 2
(π +π π)
(π +2π π)
π [(R+βR)(π +π π)+π π π]
(π +π π)π π β 1 Γ 100 [2]
=βR(π +π π)
π (π +2π π)Γ 100 [1]
(e) % error = 0.48 [2]