Sedra/Smith Microelectronic Circuits 6/E Chapter 3: Diodes

Transcription

Sedra/Smith
Microelectronic Circuits 6/E
Chapter 3: Diodes
S. C. Lin, EE National Chin-Yi University of Technology
1
【Outline】
3.1 The Ideal Diode
3.2 Terminal Characteristics of Junction Diodes
3.3 Modeling the Diode Forward Characteristic
3.4 Operation in the Reverse Breakdown Region — Zener
Diodes
3.5 Rectifier Circuits
3.6 Limiting and Clamping Circuits
3.7 Special Diode Types
2
3.1 The Ideal Diode
i
+ v −
(a) diode circuit symbol
(b) i–v characteristic
i
i
+ v −
v<0⇒i =0
(c) equivalent circuit in the
reverse direction
+ v −
i >0⇒v=0
(d) equivalent circuit in
the forward direction.
3
The two modes of operation of ideal diodes and the use of an
external circuit.
+10V
10mA
1kΩ
+10V
0mA
1kΩ
+
+
0V
−
10V
−
(a) the forward current
(b) the reverse voltage
4
Vp
vi
+ vD −
t
0
iD
vi
(a) Input waveform.
iD
+ vD = vi −
R
+
vo = vi
−
(c) Equivalent circuit when vi ≥ 0.
0
vi
iD = 0
R
+
vo = 0
−
vi ≤ 0
vi ≥ 0
Vp
R
(b) Rectifier circuit.
+ vD = 0 −
vi
+
vo
−
(d) Equivalent circuit when
vi ≤ 0
.
vo
t
(e) Output waveform.
5
Example 3.1 Fig.3.4(a), shows a circuit for charging a 12-V battery.
If vs is sinusoid with 24-V peak amplitude, (a) Find the fraction of
each cycle during which the diode conducts. (b) Also find the peak
value of the diode current and (c) the maximum reverse-bias voltage
that appears across the diode.
iD
100Ω
vs
Method 1:
12V
24V
VD
12V
αθ
(a ) 24sin α = 12 ⇒ α = 30o
2θ = 180o − 2α = 120o
6
Method 2:
24V
VD
12V
θ
θ
(a)24 cos θ = 12 ⇒ θ = 60o
Conduction angle = 2θ = 120o ▲
(b)I D (max) =
24 − 12
= 0.12A ▲
100
(c)The maximum reverse voltage across the diode:
VD (max) = 24 + 12 = 36V ▲
7
3.1.3 Another Application: Diode Logic Gates
+5V
vA
R
vB
vC
vY
(a) OR gate
vA
(b) AND gate
vB
R
vY
vC
vA
vB
vC
vY
vA
vB
vC
vY
8
Example 3.2 Assuming the diode to be ideal, Find the values of I and
V in the circuit of Fig. (a) and (b)
+10V
Sol:
Asssume that both diodes are conducting
10kΩ
I D2
I
D1
5kΩ
−10V
D2
VB = 0 and V = 0
+
V
−
ID =
2
(10 − 0)V
= 1mA
10kΩ
Writting a node equation at B (K.C.L)
0 − (−10)
I +1 =
⇒ I = 1mA
5
Results in I = 1mA,
Thus D1 is conducting as originally assumed
and the find resuit is I = 1mA and V = 0V ▲
9
Sol:
+10V
.
We asssume that both diodes are conducting
5kΩ
I D2
D1
I
D2
then VB = 0 and V = 0
ID =
+
V
−
2
(10 − 0)V
= 2mA
5kΩ
The node equation at B is
0 − (−10)
⇒ I = −1mA
10
(This is not possible)
I +2=
10kΩ
−10V
Our assumption is not correct, we start again assuming that D1 is OFF and D2 is ON.
The current I D is given by
2
10 − (−10)
= 1.33mA
15
VB = −10 + 10 ⋅ 1.33 = 3.3V
ID =
2
Thus D1 is reverse biased as assumed ∴ I = 0A and V = 3.3V ▲
10
3.2 Terminal Characteristics junction diode
i
(1) The forward-bias region
determined by v > 0
1
(2) The reverse-bias region
determined by − VZK < v < 0
−VZK
3
2
Expanded scale
0
0.5V
0.7V
v
(3) The breakdown-bias region
determined by v < −VZK
i
+ v −
Figure 3.8 The diode i–v relationship with some scales expanded
and others compressed in order to reveal details.
11
3.2.1 The Forward-bias region
(
)
i = I s ev / nV − 1
T
(3.1)
(a ) I s is saturation current , another name for I s is the scale current.
(b) I s is a constant for a given diode at a given temperature.
(c) I s is directly propotional to the across-sectional area of the diode.
(d) I s For "small-signal" diode intened for low-power application,
I s is on the order of 10-15A
(e) I s a very strong function of temperature. As a rule of thumb,
I s doubles in value for every 5oC rise in temperature.
12
VT : Thermal voltage
K ⋅T
VT =
q
(3.2)
where K = Boltzmann's constant = 1.38 × 10-23 Joules/Kelvin
T = The absolute temperature in Kelvin = 273+temperature oC
q = The magnitude of electronic charge = 1.60 × 10−19 Coulomb
At room temperature(20o C), VT ≅ 25.2mV.
At room temperature(27o C), VT ≅ 26mV.
n
n :1 ↔ 2, Depending on the material and
the physical structure of the diode.
13
the forward direction i >> I s
i
i ≅ Ise
⇒ v = nVT ln
Is
If the voltage is v1 (v2 ), the diode current I1 ( I 2 )
v / nVT
thus
i1 ≅ I s ev1 / nVT
and i2 ≅ I s ev2 / nVT
i2
= e( v −v ) / nV
i1
2
1
T
i2
i2
v2 − v1 = nVT ln = 2.3nVT log
i1
i1
(3.5)
This equation simply states that for a decade changes in current,
the diode voltage drop changes by 2.3nVT ≈ 60mV,for n = 1
14
Example: Design the circuit in below Figure to provide an output
voltage of 2.4V. Assume that diode available have 0.7V drop
at 1mA and that ΔV= 0.1V/decade changes in current.
Sol:
Refer Fig.a to obtain vo = 2.4V
+10V
each diode must have a voltage drop of 0.8V
thus
i2
v2 − v1 = 2.3nVT log
i1
R
vo
i2
i2
0.1 = 2.3 ⋅ 2 ⋅ 25 ⋅10 ⋅ log = 0.1log
i1
i1
−3
i2 = 10i1 = 10mA
The value of R is determined from
(10 − 2.4)V
R=
= 760Ω ▲
10mA
15
T2
i
>
T1
−2mV/ o C
I
v
Figure 3.9 Illustrating the temperature dependence of the diode
forward characteristic. At a constant current, the voltage drop
decreases by approximately 2 mV for every 1°C increase in
temperature.
16
Example 3.3 : A silicon diode said to be a 1-mA device displays a
forward voltage of 0.7V at a current of 1mA. Evaluate the
junction scaling constant Is in the event that n is either 1or 2,
what scaling constants would apply for a 1-A diode of the
same manufacture that conducts 1A at 0.7V
Since i ≅ I s ev / nVT ⇒ I s = i ⋅ e − ( v / nVT )
for the 1 − mA diode.
If n = 1
If n = 2
I s = 1×10−3 × e −700 / 25 = 6.9 × 10−16 A ▲
I s = 1× 10−3 × e −700 / 50 = 8.3 × 10−10 A ▲
The diode conducting 1A at o.7V corresponds to 1000 1mA in parallel
with a total junction area 1000 times greater. Thus it is also 1000 times
greater being 1pA and 1µA , respectively for n = 1 and n = 2 .
17
3.2.2 The Reverse-Bias Region
If v is negative and a few times large then VT (25mV)
in magnitude the diode current becomes
i ≅ − I s (Saturation current)
I s : It doubles for every 10o C rise in temperature
I s2 = I s1 ⋅ 2(T2 −T1 /10)
3.2.3 The Breakdown Region
⎧ Z: Zener
v < −VZK ⎨
⎩ K: Knee
In the breakdown region the reverse current increase rapidly
with the associated increase in voltage drop very small
18
3.3 Modeling the diode forward characteristic
3.3.1 The Exponential Model.
ID
R
+
VDD
VD
−
⎧VDD > 0.5V
Assuming that ⎨
(3.6)
⇒ I D = I s eVD / nVT
⎩ ID > Is
The other equation that governs circuit operation is obtained by writing a
loop equation by KVL, resulting in I D =
VDD − VD
R
(3.7)
19
3.3.2 Graphical Analysis Using the Exponential Model.
i
VDD
R
Q (Operating point)
ID
0
−1
Slop =
R
VD
VDD
v
Figure 3.11 Graphical analysis of the circuit in
Fig. 3.10 using the exponential diode model.
20
3.3.3 Iterative analysis using the exponential model
Example3.4 determine the current I D and the diode voltage VD for the circuit
in Fig.3.10 with VDD = 5V and R = 1kΩ, assume that the diode has a current
of 1mA at voltage of 0.7V and that its voltage drop changes by 0.1V for every
decade change in current.
Sol : We assume that VD = 0.7V, I D = 1mA
VDD − VD (5 − 0.7)V
=
= 4.3mA
ID =
R
1kΩ
I
V2 − V1 = 2.3 ⋅ n ⋅ VT ⋅ log 2
I1
V2 = V1 + 0.1 ⋅ log
I2
( n = 2, 2.3 ⋅ n ⋅VT ≅ 0.1V )
I1
4.3mA
= 0.763V
V2 = 0.7 + 0.1 ⋅ log
1mA
The first iteration are I D = 4.3mA, VD = 0.763V
21
We assume that VD = 0.763V, I D = 4.3mA
VDD − VD (5 − 0.763)V
=
= 4.237mA
1kΩ
R
4.237mA
V2 = 0.763 + 0.1 ⋅ log
= 0.762V
4.3mA
ID =
The second iteration are I D = 4.237mA, VD = 0.762V
We assume that VD = 0.762V, I D = 4.237mA
VDD − VD (5 − 0.762)V
ID =
=
= 4.238mA
R
1K Ω
4.238mA
V2 = 0.762 + 0.1⋅ log
= 0.762V
4.237mA
Because I D did not change by much stop here and the solution is
I D = 4.237mA and VD = 0.762V
22
3.3.5 The Constant-Voltage-Drop Model. (Rapid analysis)
iD (mA)
VD = 0.7V
⎧ iD = 0,
⎨
⎩iD = (vD − 0.7V) / R
VD
vD ≤ 0.7V
vD ≥ 0.7V
vD (V)
23
The constant-voltage-drop model of the diode forward characteristics
and its equivalent-circuit representation.
iD (mA)
12
ID
10
+
8
vD
6
−
VD = 0.7V
VD = 0.7V
4
2
0
0.2
0.4
0.6
0.8
1.0 v (V)
D
24
Example: Repeat the problem in example 3.4, utilizing the
constant-voltage-drop model whose parameters are given
in Fig 3.12 . Find I D and VD .
ID
ID
R
+
VDD
VD
−
+
vD
−
VD = 0.7V
Sol:
VDD − VD (5 − 0.7)V
ID =
=
= 4.3mA
R
1kΩ
VD = VD = 0.7V
25
3.3.7 The Diode Small-Signal Model.
iD = I S evD / nVT
(3.8)
vD (t ) = VD + vd (t )
(3.9)
26
iD (t ) = I S evD (t ) / nVT = I S e[VD + vd (t )] / nVT
= I S eVD / nVT ⋅ evd (t ) / nVT
ID
= I D ⋅ evd (t ) / nVT
(3.12)
If the signal vd (t ) is kept sufficiently small
vd (t )
such that
<< 1
nVT
Thus, the approximate expression
⎛
vd ⎞
iD (t ) = I D ⋅ ⎜1 +
⎟
⎝ nVT ⎠
(3.14)
Small-signal approximation
27
We have
iD (t ) = I D +
ID
vd
nVT
(3.15)
id
ID
vd
nVT
vd nVT
rd ≡
=
id
ID
where id =
(3.17)
(3.18)
The rd is diode smal-signal
resistance,
and called incremental resistance)
In Fig.3-13(b), which is equal to the slope of the tangent to the
i - v curve at the operating point Q is equal to the small-signal
conductance.
that is, rd =
1
⎡ ∂id ⎤
⎢ ∂v ⎥
⎣ d ⎦i
d
(3.19)
=ID
28
Example 3.6: Consider the circuit shown in Fig 3.15 a string of three
diode is used to provide a constant voltage of about 2.1V, we want to
calculate the percentage change in this regulated voltage cause by :
(a) ±10% change in the power-supply voltage
(b) Connection of a 1kΩ load resistance. Assume n=1
Sol:
with no load
(10 − 2.1)V
= 7.9mA
I=
1kΩ
nVT 25mV
=
= 3.2Ω
rd =
I
7.9mA
The total incremental resistance
r = 3rd = 9.6Ω
29
r
0.0096
= ±1
= ±9.5mV ▲
(a) ∆vo = ±1
r+R
0.0096 + 1
This implies a change of about ± 3.2mV per diode.
(b)When a load resistance of 1kΩ is connected across
the diode string. it draws a current of 2.1mA
∴∆vo = −2.1× r = −2.1× 9.6 = −20mV. ▲
this implies that the voltage across
each diode decreses by about 6.7mV.
30
3.4 Operation in the reverse breakdown region
-Zener diode
(1) rZ is the inverse of the slop of the
almost-linear i - v curve at point Q
(2) rZ : increase resistance
(3) Typically, rZ i s in the range
of a few ohms to a few ten ohms
31
Vz = Vzo + I z rz
(3.20)
32
3.4.2 Use of the Zener as a Shunt Regulator
Example 3.7: The 6.8V Zener diode in the circuit of Fig.3.19(a)
is specified to have VZ = 6.8V at I Z = 5mA, rz = 20Ω and
I ZK = 0.2mA, the V + is 10V ± 1V.
33
(a) Find Vo with no load and with V + at its nominal value.
Sol: first we must determine the value of V
zo
Vzo = Vz − I z rz = 6.8 − 5 ⋅10−3 × 20 = 6.7 V
10 − 6.7
Iz = I =
= 6.35mA
0.5 + 0.02
Thus Vo =Vz = Vzo + Iz rz = 6.83V ▲
(b) Find change in Vo resulting from the ± 1V change in V+ .
Sol: ∆ Vo = ∆V +
rz
20
= ±1 ×
= ±38.5mV
R + rz
500 + 20
line regulation ≡ ∆Vo / ∆Vi = ±38.5mV/1V = ±38.5mV/V ▲
34
(c) Find change in Vo resulting from connecting a load resistance RL .
that draws a current I L = 1mA, and hence find the load regulation
(∆Vo / ∆I L ) in mV/mA.
Sol:
∆ Vo = ∆I z ⋅ rz = 20Ω × −1mA = −20mV
load regulation ≡ ∆Vo / ∆I L = −20mV/mA ▲
(d)Find change in Vo when RL = 2kΩ.
Sol: RL = 2kΩ, I RL ≈ ∆I z
≈ −6.8V / 2kΩ = −3.4mA
∆ Vo = ∆I z ⋅ rz = −20 × 3.4 = −68mV ▲
35
(e)Find the value of Vo when RL = 0.5kΩ.
Sol:
RL = 0.5kΩ, would draw a I L = 6.8 / 0.5 = 13.6mA.This is impassible.
because I supplied through R is only 6.4mA, therefore,
RL
the zener must be cut-off. Hence Vo = V
= 5V ▲
R + RL
+
(f) What is the minmum value of RL for which the diode
still operates region in the breakdown .
Sol:
I z = I zK = 0.2mA ⇒ Vz = VzK ≈ Vzo = 6.7V
9 − 6.7
I (min) =
= 4.6mA
0.5
I (max) = (4.6 − 0.2)mA = 4.4mA, ∴ RL =
6.7
= 1.5kΩ ▲
4.4
36
3.4.3 Temperature effects, T.C or Temco
(1) VZ < 5V, dVz / dT < 0. Negative T.C
(2) VZ > 5V, dVz / dT > 0. Positive T.C
(3) VZ = 5V, dVz / dT = 0. Zero T.C
37
3.5 Rectifier Circuits
Figure 3.20 Block diagram of a dc power supply.
(1) The Half-wave rectifier.
(2) The full-wave rectifier
(a) The center tap rectifier
(b) The bridge rectifier
38
3.5.1 Half-Wave Rectifier.
(a) Half-wave rectifier.
(b) Equivalent circuit of the half-wave
rectifier with the diode replaced with
its battery-plus-resistance model.
39
vS < VDO
⎧vO = 0,
⎪
R
⎨
v
=
⎪ O R + r (vS − VDO ), vS ≥ VDO
D
⎩
(c) Transfer characteristic of the rectifier circuit.
VD
φi = sin
Vm
−1
φi
(d) Input and output waveforms, assuming that rD << R.
40
We assume vS = Vm sin α thus
1 π
VDC =
vS dα
∫
0
2π
1 π
Vm sin αdα
=
∫
0
2π
Vm
π
= (− cos α 0 )
2π
Vm
= 0.318Vm
=
π
1/ 2
⎡1
⎤
Vrms = ⎢ ∫ vS 2 dα ⎥
⎣ 2π 0
⎦
1 π 2 2
2
Vrms =
Vm sin αdα
∫
0
2π
Vm 2 π 1 1
=
( − cos 2α )dα
∫
2π 0 2 2
π
π
Vm 2 1
1
=
( α − sin 2α )
2π 2
4
0
Vm 2
=
⇒ Vrms = 0.5Vm
4
PIV = Vm (PIV : Peak inverse voltage)
VDO
φi = sin
Vm
−1
41
3.5.2 The full-wave rectifier
(a) circuit
(b) transfer characteristic assuming a constant-voltage-drop
model for the diodes
42
(c) input and output waveforms.
43
We assume vS = Vm sin α thus
VDC
⎧1 π
⎫
= 2 × ⎨ ∫ vS dα ⎬
⎭
⎩ 2π 0
2 ×1 π
=
Vm sin αdα
∫
0
2π
Vm
π
= (− cos α 0 )
π
2 × Vm
= 0.636Vm
=
π
1/ 2
⎡ 2 ×1 π 2 ⎤
Vrms = ⎢
vS dα ⎥
∫
0
⎣ 2π
⎦
2 ×1 π 2 2
2
Vrms =
Vm sin αdα
∫
0
2π
2 × Vm 2 π 1 1
=
( − cos 2α )dα
∫
2π 0 2 2
π
Vm 2 1
1
=
( α − sin 2α )
4
π 2
0
Vm 2
=
⇒ Vrms = 0.707Vm
2
PIV = 2Vm (PIV : Peak inverse voltage)
VD
φi = sin
Vm
−1
44
3.5.3 The Bridge Rectifier
Figure 3.27 The bridge rectifier: (a) circuit; (b) input and output waveforms.
45
3.5.4 The Rectifier with a Filter Capacitor-The Peak Detector
A simple circuit used to illustrate the effect of a filter capacitor
46
47
48
We assumption that RC >> T in Fig3.29(b)
The load current iL = vO / R
dv
The diode current iD = iC + iL = C I + iL
dt
(1) Derivative the Vr ( rms )
During the diode cutoff interval vo = VP − Vr
VP − Vr = VP e −T / RC
Vr = VP
V
VP
T
= P ⇒ Vr ( rms ) =
RC fRC
2 3 fRC
(2) Derivative the conduction interval ∆t
VP cos(ω∆t ) = VP − Vr
1
The angle ω∆t is very small , ∴ cos(ω∆t ) ≈ 1 − (ω∆t ) 2
2
⎡ 1
⎤
VP ⎢1 − (ω∆t ) 2 ⎥ = VP − Vr ⇒ ω∆t = 2Vr / VP
(3.30)
⎣ 2
⎦
We note that when Vr << VP the conduction angle ω∆t will be small
49
(3) Dertermine the average diode current during conduction, iDav
the charge that the diode supplies to the capacitor
QSupplied = iC ( av ) ⋅ ∆t
(a)
the charge that the capacitor loses
Qlost = C ⋅Vr
(b)
C ⋅ Vr
The(a) equal to (b) ⇒ iC ( av ) =
∆t
VP
VP
Using(3.29a ) Vr =
,
⇒C =
fRC
Vr fR
2Vr
and(3.30) ω∆t = 2Vr / VP ⇒ ∆t =
ω VP
VP
⋅V r
f V rR
2π VP VP
C ⋅Vr
we obtain: iC ( av ) =
=
⋅ = I L ⋅ π ⋅ 2VP / Vr
=
∆t
2Vr
2Vr R
2π f
iD ( av ) = iC ( av ) + I L = I L ⎡⎣1 + π 2VP / Vr ⎤⎦
VP
(3.31)
50
(4) Dertermine the peak value of the diode current , iD (max)
iD (max)
dvI
=C
+ iL
dt
dvI
d
(t = t1 = −∆t ) ⇒ C
= C (VP cos(−ω∆t ) ) = ωCVP sin(ω∆t )
dt
dt
(ω∆t )3
+
= ω ⋅ C ⋅ VP ⋅ (ω∆t )
[sin(ω∆t )] = ω∆t +
3!
2Vr
VP
= 2πf ⋅
⋅ VP ⋅
= I L ⋅ 2π ⋅ 2VP / Vr
fVr R
VP
iD (max) = iC + I L = I L ⎣⎡1 + 2π 2VP / Vr ⎦⎤
(3.32)
51
the full-wave rectifier circuit.
∵ f r = 2 fi
VP − Vr = VP e
−T / 2
RC
VP
T /2
=
Vr = VP
RC 2 fRC
VP
Vr ( rms ) =
4 3 fRC
(3.33)
52
2.4VP
Vr ( rms ) =
( f = 60Hz, RL = kΩ, C = µF )
RC
when the ripple small then 6.5% ⇒ VP = VDC
2.4VDC 2.4 I DC
Vr ( rms ) =
=
RC
C
Vr ( rms ) 2.4
r% =
%
=
VDC
RC
If r % > 6.5%
V
1
⎡ 4.17 ⎤
⋅ VP
VDC = VP − Vr = VP − P = ⎢1 −
⎥
2
4 fRC ⎣
RC ⎦
iD ( av ) = iC ( av ) + I L = I L ⎡⎣1 + π VP / 2Vr ⎤⎦
(3.34)
iD (max) = iC + I L = I L ⎡⎣1 + 2π VP / 2Vr ⎤⎦
(3.35)
53
Figure 3.27 The “superdiode” precision half-wave rectifier and its
almost-ideal transfer characteristic. Note that when vI > 0 and the
diode conducts, the op amp supplies the load current, and the
source is conveniently buffered, an added advantage. Not shown
are the op-amp power supplies.
54
3.6 limiting and clamping circuits
3.6.1 limiter circuit.
L−
L+
≤ vI ≤
K
K
vO = KvI
Ingeneral K >1
In this section have K ≤1
L+
(a ) If vI > (the upper threshold),
K
the vO is limited or clamping to the upper limiting level L+
L−
(b) If vI < (the lower limited threshold),
K
the vO is limited to the lower limiting level L− .
55
Hard limiting.
Soft limiting.
56
57
Example.3.26 Assuming the diode to be ideal, describe the transfer
characteristic of the circuit shown in Figure E3.26.
(2) − 5V ≤ vI ≤ 5V,
both diode are cut-off and
vO = vI
Figure E3.26
Sol:
(1) vI ≤ −5V,
D1 : conducts, D2 cut-off
10
10
vO = vI
−5
10 + 10 10 + 10
= 0.5vI − 2.5
(3) vI ≥ 5V,
D1 : cut-off, D2 conducts
10
10
vO = vI
+5
10 + 10 10 + 10
= 0.5vI + 2.5
58
.
vo
vO = 0.5vI + 2.5
vI
vO = 0.5vI − 2.5
59
Example : Assuming the diode to be ideal, describe the transfer
characteristic of the circuit shown in Fig.a. 0 < vi ≤ 150V
A
D2
100kΩ
vi
(1)For 0 <Vi ≤ 25V,
200kΩ
D1
+
+
−
−
vo
both diodes are cut-off
vO = 25V
(2)For 25V<Vi < 137.5V,
D1 conducts and D2 cut-off
Sol:
For VA ≥ 100V,
both diodes are conducts
2 25
vi ⋅ +
= 100 ⇒ vi = 137.5V
3 3
vO = VA =
2
25
vI + (V)
3
3
(3)For 137.5V< Vi < 150V,
both diodes are conducts
vO = 100V
60
vo
vO =
2
25
vi +
3
3
2
3
vi
61
3.6.2 The Clamped Capacitor or DC Restorer
Figure 3.32 The clamped capacitor or dc restorer with a
square-wave input and no load.
62
Figure 3.33 The clamped capacitor with a load resistance R.
63
3.6.3 Voltage Doubler
Figure 3.34 Voltage doubler: (a) circuit; (b) waveform of the
voltage across D1.
64
65

Sedra/Smith Microelectronic Circuits 6/E Chapter 3: Diodes

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