VNU-University of Engineering and Technology
Faculty of Electronics and Telecommunications
Lecture #1: Diodes & applications
8/28/2023
1
Outline
1. Introduction
2. The ideal diode
3. The real diode
4. Applications
5. Other diode
Textbook: Adel. S. Sedra, Kenneth C. Smith. Microelectronic Circuits. Oxford University
Press. 2011/2014 (Chapter 4).
References: Thomas L. Floyd, Electronic devices, 9th edition, Prentice Hall (Section 2).
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1. Introduction
• Diode is a semiconductor device which conduct the
current in one directly only. The name diode
comes from dielectrodes, that is, 2 electrodes.
• Two terminals: Anode (A+) and Cathode (K-).
• When the positive polarity is at the anode, the
diode is forward biased and is conducting.
• When the positive polarity is at the cathode, the
diode is reversed biased and is not conducting.
• If the reverse-biasing voltage is sufficiently large,
the diode is in reverse-breakdown region and large
current flows though it.
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2. The Ideal Diode
2.1 Current-Voltage characteristic
• The ideal diode may be considered the most
fundamental nonlinear circuit element
(b) i-v characteristic
(a) Diode circuit symbol
(c) Equivalent circuit in the
reserve direction
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(d) Equivalent circuit in the
forward direction
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2. The Ideal Diode
2.2 Some applications – The rectifier
(1) Input waveform
(2) Rectifier circuit
(3) Output waveform
Another example is in our cars: the
alternator charges the battery when the
engine is running, but when the engine
stops, a diode prevents the battery from
discharging through the alternator.
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Exercise 1
For the circuit as shown:
1) Sketch the transfer characteristic 𝑣𝑜 versus 𝑣𝐼
2) Sketch the waveform of 𝑣𝐷
3) Find the peak value of 𝑖𝐷 and the DC component 𝑣𝑜 if 𝑣𝐼 has a peak value
of 10 V and R = 1 k
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Exercise 1 (Solution)
1) The transfer characteristic 𝒗𝒐 versus 𝒗𝑰
2) The waveform of 𝒗𝑫
3) Find the peak value of 𝒊𝑫 and the DC component 𝒗𝒐
𝑣
We have: 𝑣𝐼 = 𝑣𝐷 + 𝐼𝐷 𝑅. 𝐼𝐷 max when diode is conducting => 𝑣𝐷 = 0 → 𝐼𝐷 = 𝐼 = 10 (𝑚𝐴)
DC component 𝑣𝑜 : 𝑣𝑜𝐷𝐶 =
𝑣𝑜𝐷𝐶
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𝑣𝑝
𝑇
𝑠𝑖𝑛𝜔𝑡𝑑𝑡
‫׬‬
𝑇 0
𝑅
𝑣𝑝 𝑇/2
𝑣𝑝
𝑣𝑝
𝑣𝑝
𝑇
= න 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 = −
𝑐𝑜𝑠𝜔 − 𝑐𝑜𝑠0 = −
𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠0 =
= 3.18
𝑇 0
𝜔𝑇
2
2𝜋
𝜋
(𝑉)
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2. The Ideal Diode
2.2 Some applications – Diode Logic Gates
The output will be high if one or
more of the inputs are high
vY = vA + vB + vC ⟹ OR gate
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The output will be high if all of
the inputs are high
vY = vA.vB.vC ⟹ AND gate
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Exercise 2
• Assuming the diodes to
be ideal, find the values
of I and V.
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Exercise 3: Find the values of I and V in the circuits
shown in below Fig. Ex3
Fig.: Ex3
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Exercise 4
Figure Ex4 shows a circuit for an AC voltmeter. It utilizes a moving-coil meter that
gives a full-scale reading when the average current flowing through it is 1 mA. The
meter has a 50 Ω resistance. Find the value of R that results in the meter indicating
a full-scale reading when the input sine-wave voltage 𝑣𝐼 is 20 V peak-to-peak.
(Hint: The average value of half-sine waves is Vp /π.)
Figure Ex4
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3. The Real Diode
3.1 The I-V characteristic
• The characteristic curve consists of three
distinct regions:
✓The forward-bias region, by v > 0
✓The reverse-bias region, by v < 0
✓The breakdown region, by 𝑣 < −𝑉𝑍𝐾
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Knee
voltage
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3. The Real Diode
3.1 The I-V characteristic
• The Forward-Bias Region:
𝑖 = 𝐼𝑆 (𝑒 𝑣Τ𝑉𝑇 − 1)
✓𝐼𝑆 is constant (~10−15 𝐴) for a given diode at a given
temperature (Saturation current).
𝑘𝑇
✓𝑉𝑇 : thermal voltage 𝑉𝑇 = , Note: 𝑉𝑇 = 25.3 𝑚𝑉
𝑞
(≅ 25𝑚𝑉) at room temperature (20𝑜 𝐶)
✓For 𝑖 ≫ 𝐼𝑆 → 𝑖 = 𝐼𝑠 𝑒 𝑣Τ𝑉𝑇
or
Where 2.3𝑉𝑇 = 60 𝑚𝑉
✓ Voltage drop: 0.7 V
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Exercise 5
A silicon diode said to be a 1 mA device displays a forward voltage of 0.7 V at a
current of 1 mA. Evaluate the junction scaling constant 𝐼𝑆 . What scaling constants
would apply for a 1A diode of the same manufacture that conducts 1 A at 0.7 V?
• Solution:
✓ We have: 𝑖 = 𝐼𝑠 𝑒 𝑣/𝑉𝑇 => 𝐼𝑠 = 𝑖𝑒 −𝑣/𝑉𝑇
✓ Diode 1 mA: Is = 10-3 × e-700/25 = 10-3 × e-28 (A) = 6.9 × 10-16 A
✓ Diode 1 A: Is = 1 × e-700/25 = e-28 (A) = 6.9 × 10-13 A
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Exercise 6: Find the change in diode voltage if the current changes from 0.1 mA to
10 mA.
• Solution: we have V2 – V1 = 2.3 × VT × log(I2/I1) = 60 × log (10/0.1) = 120 (mV)
Exercise 7: A silicon junction diode has v = 0.7 V at i = 1 mA. Find the voltage drop
at i = 0.1 mA and i = 10 mA.
Solution: we have V2 – V1 = 2.3 × VT × log(I2/I1)
o i = 0.1 mA: V2 – 700 = 60 × log(0.1/1) => V2 – 700 = – 60 =>V2 = 640 mV = 0.64 V.
o i = 10 mA: V2 – 700 = 60 × log(10/1) => V2 = 700 + 60 = 760 (mV) = 0.76 V.
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3. The Real Diode
3.1 The I-V characteristic
• The Reverse-bias Region: If 𝑣 < 0
and a few time larger than 𝑉𝑇 (~25
mV) in magnitude, the diode current
becomes 𝒊 ≅ −𝑰𝑺
• The Breakdown Region: 𝑣 < −𝑉𝑍𝐾
The reverse current increase rapidly
with the associated increase in voltage
drop very small
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Knee
voltage
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• The exponential model: The most accurate description of the diode
operation in the forward region is provided by the exponential model.
𝐼𝐷 = 𝐼𝑆 𝑒 𝑉𝐷 Τ𝑉𝑇
𝑉𝐷𝐷 − 𝑉𝐷
𝐼𝐷 =
𝑅
(Kirchhoff loop)
Two ways for obtaining the solution of the two above equations:
✓ Graphical Analysis using the exponential model
✓ Iterative analysis using the exponential model
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• Graphical analysis using the exponential model
𝐼𝐷 = 𝐼𝑆 𝑒 𝑉𝐷 Τ𝑉𝑇
𝑉𝐷𝐷 − 𝑉𝐷
𝐼𝐷 =
𝑅
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
• Iterative analysis using the exponential model
𝐼𝐷 = 𝐼𝑆 𝑒 𝑉𝐷 Τ𝑉𝑇
𝑉𝐷𝐷 − 𝑉𝐷
𝐼𝐷 =
𝑅
Two equations can be solved using a simple iterative procedure, as illustrated in
the following exercise.
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Exercise 8
Determine the current 𝐼𝐷 and the diode voltage 𝑉𝐷 for the circuit in
the figure with 𝑉𝐷𝐷 = 5 V and R = 1 k. Assume that the diode has
a current of 1 mA at a voltage of 0.7 V.
𝑉𝐷𝐷 −𝑉𝐷
𝑅
5−0.7
1𝑘
• Solution: 𝐼𝐷 =
=
= 4.3 (𝑚𝐴)
✓ We then use the diode equation to obtain a better estimate for 𝑉𝐷 using equation
𝐼
𝑉2 − 𝑉1 = 2.3𝑉𝑇 log 2 .
𝐼1
with 𝑉1 = 0.7 𝑉, 𝐼1 = 1 mA, and 𝐼2 = 4.3 mA results in 𝑉2 = 0.738 V.
=> Thus the results of the first iteration are 𝐼𝐷 = 4.3 mA and 𝑉𝐷 = 0.738 V.
✓ The second iteration proceeds in a similar manner:
𝐼𝐷 = 4.262 𝑚𝐴; 𝑉𝐷 = 0.738 𝑉
Since these values are very close to the values obtained after the first iteration, no
further iterations are necessary.
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
To speed up the analysis process, we must
find a simpler model for the diode forward
characteristic.
• The Constant Voltage Drop Model
✓It is the simplest and most widely used
diode model.
✓When forward biased, the diode has a
voltage drop that varies in a narrow
range, 0.6 V to 0.8 V
✓This model assumes 𝑉𝐷 = 0.7 𝑉
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Exercise 9
For the circuit in beside figure, find 𝐼𝐷 and 𝑉𝐷 for the case 𝑉𝐷𝐷 = 5 V and
R = 10 k. Assume that the diode has a voltage of 0.7 V at 1 mA current.
Use (a) iteration and (b) the constant-voltage-drop (CVD) model with 𝑉𝐷 =
0.7 V.
a) Solution for iteration:
b) Solution for CVD model
✓ Iteration 1: VD = 0.7V, 𝐼𝐷 =
𝑉𝐷𝐷 −𝑉𝐷
𝑅
=
5−0.7
10𝑘
= 0.43 (𝑚𝐴).
𝐼
𝑉2 = 𝑉1 + 0.06 × log 𝐼2 = 0.7 + 0.06 × log
1
✓ Iteration 2: VD = 0.678 V, 𝐼𝐷 =
𝑉𝐷𝐷 −𝑉𝐷
𝑅
𝐼
=
5−0.678
10𝑘
= 0.678 (𝑉)
= 0.432 (𝑚𝐴).
=> 𝑉2 = 𝑉1 + 0.06 × log 𝐼2 = 0.678 + 0.06 × log
1
0.43
1
𝑉𝐷𝐷 − 𝑉𝐷 5 − 0.7
=
= 0.43 (𝑚𝐴)
𝑅
10𝑘
=> ID = 0.43 mA, VD = 0.7 V
𝐼𝐷 =
0.432
0.43
= 0.678 (𝑉)
Summary: ID = 0.43 mA, VD = 0.678 V
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3. The Real Diode
3.2 Modeling the Diode Forward Characteristics
The small signal model: vD(t) = VD + vd(t)
V + v ) /V
iD ( t ) = I S ev /V → iD ( t ) = I S e(
= I S eV /V e v /V
D
D
T
d
T
D
T
d
T
I D = I S eVD /VT → iD (t ) = I D evd /VT
If 𝑣𝑑 (𝑡) satisfies
 iD (t ) = I D +
𝑣𝑑
𝑉𝑇
ID
vd
VT
≪ 1,
iD (t )
 v 
I D 1 + d 
 VT 
or iD = I D + id
with id =
ID
vd
VT
VT
The dynamic resistance: rd =
ID
→ This is the small signal model of the diode, which applies for
signal that has amplitude smaller than 5mV
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Exercise 10
Find the value of the diode small-signal resistance 𝑟𝑑 at bias currents of 0.1 mA,
1 mA, and 10 mA.
• Solution: The dynamic resistance 𝑟𝑑 =
𝑉𝑇
𝐼𝐷
25
= 250 (Ω)
0.1
25
1 𝑚𝐴 → 𝑟𝑑 = = 25 (Ω)
1
25
10 𝑚𝐴 → 𝑟𝑑 = = 2.5 (Ω)
10
o 𝐼𝐷 = 0.1 𝑚𝐴 → 𝑟𝑑 =
o 𝐼𝐷 =
o 𝐼𝐷 =
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Exercise 11
Consider the circuit shown in the figure, with R = 10 k. The power supply V+ has a DC value of 10 V
on which is superimposed a 60 Hz sinusoid of 1 V peak amplitude. (This “signal” component of the
power-supply voltage is an imperfection in the power-supply design. It is known as the power-supply
ripple). Calculate both the dc voltage of the diode and the amplitude of the sine-wave signal appearing
across it. Assume the diode to have a 0.7 V drop at 1 mA current.
a) Circuit for the Example
b) Circuit for calculating the dc operating point
c) Small signal equivalent circuit
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Exercise 11 (solution)
• Considering dc quantities only, we assume 𝑉𝐷 = 0.7 V and calculate
(10−0.7)
the diode dc current: 𝐼𝐷 =
= 0.93 (𝑚𝐴)
10𝑘
• Since this value is very close to 1 mA, the diode voltage will be very
close to the assumed value of 0.7 V. At this operating point, the
diode incremental resistance 𝑟𝑑 is:
𝑉𝑇
25
𝑟𝑑 =
=
= 26.9 (Ω)
𝐼𝐷 0.93
• The signal voltage across the diode can be found from the smallsignal equivalent circuit in Fig.(c). Here 𝑣𝑠 (@ 60 Hz) 1 V peak
sinusoidal component of V+, and 𝑣𝑑 is the corresponding signal
across the diode. Using the voltage-divider rule provides the peak
amplitude of 𝑣𝑑 as follows:
vd ( peak ) = Vˆs
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rd
0.0269
=1
= 2.68 (mV )
R + rd
10 + 0.0269
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3. The Real Diode
3.3 Zener diode Operation in the Reverse Region
• In breakdown region, a reverse bias (𝑉𝑍 ) beyond
the knee voltage (𝑉𝑍𝐾 ) leads to a large reverse
current (𝐼𝑍 )
(knee current)
• 𝑟𝑧 : dynamic resistance, is the inverse of the slope
of the almost-linear I-V curve at the Q-point.
Typically, it is in range of a few ohms to a few ten 
• ∆𝐼 and ∆𝑉 on Zener: V = rzI
• 𝑉𝑍0 denotes the point at which the straight line of
1
slope intersects the voltage axis.
(maximum current)
𝑟𝑧
• Voltage 𝑉𝑧 :
VZ = VZ 0 + rz I Z
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Exercise 12
The 6.8-V zener diode in the circuit of Fig. Ex12(a) is
specified to have 𝑉𝑍 = 6.8 𝑉 at 𝐼𝑍 = 5 𝑚𝐴, 𝑟𝑧 = 20 Ω, and
𝐼𝑍𝐾 = 0.2 𝑚𝐴. The supply voltage 𝑉 + is nominally 10 V but
can vary by ±1 𝑉.
a) Find 𝑉𝑂 with no load and with 𝑉 + at its nominal value.
b) Find the change in 𝑉𝑂 resulting from the ±1 𝑉 charge in 𝑉 + .
∆𝑉
Note that 𝑂+, usually expressed in mV/V, is known as line
∆𝑉
regulation.
c) Find the change in 𝑉𝑂 resulting from connecting a load
resistance 𝑅𝐿 that draws a current 𝐼𝐿 = 1 mA, and hence find
∆𝑉
the load regulation ( 𝑂) in mV/mA.
∆𝐼𝐿
d) Find the change in 𝑉𝑂 when 𝑅𝐿 = 2 k.
e) Find the value of 𝑉𝑂 when 𝑅𝐿 = 0.5 k.
Fig. Ex12 (a) Circuit for example; (b) The
circuit with the Zener diode replaced with
its equivalent circuit model.
f) What is the minimum value of 𝑅𝐿 for which the diode still
operates in the breakdown region?
28
Exercise 12 (Solution)
• 𝑉𝑍 = 𝑉𝑍𝑂 + 𝐼𝑍 𝑟𝑧 ; 𝑉𝑍 = 6.8 V, 𝐼𝑍 = 5 mA,
and 𝑟𝑧 = 20 Ω => 𝑉𝑍𝑂 = 6.7 𝑉
a) 𝐼𝑍 = 𝐼 =
𝑉 + −𝑉𝑍𝑂
𝑅+𝑟𝑧
=
10−6.7
500+20
= 6.35 (𝑚𝐴)
→ 𝑉𝑂 = 𝑉𝑍𝑂 + 𝐼𝑍 𝑟𝑧 = 6.7 + 6.35×0.02 = 6.83 (𝑉)
b) (∆𝑉 + = ±1𝑉).
∆𝑉𝑂 = ∆𝑉 +
𝑟𝑧
𝑅+𝑟𝑧
=
20
±1
500+20
= ±38.5 𝑚𝑉 . So line regulation: 38.5 (mV/V)
c)
Because 𝐼𝐿 = 1 mA, the zener current 𝐼𝑍 will decrease by 1 mA.
∆𝑉𝑂 = 𝑟𝑧 ∆𝐼𝑍 = 20 × −1 = −20 (𝑚𝑉)
∆𝑉𝑂
The load regulation is:
= −20 (𝑚𝑉/𝑚𝐴).
∆𝐼𝐿
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Exercise 12 (Solution)
6.8𝑉
d) 𝑅𝐿 = 2𝑘 → 𝐼𝐿 =
= 3.4 (𝑚𝐴)
2𝐾
→ ∆𝐼𝑍 = −3.4 𝑚𝐴 → ∆𝑉𝑂 = 20 × −3.4 = −68 (𝑚𝑉)
6.8
e) 𝑅𝐿 = 0.5𝑘 -> 𝐼𝐿 =
= 13.6 (𝑚𝐴).
0.5
This is not possible because I = 6.35 mA (for 𝑉 + = 10 𝑉). So the Zener must be cut off.
If this is indeed the case, then 𝑉𝑂 is determined by the voltage divider formed by 𝑅𝐿 and R (Fig. a):
𝑅
𝑉𝑂 = 𝑉 + × 𝐿 = 5 (𝑉).
𝑅+𝑅𝐿
Since this voltage is lower than the breakdown voltage of the zener, the diode is indeed no longer
operating in the breakdown region.
f) For the zener to be at the edge of the breakdown region, 𝐼𝑍 = 𝐼𝑍𝐾 = 0.2 𝑚𝐴 and 𝑉𝑍 ≅ 𝑉𝑍𝐾 = 6.7 𝑉.
At this point the lowest (worst-case) current supplied through R is I = (9 − 6.7)/0.5 = 4.6 (mA)
→ load current is 𝐼𝐿 =4.6 - 0.2 = 4.4 mA. The corresponding value of 𝑅𝐿 is
6.7
𝑅𝐿 =
≅ 1.5𝑘 ()
4.4
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4. Some application circuits using diodes
4.1 Rectifier circuits
Fig.: Block diagram of a DC power supply
• Half-Wave Rectifier
• Full-Wave Rectifier
• Bridge Rectifier
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Ripple
Still contains a time-dependent
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
PIV (Peak Inverse Voltage) = 𝑽𝒔
the diode must be able to withstand without breakdown
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Exercise 13
For the half-wave rectifier circuit (figure), show the following: (a) For the half-cycles during
which the diode conducts, conduction begins at an angle 𝜃 = sin−1 (𝑉𝐷 /𝑉𝑆 ) and terminates at
(𝜋 − 𝜃), for a total conduction angle of (𝜋 − 2𝜃). (b) The average value (dc component) of 𝑉𝑜 is
𝑉𝑜 ≅ 1Τ𝜋 𝑉𝑠 − 𝑉𝐷ൗ2, (c) The peak diode current is (𝑉𝑠 − 𝑉𝐷 )Τ𝑅. Find numerical values for these
quantities for the case of 12-V (rms) sinusoidal input, 𝑉𝐷 =0.7 V, and R = 100 Ω. Also, give the
value for PIV.
• Solution: rms (Root mean square) of sinusoidal input is 12V => 𝑉𝑠 = 12 × 2 (V)
✓ The conduction angle(𝝅 − 𝟐𝜽)
𝜃 = sin−1
𝑉𝐷
𝑉𝑆
= sin−1
0.7
12 2
= 2.4𝑜 => The conduction angle: 𝝅 − 𝟐𝜽 = 175.2𝑜
𝑉𝑠
𝑉
12 2
0.7
− 𝐷=
− = 5.05 (𝑉)
𝜋
2
𝜋
2
12 2−0.7
= 0.1267 (A); 𝑃𝐼𝑉 = 𝑉𝑠 = 12 2 (𝑉)
100
✓ The average value (DC component) of 𝑽𝒐 : 𝑉𝑜 ≅
✓ The peak current: 𝑰𝑫 = (𝑉𝑠 − 𝑉𝐷 )Τ𝑅 =
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= 16.97 𝑉
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4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
Let the input vI be a sinusoid with a peak
value Vp, and assume the diode to be ideal.
Fig.: Input & output
waveforms assuming
an ideal diode.
Real Diode?
Fig.: A simple circuit used to illustrate
the effect of a filter capacitor.
The circuit provides a dc voltage output equal to
the peak of the input sine wave.
There is no way for the capacitor to discharge.
34
4. Some application circuits using diodes
4.1 Rectifier circuits
• Half-Wave Rectifier
dvI
iD = iC + iL = C
+ iL
dt
Ideal diode
1
VO = V p - Vr
2
T
is peak-to-peak ripple voltage
Vr V p
CR
𝜔∆𝑡 =
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2𝑉𝑟 Τ𝑉𝑝 , where 𝜔 =
2𝜋
𝑇
35
4. Some application circuits using diodes
4.1 Rectifier circuits
• Full-wave Rectifier
PIV (Peak Inverse Voltage) = 𝟐𝑽𝒔 −𝑽𝑫
the diode must be able to withstand without breakdown
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Exercise 14
For the full-wave rectifier circuit as the beside figure, show the following: (a) The
output is zero for an angle of 2𝑠𝑖𝑛−1 (𝑉𝐷 Τ𝑉𝑠 ) centered around the zero-crossing points
2𝑉
of the sin-wave input. (b) The average value (dc component) of 𝑣𝑜 is 𝑉𝑜 ≅ 𝑠 − 𝑉𝐷 . (c)
𝜋
The peak current through each diode is (𝑉𝑠 − 𝑉𝐷 )/𝑅.
Find the fraction (percentage) of each cycle during which 𝑣𝑜 > 0, the value of 𝑉𝑜 , the
peak diode current, and the value of PIV, all for the case in which 𝑣𝑠 is a 12-V (rms)
sinusoid, 𝑉𝐷 = 0.7 𝑉, and R = 100 Ω
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Exercise 14 (solution)
rms (Root mean square) of sinusoidal input is 12 V => 𝑉𝑠 = 12 2 (V)
a)
The 𝑣𝑜 = 0 for an angle 𝜃 = 2 ×
−1 𝑉𝐷
sin
𝑉𝑆
= 2 × sin
b) The average value (dc component) of 𝑣𝑜 is 𝑉𝑜 ≅
c)
2𝑉𝑠
𝜋
−1
0.7
12 2
− 𝑉𝐷 =
= 4.8𝑜
2×12 2
−
𝜋
0.7 = 10.1 (V)
The peak current through each diode is (𝑉𝑠 − 𝑉𝐷 )/𝑅 = (12 2 - 0.7)/100 = 163 (mA)
The fraction (percentage) of each cycle during which 𝑣𝑜 > 0:
2𝜋−2𝜃
2𝜋
= 97.4%
The value of 𝑉𝑜 = 𝑉𝑠 − 𝑉𝐷 = 12 × 2 − 0.7 = 16.27 (V)
PIV = 2Vs − 𝑉𝐷 = 2 × 12 2 − 0.7 = 32.2 (V)
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4. Some application circuits
using diodes
4.1 Rectifier circuits
• The Bridge Rectifier
PIV (Peak Inverse Voltage) = 𝑽𝒔 −𝟐𝑽𝑫 + 𝑽𝑫 = 𝑽𝒔 − 𝑽𝑫
the diode must be able to withstand without breakdown
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Noted: the PIV is about half the value for the full-wave rectifier
with a center-tapped transformer. This is another advantage of
the bridge rectifier.
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Exercise 15
For the bridge rectifier circuit as the beside figure, use the constantvoltage-drop diode model to show that (a) the average (or dc
2𝑉𝑠
component) of the output voltage is 𝑉𝑜 ≅
− 2𝑉𝐷 (b) the peak diode
𝜋
current is (𝑉𝑠 − 2𝑉𝐷 )/𝑅.
Find numerical values for the quantities in (a) and (b) and the PIV for
the case in which 𝑣𝑠 is a 12 V (rms) sinusoid, 𝑉𝐷 = 0.7 𝑉, and
R = 100 Ω.
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Exercise 15 (solution)
• rms (Root mean square) of sinusoidal input is 12V => 𝑉𝑠 = 12 2
a) The average (or dc component) of the output voltage is
𝑉𝑜 ≅
2𝑉𝑠
𝜋
− 2𝑉𝐷 => 𝑉𝑜 ≅
a) The peak diode current is:
2×12 2
− 2 × 0.7 = 9.4 (𝑉).
𝜋
𝑉𝑠 −2𝑉𝐷
12 2−2×0.7
=
= 156 (𝑚𝐴)
𝑅
100
𝑃𝐼𝑉 = 𝑉𝑠 − 𝑉𝐷 = 12 2 − 0.7 = 16.27 (V)
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4. Some application circuits using diodes
4.2 Limiting Circuits
• Limiter/clipper circuits
✓
𝐿−
𝐾
≤ 𝑣𝐼 ≤
✓ If 𝑣𝐼 >
✓ If 𝑣𝐼 <
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𝐿+
𝐾
𝐿+
, 𝑣𝑜
𝐾
𝐿−
, 𝑣𝑜
𝐾
𝑣𝑜 = 𝐾𝑣𝐼
is limited to 𝐿+
is limited to 𝐿−
Transfer
characteristic for
a double limiter
circuit
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4. Some application circuits using diodes
4.2 Limiting Circuits
• Limiter/clipper circuits
double-anode zener
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43
Exercise 16
Assuming the diodes to be ideal, describe the transfer
characteristic of the circuit shown in the beside figure.
Solution:
𝐷1
𝐷2
2 diodes are cut-OFF for − 5 ≤ vI ≤ +5 & vo = vI
For vI ≤ − 5 V: D1 is ON, D2 is OFF
vI 
1

vo = -5 + (+ vI ) + 5 =  -2.5 -  (V )
2
2

vI 
1

For vI ≥ − 5 V: D2 is ON, D1 is OFF vo = -5 + (vI - 5) =  2.5 +  (V )
2
2

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4. Some application circuits using diodes
4.3 Clamping Circuits
• The Clamped Capacitor or DC Restorer
A Clamper Circuit is a circuit that adds a DC level to an AC signal. As the DC level gets
shifted, a clamper circuit is called as a Level Shifter. It shifts the waveform to a desired DC
level without changing the shape of the applied signal (but only shifts the amplitude of the
signal).
Types of Clampers: Clamp circuits are categorised by their operation: negative or positive,
and biased or unbiased.
• Positive Clamper
✓ Positive clamper with positive Vb
✓ Positive clamper with negative Vb
• Negative Clamper
✓ Negative clamper with positive Vb
✓ Negative clamper with negative Vb
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4. Some application circuits using diodes
4.3 Clamping Circuits
Positive Clamper
Initially when the input is given, the
capacitor is not yet charged and the
diode is reverse biased.
C1
Vi
D1
R1
VO
@ t0 the −25 V input signal appears across R1 & D1 (the C is a short at the
first instant). The initial voltage across R1 & D1 causes a voltage spike in the
output. The charge time of C1 through D1 is almost instantaneous, the
duration of the pulse is so short → it has a negligible effect on the output.
@ t1 the D1 is off (as open circuit) → KVL: VO = Vi + VC1 = 50 V. From t1 to t2, C1 has small discharge thru R1
so it has the voltage VC1 about +23 V and the VO drops from +50 V to +48 V.
@ t2 the Vi changes from +25 V to −25 V. The input is now series opposing with the +23 V across C1. This
leaves VO = −25 + 23 = −2 V. D1 is on. From t2 to t3, C1 charges from +23 V to +25 V; VO reaches to 0 volts.
@ t3 the Vi and VC1 (voltage of C1) are again series adding. → VO is again +50 V.
During t3 and t4, C1 discharges 2 V through R1. Then circuit operation from t3 to t4 is the same as it was
from t1 to t2.
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Lecture # 1 - Diodes
46
4. Some application circuits using diodes
4.3 Clamping Circuits
Negative Clamper
Initially when the input is given, the capacitor
is not yet charged and the diode is reverse
biased.
In this case, you must explain
how does the circuit works
by yourself at home.
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Lecture # 1 - Diodes
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4. Some application circuits using diodes
4.3 Clamping Circuits
❑ Positive Clamper
For the input signal is AC
sin-wave, you must explain
how does the circuit works
by yourself at home.
❑ Negative Clamper
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Lecture # 1 - Diodes
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4. Some application circuits using diodes
4.3 Clamping Circuits
❑ Positive Clamper with positive biasing Vb
The positive clamper can be biased with another
voltage source to further shift the input signal
waveform. The positive biasing further shifts up
the waveform by the amount of the biasing
voltage.
❑ Positive Clamper with negative biasing Vb
The negative biasing lower down the waveform
by the amount of the biasing voltage.
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Lecture # 1 - Diodes
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4. Some application circuits using diodes
4.3 Clamping Circuits
❑ Negative Clamper with positive biasing Vb
Positive Biasing
The positive biasing of the negative clamper adds a
positive or upward shift by the amount of biasing
voltage to the negative clamped waveform. It shifts the
waveform up to the positive level due to positive
basing.
❑ Negative Clamper with negative biasing Vb
The negative biasing of the negative clamper further
shifts downward the input signal waveform.
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5. Some other types of diodes
5.1 Photo diode
Photo Diode
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5. Some other types of diodes
5.2 Light-emitting Diode (LED)
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