Electronics I – Chapter 3
Diode Circuits
Outline
• Diode implementation of logic gates
• Rectifiers (Half wave rectifier, Full wave rectifier)
3.1. Ideal Diode
3.1.2. Diode Implementation of OR Gate
To know whether the diodes are on or off:
1. Make a set of assumptions e.g. (D1 is on, D2 is off)
2. Replace the diode with the convenient model, apply circuit laws e.g. KVL, KCL
3. Check your set of assumptions by following the chart below
Assume
Diode on
a. Replace diode with short circuit (case of ideal model) or short
circuit+ voltage source (case of constant voltage model)
b. Find the value of ID
If ID is positive è correct assumption è move to next diode
If ID is negative è wrong assumption è new set of assumption
Diode off
a. Replace diode with open circuit
b. Find the value of VD
If VD negative è correct assumption èmove to next diode
If VD is positive è wrong assumption è new set of assumption
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3.1. Ideal Diode
3.1.2. Diode Implementation of OR Gate Example 3.6
For every input combination
(4 in total), there are 4
possible set of assumptions:
VA
VB
OR
Gate
Vout
0
0
0
0
3
3
1. D1 off, D2 off
2. D1 off, D2 on
3. D1 on, D2 Off
4. D1 on, D2 On
3
0
3
3
3
3
The circuit above shows an example of diode-implemented OR gate.
Example: VA= 0V and VB= 3V
Possible set of
assumptions:
1. D1 off, D2 off
2. D1 off, D2 on
3. D1 on, D2 Off
4. D1 on, D2 On
Set of assumptions 1:
(D1 off, D2 off), from slide 3:
a. Replace diode with open circuit
b. Find the value of VD1
VD1= 0 è correct assumption èmove
to D2
VD2= VB =3 V> 0 è wrong
assumption è new set of assumption
Set of assumptions 4:
(D1 on, D2 on), from slide 3:
a. Replace diode with short circuit
b. Before calculating ID, we notice that
Vout= 0 v and 3 V simultaneously
è conflict è wrong assumption è
new set of assumption
5
Possible set of assumptions:
D1 off, D2 off
D1 off, D2 on
D1 on, D2 Off (VA=0,
VB=3V)
D1 on, D2 On
+ VD1
-
-
Set of assumption 3:
(D1 on, D2 off)
D1 onà replace D1 with closed switch à goal is to check ID1 (ID1
should be positive)
D2 Off à replace D2 with open switch (ideal diode) à check VD2
(for the assumption to be correct VD2 should be negative)
Blue loop: KVL
-VA+ VD1+ Vout= 0 VD1= Vwire= Vshortcircuit= 0
-0 + 0+ Vout =0
Vout= VA=0V and ID1= IR = Vout/RL= 0 ≥ 0
now check D2 off
ID1
+ VD2
VA=0V and VB=3V
IR
Orange loop to check VD: KVL
-VB+ VD2+ Vout= 0
-3+ VD2+ 0 =0
VD2=3V> 0 (positive)
Thus, it is a wrong assumption
(D1 on, D2 off) is a wrong assumption
6
Possible assumptions:
D1 off, D2 off
D1 off, D2 on (VA=0, VB=3V)
D1 on, D2 Off
D1 on, D2 On
+ VD1
-
ID2
IR
+ VD2
-
VA=0V and VB=3V
Set of assumption 2:
(D1 off, D2 on)
D1 offà replace D1 with open switch à goal is to check VD1 (for the
assumption to be correct VD1 should be negative)
D2 On à replace D2 with closed switch (ideal diode) à check ID2
(ID2 should be positive)
Yellow loop: KVL
-VB+ VD2+ Vout= 0 VD2= Vwire= Vshortcircuit= 0
-3 + 0+ Vout =0
Vout= VB=3V> 0 and ID2= IR = Vout/RL= 3/1k= 3 mA ≥ 0 (positive)
now check D1 off
Green loop to check VD: KVL
-VA+ VD1+ Vout= 0
0+ VD1+ 3 =0
VD1=-3V< 0 (negative)
Thus, it is true that D1 is off
(D1 off, D2 on) is the correct assumption
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1.
Solve the circuit using constant voltage model.
5V
1 kΩ
2. Find V1, V2 in the circuit using constant voltage model.
4 kΩ
10 V
Si
v1
v2
3 kΩ
-5 V
3. Determine the value of I and V in the circuit.
a)Using Ideal model
b)Using constant voltage model
4. Find the voltage VA in the circuit. Use constant voltage model.
3.5. Applications of Diode
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3.5.1. Half-Wave Rectifier
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3.5.1. Half-Wave Rectifier
14
• A very common application of diodes is half-wave rectification, where either the positive or negative half of the
input is blocked. But, how do we generate a constant output?
3.5.1. Half-Wave Rectifier
Diode-Capacitor Circuit: Constant Voltage Model
15
• If the resistor in half-wave rectifier is replaced by a capacitor, a fixed voltage output is obtained
since the capacitor (assumed ideal) has no path to discharge.
3.5.1. Half-Wave Rectifier
Diode-Capacitor With Load Resistor
16
VR
Ø A path is available for capacitor to discharge. Therefore, Vout will not be constant and a ripple exists.
Ø This is valid when the time constant RC >>>T.
3.5.1. Half-Wave Rectifier
Behavior for Different Capacitor Values
17
• For large RLC1, Vout has small ripple.
3.5.1. Half-Wave Rectifier
Peak to Peak amplitude of Ripple
t3- t1 ≈ T = 1/fin
𝐼"#
𝑉! =
𝑐𝑓
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂 =
𝐷𝐶 𝑜𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟
≈ 40.6%
𝐴𝐶 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟
• The ripple amplitude is the decaying part of the exponential.
• Ripple voltage becomes a problem if it goes above 5 to 10% of the output voltage.
18
VR
3.5.1. Half-Wave Rectifier
Maximum Diode Current
19
Ip
C1
Vp
in
2VR
Vp
Vp
RL
Vp
RL
( RL C1
in
2VR
Vp
1)
Ø The diode has its maximum current at t1, since that’s when the slope of Vout is the greatest.
Ø This current has to be carefully controlled so it does not damage the device.
Ø PIV= 2Vp- VD,on; PIV= Peak Inverse Voltage.
3.5.1. Full-Wave Rectifier
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3.5. Applications of Diodes (2/6)
Problem 3.37
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