Ideal diode : a two terminal device having the circuit symbol shown,
 Very simplified model of real diode
 Useful for fast approximate analysis
 Helpful introduction to diode circuit analysis
+  -
Anode Cathode i
 Polarity of voltage
 Direction of current
I i
Reverse
Bias
Forward
Bias
Reverse bias
Forward bias
0
V
Real diode characteristics drawn to scale: Ideal diode characteristics
The characteristics are very close to each other, the ideal diode can represent a very good approximation to the actual diode


Current-voltage characteristics ( i  )
Two regions
1.
Reverse Bias:  < 0 , i = 0
Anode
2.
Forward Bias:  = 0 , i > 0
+  i
The positive terminal is "Anode"
The negative terminal is "Cathode"
The direction of current flow explains the choice of the arrow like circuit symbol. It shows the normal direction of current flow.
 I-V characteristics is highly nonlinear :
Reverse
Bias
 Two straight-line segments → piecewise linear
Cathode
0 i
Forward
Bias


1- V < 0 → Reverse bias region
No current flows ( I = 0 )
Diode like open circuit
Diode is cut-off , off
Equivalent circuit
2- I > 0 → Forward bias region
No voltage drop across diode( V = 0 )
Diode like short circuit
Diode is turn-on , on
I I
+ V -
V < 0 I = 0
Reverse Biased
Supports any voltage with no current flow
I > 0
+ V -
V = 0
Forward Biased
Passes any current with zero voltage drop
In both cases external circuit limits: The forward current when "ON"
The reverse voltage when "OFF"

Diode is "ON" it is short circuit
The current I = 10 mA
The voltage drop V = 0V
Diode is "cut-off" it is open circuit "OFF"
The current I = 0 mA
The voltage drop V = -10 V
Note the polarity of the voltage drop across the diode.
10 V
10 mA
1 k 
+
0V
-
10 V
0 mA
1 k 
-
-10V
+
10 V
10 mA
1 k 
+
0V
-
10 V
0 mA
1 k 
-
-10V
+

Rectifier is a fundamental application of the diode
The circuit shows a half-wave rectifier
.
 v
I
≡ sinusoidal as shown; it alternates between positive and negative values; average = 0 .
 v
I
> 0 diode is forward (short circuit ) v
O
= v
I
;
 v
I
< 0 diode is reverse (open circuit) v
O
= 0 .
 v
O
has the waveform shown, it is unidirectional
 v
O
has a finite average or dc component.
 The circuit is said to have rectified the signal, and hence is called rectifier.
+
v
I
+ v
D
i
D
R
+ v o
-
+ v
D
=0
i
D
R v
I
 0
+ v
D -
+
v
I
 0 i
D
= 0
R
+ v o
= v
I
-
+ v o
= 0
-

Example 3.1
The circuit is for charging a (E = 12V) battery. Input V
S
is sinusoid with 24 V peak value.
 For v s
>12V the diode is on "short circuit" → current flows: 𝐼
𝐷
=
𝑉
𝑆
− 𝐸
𝑅
 The peak value is: 𝐼
𝐷
=
24− 12
100
= 0.12𝐴
 The maximum reverse voltage across diode: when V
S
= -24 V
⇒ V
Reverse
= 24 + 12 = 36 V
 conducting during an angle = 2 
 where  is given by:
24
* cos  = 12   = 60°  2  = 120° .
+ v s
-
100 
R i
D
E
12 V
+
-

Diodes with the resistors can be used to implement logic functions:
Consider positive logic:
V ~ 0V  logic 0
V ~ 5V  logic 1
Circuit 1: has three inputs A , B , C , any diode connected to +5V will conduct
 clamping the output Y to a value +5V,  Y = high (1) if any input is high  Y is high: logic " OR " function:
Y = A + B + C (logic “ OR ”)
Circuit 2: Y = Low (0) if one or more inputs are low (0), this is the logic " AND " function.
Y = A∙B∙C (logic “ AND ”)
A
B
C
Y
R
A
B
C
Circuit 1
5 V
R
Y
Circuit 2

Example 3.2
D
1
and D
2
are "ON" or "OFF": 4 possibilities
 Assume a state for the diodes and analyse the circuit
 Check whether results are consistent with assumption:
 Diode assumed "ON" check the direction of current
 Diode assumed "OFF" check the polarity of voltage drop
For this circuit assume D
1
and D
2
are "ON" short circuit →
V
B
= 0V and V = 0V ⇒ 𝐼
𝐷2
=
10−0
10
= 1 𝑚𝐴 and 𝐼
5𝑘Ω
=
0−(−10)
5 at node B: 𝐼 + 𝐼
𝐷2
= 𝐼
5𝑘Ω
⇒ 𝐼 = 1𝑚𝐴
Results: I = 1mA and V = 0V
and D
1
is "ON" as assumed
= 2 𝑚𝐴

assume D
1
and D
2
are "ON" short circuit as previously →
V
B
= 0V and V = 0V
⇒ 𝐼
𝐷2
=
10−0
5
= 2 𝑚𝐴 at node B: 𝐼 + 𝐼
𝐷2
= 𝐼
10𝑘Ω and
⇒
𝐼
10𝑘Ω
𝐼 = −1𝑚𝐴
=
;
0−(−10)
10
= 1 𝑚𝐴 assumption for D
1
is not correct
Start again with assumption D
1
⇒ 𝐼
𝐷2
=
10−(−10)
15
= 1.33 𝑚𝐴
"OFF" and D and at node B the voltage is: 𝑉
𝐵
2
"ON" → two resistor in series:
= −10 + 10 × 1.33 = +3.3 𝑉
D
1
is indeed reverse biased as assumed:
Results: I=0 and V=3.3 V