Electronics (1)

EC238-Spring 2024
Dr. Rana Aly Onsy
Lecture – 4
PN Junction Diodes (Cont’d)
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Diode DC models (Cont’d)
Solved Example:
Using the constant voltage drop model (CVD) of the diode, and
taking 𝑉𝑉𝐷𝐷𝐷𝐷 = 0.7 𝑉𝑉, find 𝑉𝑉𝑜𝑜 and 𝐼𝐼𝑜𝑜 . Given that R = 1KΩ
Solution:
𝐼𝐼𝑜𝑜 𝐼𝐼𝑜𝑜
1. Assume all diodes are on.
2. Replace each diode with a battery (0.7 V each).
𝑉𝑉𝑜𝑜 = 3 ∗ 𝑉𝑉𝐷𝐷𝐷𝐷 = 3*0.7 = 2.1 V.

10−𝑉𝑉𝑜𝑜 10 −2.1
𝐼𝐼𝑜𝑜 = = = 7.9 mA
𝑅𝑅 1 𝐾𝐾

3. Check the assumption: since 𝐼𝐼𝑜𝑜 > 0 (positive),

then the assumption is correct.
Dr. Nehad Mansour’s lectures
Diode DC models (Cont’d)
Solved Example (cont’d):
Repeat the previous problem replacing the 10 volt DC supply with a 2 volt DC
supply.
Solution:
1. Assume all diodes are on.
2. Replace each diode with a battery (0.7 V each).
𝑉𝑉𝑜𝑜 = 3 ∗ 𝑉𝑉𝐷𝐷𝐷𝐷 = 3*0.7 = 2.1 V.

2−𝑉𝑉𝑜𝑜 2 −2.1
𝐼𝐼𝑜𝑜 = = = - 0.1 mA
𝑅𝑅 1 𝐾𝐾
3. Check the assumption: since 𝐼𝐼𝑜𝑜 < 0 (negative),
then the assumption is wrong. The Diodes should be off (open circuit).
𝑉𝑉𝑜𝑜 = 2 V (equal to the DC supply) and 𝐼𝐼𝑜𝑜 = 0 A (no current is flowing).
Solving Steps for DC diode Circuits
1. Determine the DC model of the diode
2. Assume the state of the diode whether it is ON or OFF
3. Draw the circuit once more replacing the diode with its model
according to the assumed state.
4. Solve the circuit to get the required parameters (current and
voltage).
5. Check the assumption made at 2.
6. If the assumption is verified, then your problem is solved.
7. If the assumption proved to be wrong, start all over again making
the opposite assumption.
Graphical Solution (Exact)
Find 𝐼𝐼𝐷𝐷 and 𝑉𝑉𝐷𝐷 Solving two equations together graphically:
First equation (KVL):
𝑉𝑉 −𝑣𝑣
𝑖𝑖𝐷𝐷 = 𝐷𝐷𝐷𝐷 𝐷𝐷
𝑅𝑅
Second equation (Exact diode current equation, Shockley equation):
𝑣𝑣𝐷𝐷
𝑖𝑖𝐷𝐷
𝑖𝑖𝐷𝐷 = 𝐼𝐼𝑠𝑠 𝑒𝑒 ƞ𝑉𝑉𝑇𝑇

DC Supply

𝑣𝑣𝐷𝐷
From Dr. Nehad Mansour’s lectures for the course NANENG 322
Diode AC (small signal) model
• A small signal model means that this model is valid only when the voltage variation (∆𝑉𝑉𝐷𝐷𝐷𝐷 )
around the DC voltage battery 𝑉𝑉𝐷𝐷𝐷𝐷 is small.
• The small voltage variation resulted in a small current
variation ∆𝐼𝐼𝐷𝐷 and a small voltage variation across the
diode ∆𝑉𝑉𝐷𝐷 .
• The voltage across the diode is the sum of the DC
voltage 𝑉𝑉𝐷𝐷 and the time varying signal ∆𝑉𝑉𝐷𝐷 .

Sedra & Smith text book

Diode AC (small signal) model
(cont’d)
• Neglecting R in calculations
𝑣𝑣𝑑𝑑 (𝑡𝑡)
𝑖𝑖𝐷𝐷 𝑣𝑣𝐷𝐷

𝐼𝐼𝐷𝐷 + 𝑖𝑖𝑑𝑑 (𝑡𝑡)

From Dr. Nehad Mansour’s slides

Solving Steps for AC diode Circuits
1. First deactivate all AC sources (AC voltage sources are considered to
be short circuit and AC current sources are considered open circuit).
2. Determine the DC current 𝐼𝐼𝐷𝐷 by dealing with the circuit as if it is a DC
diode circuit (as described in slide 4).
ƞ𝑉𝑉𝑇𝑇
3. Get the AC model resistance 𝑟𝑟𝑑𝑑 from 𝑟𝑟𝑑𝑑 = (proved in slide 7).
𝐼𝐼𝐷𝐷
4. Solve the original circuit after replacing the diode by it’s AC small
signal model resistance 𝑟𝑟𝑑𝑑 and deactivating the DC sources.
5. The required voltage or current is the addition of both DC and AC
contributions.
Solved Example
𝑉𝑉 + is the power supply of the circuit shown. It is a 1V peak

amplitude, 60Hz sinusoidal signal which is superimposed on a DC

voltage of 10V. Calculate both the DC voltage of the diode and

the amplitude of the sin-wave signal appearing across it.

Use the constant voltage drop model for the diode,

where 𝑉𝑉𝐷𝐷𝐷𝐷 = 0.7 𝑉𝑉. Given that R = 10KΩ, 𝑉𝑉𝑇𝑇 = 25mV and ƞ =1.
Sedra & Smith text book
Solution
1. DC analysis (the AC voltage source is shorted):
Assume the diode is ON
𝑉𝑉 + − 𝑣𝑣𝐷𝐷
𝐼𝐼𝐷𝐷 =
𝑅𝑅

𝑣𝑣𝐷𝐷 = 0.7V

10−0.7
𝐼𝐼𝐷𝐷 = = 0.93mA
10𝐾𝐾

A positive current means a correct assumption

Solution (Cont’d)
2. Calculating the diode small signal model 𝑟𝑟𝑑𝑑 :
ƞ𝑉𝑉
𝑟𝑟𝑑𝑑 = 𝑇𝑇
𝐼𝐼𝐷𝐷

1∗25𝑚𝑚
𝑟𝑟𝑑𝑑 = = 26.9Ω
0.93𝑚𝑚

3. Small signal analysis:

Deactivate the 10V DC voltage source and place only the AC
source 𝑉𝑉𝑠𝑠 . Replace the diode with its AC model as shown in
the figure.
Get 𝑣𝑣𝑑𝑑 using voltage division
𝑟𝑟𝑑𝑑
𝑣𝑣𝑑𝑑 = 𝑣𝑣𝑠𝑠 ∗
𝑅𝑅 + 𝑟𝑟𝑑𝑑
Solution (Cont’d)
26.9
𝑣𝑣𝑑𝑑 = 1 ∗
10𝐾𝐾 + 26.9

𝑣𝑣𝑑𝑑 = 2.68𝑚𝑚𝑚𝑚

The Final output voltage is the summation of both the

output voltages due to both the DC and AC input voltages:
𝑉𝑉𝑜𝑜 = 𝑉𝑉𝐷𝐷𝐷𝐷 + 𝑣𝑣𝑑𝑑
𝑉𝑉𝑜𝑜 = 0.7 + 2.68𝑚𝑚 ∗ sin(2 ∗ 𝜋𝜋 ∗ 60 ∗ 𝑡𝑡)
Diode Applications

• Rectifier Circuits
• Voltage doublers
• Level Shifters
• Limiting Circuits
Half Wave Rectifier

https://www.electronics-tutorials.ws/diode/diode_5.html
Half Wave Rectifier (Cont’d)
During positive half cycle:
Assuming ideal diode (short circuit when forward biased)
When 𝑽𝑽𝒊𝒊𝒊𝒊 ≥ 𝟎𝟎, the diode will be forward biased (the anode is more positive than the cathode).
The diode will be modelled as a short circuit and the current will flow to the output resistance 𝑅𝑅𝐿𝐿 .
𝑽𝑽𝒐𝒐𝒐𝒐𝒐𝒐 = 𝑽𝑽𝒊𝒊𝒊𝒊

https://how2electronics.com/half-wave-rectifier-basics-circuit-
working-applications/#google_vignette
Half Wave Rectifier (Cont’d)
During negative half cycle:
Assuming ideal diode (open circuit when reverse biased).
When 𝑽𝑽𝒊𝒊𝒊𝒊 < 𝟎𝟎, the diode will be reverse biased (the cathode is more positive than the anode).
The diode will be modelled as an open circuit and the current will not flow (I = 0).
𝑽𝑽𝒐𝒐𝒐𝒐𝒐𝒐 = 𝟎𝟎

https://how2electronics.com/half-wave-rectifier-basics-circuit-
working-applications/#google_vignette
Half Wave Rectifier (Cont’d)
Transfer Characteristics of the shown half wave rectifier:

The output voltage will be following

the input voltage when 𝑽𝑽𝒊𝒊𝒊𝒊 ≥ 𝟎𝟎 +
𝑽𝑽𝒊𝒊𝒊𝒊 𝑽𝑽𝒐𝒐𝒐𝒐𝒐𝒐
The output voltage will be 0
-
when 𝑽𝑽𝒊𝒊𝒊𝒊 < 𝟎𝟎

https://circuitdigest.com/electroni
c-circuits/half-wave-and-full-wave-
precision-rectifier-circuit-using-op-
amp
Half Wave Rectifier (Cont’d)
• A half wave rectifier has a conduction angle equals to 180° (half cycle).
• The frequency of the output is the same as the frequency of the input.
• The peak inverse voltage (PIV) that the diode must be able to withstand without breakdown
is determined by the largest reverse voltage that is expected to appear across the diode.
PIV ≥ 𝑉𝑉𝑚𝑚 , where 𝑉𝑉𝑚𝑚 is the maximum voltage across the diode at reverse biasing.
• The average voltage (DC value) of the output can be calculated by:

𝑉𝑉𝑚𝑚
𝑉𝑉𝐷𝐷𝐷𝐷 = , 𝑉𝑉𝑚𝑚 is the peak value of the output
𝜋𝜋
𝑉𝑉𝑚𝑚
𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 = , 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 is the root mean square value
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Solved Example
Evaluate the output 𝒗𝒗𝒐𝒐 and find out the DC magnitude
of the output for the circuit design shown in the figure.
Then plot the output waveform. Assume the diode to be
Ideal.
Solution:
For the positive half cycle (𝑉𝑉𝑖𝑖𝑖𝑖 ≥ 0):
The diode will be off (as the n-side is more positive than the
P-side). The diode is considered to be reverse biased, thus it
is modelled as an open circuit.
No current will pass and 𝒗𝒗𝒐𝒐 = 0
Solved Example(cont’d)
For the negative half cycle (𝑉𝑉𝑖𝑖𝑖𝑖 < 0):
The diode will be ON (as the p-side is more
positive than the n-side).
The diode is considered to be forward biased,
thus it is modelled as a short circuit.
Applying KVL (anticlockwise):
𝑉𝑉𝑖𝑖𝑖𝑖 - 𝑉𝑉𝑜𝑜 = 0
𝑉𝑉𝑜𝑜 = 𝑉𝑉𝑖𝑖𝑖𝑖
𝑉𝑉 −20 +
𝑉𝑉𝐷𝐷𝐷𝐷 = 𝑚𝑚 = = -6.366 volts 𝑉𝑉𝑖𝑖𝑖𝑖
𝜋𝜋 𝜋𝜋
The negative sign indicates -
that the polarity is opposite
to that of 𝑉𝑉𝑜𝑜
https://www.homemade-circuits.com/diode-rectification-half-
wave-full-wave-piv/
Half Wave Rectifier (Cont’d)
Half Wave Rectifier (Cont’d)
In case of CVD model, the transfer characteristics is as shown:
Half Wave Rectifier (Cont’d)
Thank You for your attention and see you next lecture

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