RecallLecture 3. Current generated due to two main factors Drift – movement of carriers due to the existence of electric field Diffusion – movement of carriers due to gradient in concentrations. RecallLecture 3. Introduction of PN junction Space charge region/depletion region
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Vbi
i
vD
The I V characteristics of the ideal diode.
Reverse bias
Conducting state
V = 0V
Reverse biased
open circuit
Conducting state
short circuit
i
vD
Represented as a battery of voltage = V
=
V
Reverse bias
Conducting state
Reverse biased
open circuit
V
VD =V for diode to turn on.
Hence during conducting state:
vD
Represented as a battery of voltage = V and forward resistance, rfin series
+ VD 
=
rf
V
CIRCUIT REPRESENTATION OF DIODE – Piecewise Linear Model
Reverse bias
Conducting state
Reverse biased
open circuit
V
VD ≥V for diode to turn on.
Hence during conducting state:
Diode Circuits: DC Analysis and Models
Example
Consider a circuit with a dc voltage VPSapplied across a resistor and a diode.
Applying KVL, we can write,
or,
The diode voltage VD and current ID are related by the ideal diode equation: (IS is assumed to be known for a particular diode)
Equation contains only one unknown, VD:
Diode Circuits: Direct Approach
Question
Determine the diode voltage and current for the circuit.
Consider IS = 1013 A.
and
ITERATION METHOD
Example
Determine the diode voltage and current using a piecewise linear model.
Assume piecewise linear diode parameters of Vf = 0.6 V and rf = 10 Ω.
Solution:
The diode current is determined by:
I
+
VO

I = 0.2325mA
Vo = 0.35V
Find I and VOfor the circuit shown below if the diode cut in voltage is V = 0.7V
I
+
VO

I = 0.372mA
Vo = 0.14V
Determine ID if V = 0.7V
R = 4k
b)If VPS = 8V, what must be the value of R to get ID equal to part (a)
The total input voltage vI = dcVPS + acvi
iD= IDQ + id
vD = VDQ+ vd
IDQ and VDQ are the DC diode current
and voltage respectively.
If vd << VT, the equation can be expanded into linear series as:
The DC diode current Is:
Diode Circuits: AC Equivalent Circuit
The relation between the diode current and voltage can be written as:
VDQ = dc quiescent voltage
vd = ac component
The 1 term in the equation is neglected.
The equation can be written as:
iD = ID [ 1 + vd/VT]
iD = ID + ID vd / VT = ID + id
where id =ID vd / VT
using Ohm’s law:
I = V/R hence, id = vd / rd compare with id =ID vd / VT
which reveals that rd = VT / ID
CONCLUSION: During AC analysis the diode is equivalent to a resistor, rd
VDQ = V
IDQ
Example 1
Analyze the circuit (by determining VO & vo ).
Assume circuit and diode parameters of
VPS = 5 V, R = 5 kΩ, Vγ = 0.6 V & vi = 0.1 sin ωt
rd
id
VPS = 10V, R = 20k, V = 0.7V, and vi = 0.2 sin t. Determine the quiescent current, IDQ and the time varying current.