Transcript Multi-functional Packaged Antennas for Next

Ideal Diode Model
Analysis of Ideal Diode Circuits:
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12 out of 33 students got higher marks in Quiz 2 than HW2
Grade in Quiz2 - Grade in Hw2
100
75
50
25
0
-25
-50
-75
-100
BDR
FM
HD
MR
PM
Identity of Students
SA
Example: Circuit Solution by assumed diode states
Solution:
1. Assume that the D1 is on, and D2 is off, and determine the states of the diodes
2. Assume D2 is on and D1 is off, and determine the states of the diodes
3. Find out which assumption is correct, and proceed to solve the circuit from there
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More examples
Find V and I as marked.
(a) D1 ON, D2 OFF. V = 10 volts, I = 0
(b) D1 ON, D2 OFF. V = 6 volts, I = 6 mA
(c) D1 ON, D2 ON. V = 30 volts, I = 33.6 mA
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Half wave rectifiers
The current is conducted by the diode when it has a positive voltage across it, i.e. it is forward biased.
Only then the voltage appears across the load.
In reverse bias, the diode blocks the voltage, i.e. acts as an open circuit. Thus the voltage across the
load becomes zero.
For ideal diodes, the voltage drop in forward bias is zero. For real diodes the voltage drop is usually
few tenths of a volt. The drop for a Si diode is usually assumed to be 0.7 V in forward bias.
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Half-wave rectifier with smoothing capacitor
Figure 3.2 Half-wave rectifier with smoothing capacitor
Determining the value of
the smoothing capacitor
(3.2)
(3.3)
(3.4)
IL is the load current, and T is the
time period of the ac voltage. Vr
is the peak-to-peak ripple voltage
and C is the capacitance
The average voltage supplied to the
load is approximately midway between
the maximum and minimum voltages,
making VL as
(3.5)
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Full wave rectifier circuits
1.
In a full wave rectifier circuit, usually
a transformer is used to isolate the
load from the ac power line, and
provide out of phase input voltages.
2.
It also allows the designer to
choose the input to the rectifier
circuit from the turns ratio.
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Diode-bridge full-wave rectifier
Note that the capacitor is half
compared to a half-wave rectifier
circuit, as the capacitor needs to
discharge for only half the time.
The output voltage can be further “smoothed out” by using a regulator circuit.
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Diode circuit problem
Ans: Peak voltage is 10 V, turns ratio = 15.6, C = 833 uF
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