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Every basic thing about clippers and clampers.
Typology: Assignments
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V (^) in
If, V (^) i > V (^) d, diode ON, V 0 =V (^) i - V (^) d
V (^) i R V^0
V (^) d
Where, V (^) d=0 V (for ideal diode) V (^) d=0.7 V (practical diode or silicon diode)
Red line represent Practical diode and Blue line for Ideal diode
Input and output waveform
Where, V (^) d=0 V (for ideal diode) V (^) d=0.7 V (practical diode or silicon diode)
Parallel configuration
i
The parallel variety has the diode in a branch parallel to the load.
If ,V (^) i > V (^) d, positive half cycle diode ON, V 0 =V (^) d
i
Vd
V (^) i^ V^0 V (^) R
i >V^ R +V^ d
R
d
The output voltage V 0 =V (^) i -V (^) R -V (^) d
V (^) m
V (^) R+ V (^) d
Diode on
Diode off
V (^) i
T/2 (^) T
V (^) m –V (^) R -V (^) d
V (^) o
T/2 T
V (^) i =V (^) R diode change the state
Fig. 1 Solution: For ideal diode V (^) d = Step 1: The output is again directly across the resistor R. Step 2: The positive region of Vi and the dc reference voltage (V (^) R ) are both applying “pressure” to turn the diode on. The result is that we can safely assume the diode is in the “on” state for the entire range of positive voltages for Vi. Once the supply goes negative, it would have to exceed the reference voltage of 5 V before it could turn the diode off.
Fig. 1
Practice problem: Find the output voltage for the network examined in above example if the applied signal is the square wave as,
Fig. 3
Ans.
Solution:
- Step 1: In this example the output is defined across the series combination of the 4-V supply and the diode, not across the resistor R.
Practice problem: Repeat above using a silicon diode with V (^) d = 0.7 V.
Ans