Transcript Recall-Lecture 7 - International Islamic University Malaysia

Recall Lecture 8
• Standard Clipper Circuit
– Step 1: Find the clip value by doing KVL at the output
branch
– Step 2: Set the conditions to know whether diode is on or
off – sketch your output waveform
• Clipper in series
– clips at zero. It is similar to half wave where the diode only
turns on during one of the cycle.
Clamper
Clampers
● Clamping shifts the entire signal voltage
by a DC level.

Consider, the sinusoidal input voltage
signal, vI.

1st 900, the capacitor is charged up to
the peak value of VI which is VM.

Then, as VI moves towards the –ve
cycle,


the diode is reverse biased.

Ideally, capacitor cannot discharge,
hence Vc = VM
By KVL, we get
NOTE: The input signal is shifted by a dc
level; and that the peak-to-peak value is
the same
Clampers
●
A clamping circuit that includes an independent voltage source VB.
Peak value VM

STEP 1: Knowing what value that the capacitor is charged to. And from
the polarity of the diode, we know that it is charged during positive
cycle. Using KVL,


VC + VB – VS = 0  VC = VM – VB
STEP 2: When the diode is reversed biased and VC is already a constant
value

VO – VS + VC = 0  VO = VS – VC.
EXAMPLE – clampers with ideal diode
For the circuit shown in figure below, sketch the
waveforms of the output voltage, vout. The input voltage
is a sine wave where vin = 20 sin t. Assume ideal diodes.
Vin
What if the diode is non-ideal?
Vi
C
+
+
Vi
Vo
-
5V
The diode is a non-ideal
with V = 0.7V
-
10
t
-4.3
-10
-14.3
-24.3
Step 1: VC + V - VB – Vi = 0  VC = 10 + 5 – 0.7 = 14.3V
Step 2: VO – Vi + VC = 0  VO = Vi – 14.3.
Multiple Diode Circuits
Final Exam SEM I 2013/2014
DIODE
ID
VD
OFF
0
VD < V
ON
ID > 0
VD = V
REMEMBER THAT:
A pn junction diode will conduct when the p-type material is more
positive than the n-type material
OR GATE
Vo = voltage
across R
V1
V2
VO
D1 and D2 off; no current flow,
0
0
0
D1 off, D2 on, current flow,
Vo – V2 + V = 0
0
5V ( 1 )
4.3V
D1 on, D2 off, current flow,
Vo – V1 + V = 0
5V ( 1 )
0
4.3V
Both on, using both loops will
give the same equation
5V ( 1 )
5V ( 1 )
4.3V
AND GATE
Vo = node
voltage
V1
V2
VO
Both on, using both loops will
give the same equation
0
0
0.7
D1 on, D2 off
0
5V ( 1 )
0.7
D1 off, D2 on
5V ( 1 )
0
0.7V
Both are off; open circuit no
current flowing through R since
no GND destination
5V ( 1 )
5V ( 1 )
5V