# Diode sketch ```Diodes Waveform shaping Circuits
Lecture notes: page 2-20 to 2-31
Sedra & Smith (6th Ed): Sec. 4.5 & 4.6
Sedra & Smith (5th Ed): Sec. 3.5 & 3.6
Two-port networks as building blocks
 Recall: Transfer function of a
two-port network can be found
by solving this circuit once.
 Concept of input resistance can be used to find vi/vsig (will be
discussed in transistor amplifier section)!
 We focus on finding transfer function, vo vs vi (circuit below)
o “Open-loop” Transfer function (RL → ∞ or io = 0)
Rectifier Circuit
KCL :
io = iD
KVL :
vi = vD + vo → vo = vi − vD
Ω Law : iD = vo / RL
Diode OFF : iD = 0 and vD < VD 0
vo = RL iD = 0
vD < VD 0
→ vi − vo = vi < VD 0
Diode ON :
vD = VD 0
and
iD ≥ 0
vo = vi − vD = vi − VD 0
iD = vo / RL ≥ 0
→ vo = vi − vD ≥ 0 → vi ≥ VD 0
For vi ≥ VD 0 , Diode ON
and
For vi < VD 0 , Diode OFF and
vo = vi − VD 0
vo = 0
Rectifier Circuit: vo is the positive portion vi
For vi ≥ VD 0 , Diode ON
and
For vi < VD 0 , Diode OFF and
vo = vi − VD 0
vo = 0
Application of Rectifier Circuit: AC to DC
convertor for power supply
Half-wave rectifier (only converts half of AC input to DC value)
Full-wave rectifier (converts all of AC input to DC value)
Each pair of diodes conduct only for half
of the cycle
Clipper or Limiter Circuit
(open-loop transfer function)
vD = VD 0
Diode OFF : iD = 0 and vD < VD 0
Diode ON :
vi = R × 0 + vo → vo = vi
vo = VD 0
vD < VD 0
iD = (vi − VD 0 ) / R ≥ 0
→ vi < VD 0
For vi ≥ VD 0 , Diode ON
and
For vi < VD 0 , Diode OFF and
vo = VD 0
vo = vi
and
iD ≥ 0
→ vi ≥ VD 0
Clipper Circuit
does not allow vo > VD0 to go through
For vi ≥ VD 0 , Diode ON
and
For vi < VD 0 , Diode OFF and
vo = VD 0
vo = vi
 Impact of RL is discussed as an exercise problem
Rectifier & clipper circuits are the same
but vo is taken at different locations
Half-wave
Rectifier
Clipper
Clipper circuit limits vo when the diode is ON
vo limited to ≤ VD0 + VDC
vo limited to ≤ VD0 + VZ
Bottom portion of signal can also be clipped
vo limited to ≥ − VD0 − VDC
vo limited ≥ − VD0 −VZ
Both top & bottom portions of the signal
can be clipped simultaneously
vo limited to ≤ VD0 + VDC1 and ≥ − VD0 − VDC2
vo limited to ≤ VD0 + VZ1
and ≥ − VD0 − VZ2
“Ideal” Peak Detector Circuit
 Because vc cannot change suddenly, the
state of diode will depend not only on vi
but also on the “history” of the circuit
(e.g., dvi/dt , vc at certain times,)
Diode OFF : iD = 0 and vD < VD 0
 Capacitor does not charge or discharge!
 vc (t) = vc0 where vc0 is the capacitor voltage at
the moment diode turned OFF!
vo = vc 0 = const.
vD = vi − vc < VD 0 → vi < vc 0 + VD 0
“Ideal” Peak Detector Circuit
(open-loop transfer function)
Diode ON :
vD = VD 0
and
iD ≥ 0
vo = vc = vi − VD 0
dvc
d (vi − VD 0 )
dv
=C
=C i
dt
dt
dt
dvi
≥0
iD = ic ≥ 0 →
dt
iD = ic = C
For dvi /dt ≥ 0 & vi = vc + VD 0 : Diode ON , vo = vc = vi − VD 0
For vi < vc + VD 0 :
Diode OFF, vo = vc 0 = const
 Because state of diode depends on vc , we cannot produce
a universal plot vo vs vi
Response of the “Ideal” Peak Detector (1)
For dvi /dt ≥ 0 & vi = vc + VD 0 : Diode ON , vo = vc = vi − VD 0
For vi < vc + VD 0 :
 Start at t = 0 with vc= 0
 For t > 0, dvi/dt > 0.
 For vi < vc0 + VD0 = VD0 ,
diode remains OFF.
o vo = vc0 = 0
Diode OFF, vo = vc 0 = const
 When vi = vc0 + VD0 = VD0 , diode
turns ON (since dvi/dt > 0)
 Capacitor starts to charge and vc
tracks vi
o vo = vc = vi - VD0
Response of the “Ideal” Peak Detector (2)
For dvi /dt ≥ 0 & vi = vc + VD 0 : Diode ON , vo = vc = vi − VD 0
For vi < vc 0 + VD 0 :
 Cap continue to charge until
vi = V + (vc = V + - VD0 )
 Afterward vi starts to
decrease (dvi/dt < 0) and
diode turns OFF.
o vo = vc0 = V + − VD0
Diode OFF, vo = vc 0 = const
 Even when vi starts to increase (dvi/dt > 0)
diode remains OFF as vo < vc0 + VD0
o vc0 + VD0 = V + − VD0 +VD0 = V + !
 Diode turns ON vi = V + and immediately
turns OFF vi starts to decrease (dvi/dt < 0)
Response of the “Ideal” Peak Detector (3)
 vo is the “peak” value of input waveform (V + – VD0 ): “Peak Detector”
o Note vo did not “drop” after the peak was decreased in the 3rd cycle.
Exercise: Show that if the diode direction is reversed, circuit detects the
“negative” peak value, −V − (i.e., lowest voltage of the wave form which
should be negative)
Practical Peak Detector Circuit (1)
 A resistor is added in parallel
to the capacitor! (It can be
Diode OFF : iD = 0 and vD < VD 0
 Capacitor discharges into the resistor
with a time constant of τ = RC
vo = vc (t ) = vc 0 exp[ - (t − t0 )/τ ]
vD = vi − vc < VD 0 → vi < vc (t ) + VD 0
Practical Peak Detector Circuit (2)
Diode ON :
vD = VD 0
and
iD ≥ 0
vo = vc = vi − VD 0
dvc
d (vi − VD 0 )
dv
=C
=C i
dt
dt
dt
dvi
≥0
iD = ic ≥ 0 →
dt
iD = ic = C
For dvi /dt ≥ 0 &, vi = vc + VD 0 : Diode ON , vo = vc = vi − VD 0
For vi < vc + VD 0 :
Diode OFF, vo = vc (t ) = vc 0 exp[ - (t − t0 )/τ ]
Response of the Practical Peak Detector (1)
For dvi /dt ≥ 0 &, vi = vc + VD 0 : Diode ON , vo = vc = vi − VD 0
For vi < vc + VD 0 :
 Start at t = 0 with vc= 0
 For t > 0, dvi/dt > 0.
 For vi < vc0 + VD0 = VD0 ,
diode remains OFF.
o vo = vc0 = 0
Diode OFF, vo = vc (t ) = vc 0 exp[ - (t − t0 )/τ ]
 When vi = vc0 + VD0 = VD0 , diode
turns ON (since dvi/dt > 0)
 Capacitor starts to charge and vc
tracks vi
o vo = vc = vi - VD0
Response of the Practical Peak Detector (2)
For dvi /dt ≥ 0 &, vi = vc + VD 0 : Diode ON , vo = vc = vi − VD 0
For vi < vc + VD 0 :
 Cap continue to charge until
vi = V + (vc = V + - VD0 )
 Afterward vi starts to decrease
(dvi/dt < 0) and diode turns
OFF. Capacitor discharges:
vo = vc (t ) = vc 0 exp[ - (t − t0 )/τ ]
Diode OFF, vo = vc (t ) = vc 0 exp[ - (t − t0 )/τ ]
 Even when vi starts to increase (dvi/dt > 0)
diode remains OFF as long as vo < vc + VD0
 Diode turns ON when vi = vc + VD0 and
charges capacitor until vi = V + is reached)
Response of the Practical Peak Detector (3)
 Shape of output signal depends on the ratio of τ/T
 “ideal” peak detector: τ/T → ∞
 “Good” peak detector: τ/T >> 1
 As τ/T decreases, the circuit departs from a peak detector.
 For τ/T << 1, capacitor discharges very fast and circuit resembles a rectifier
circuit
Decreasing τ/T
Peak detector is used in AM receivers
Carrier wave amplitude is
modulated with the sound data
(sound signal is the “envelop” of
the carrier wave)
Tcarrier << τ = RC << Tsound
 A clipper circuit with a load RL is similar to the open-loop clipper with R → R || RL
Examples of Design Choices:
 As a peak detector (want τ/T → ∞) R is NOT needed and we should set
C RL to be large (>>T).
o Peak detector circuit is used to “smooth” out the output voltage of a
rectifier for the power supply circuit (Need a large C!).
 For applications such as AM receiver when the peak detector is used as
separate the signal from a carrier, R and C should be chosen such that
Tcarrier << τ = RC << Tsound
and
R << RL
Clamp Circuit
“Ideal” peak detector:
vo = vc = V + − VD0
Clamp circuit: vo = vD
vc = V + − VD 0
vo = vD = vi − vc = vi − (V + − VD 0 )
vo is equal to vi but shifted
“downward” by − (V + − VD0)
 If amplitude of vi (V + ) changes, the shift would
changes and vo becomes distorted!
 Capacitor charges when
the diode is ON:
 Capacitor charges when the diode
is ON:
vc = V + − VD0
 Capacitor remains charged
when diode is OFF.
vc = V + − VD0
 Capacitor discharges into RL
when diode is OFF.
 As long as τ = RLC >> T
capacitor discharges little and
clamp circuits works fine!
Voltage shift in a clamp circuit can be adjusted!
vA = vi − VDC
V + : peak of vi
V +A : peak of vA
V +A = V + − VDC
Peak detector circuit:
vc = V +A − VD0
vc = V + − VDC − VD0
vc = V + − VDC − VD 0
vo = vi − vc = vi − (V + − VDC − VD 0 )
 vo is equal to vi but shifted
“downward” by − (V + − VDC − VD0)
vo = vi − (V + − VZ − VD 0 )
Clamp circuit can also introduce a “positive” shift
Peak detector (diode is reversed):
vo = vc = − (V − − VD0)
Clamp circuit (diode reversed):
v o = vD
vc = − (V − − VD 0 )
vo = vD = vi − vc = vi + (V − − VD 0 )
vo is equal to vi but shifted
“upward” by (V − − VD0)
The positive shift can also be adjusted.
vo = vi + (V − − VDC − VD 0 )
vo = vi + (V − − VZ − VD 0 )
How to find response of clipper or clamp circuits:
 Assume diode is ON and calculate vc .
o If vc = +vi …, replace vi with V+ (peak positive value)
o If vc = −vi …, replace vi with −V− (peak negative value)
 If clipper, vo = vc . If Clamp, use KVL to find vo (e.g., , vo = vi − vc )
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