Slide điện tử từ trường lecture 3 diode circuits
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Lecture 03
Diode circuits
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topics
• Rectifier circuit
• Limiting and clamping circuits
• Small signal model
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Diode logic gates (ideal diode)
Y = A+ B +C
Y = ABC
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Example (ideal diode)
Assume D1 and D2 are ideal diodes and conducting
10 − 0
I D2 =
= 1mA
10k
0 = 5k × ( I + I D 2 ) − 10v
⇒ I = 1mA
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Example (ideal diode)
Assuming that D1 and D2 are ideal diodes and conducting
10 − 0
I D2 =
= 2mA
5k
0 = 10k × ( I + I D 2 ) − 10v
⇒ I = −1mA
Contradictory result
Assuming that D1 is off and D2 is on
10 − (−10)
I D2 =
= 1.33mA
15k
VB = 10k × I D 2 − 10v = 3.3v
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ACỈDC Rectifier
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Rectifier parameters:
• Crest factor (C.F.)
• Form factor (F.F.)
• Ripple factor (R.F.)
Vmax
C .F . =
Vrms
Vrms
F .F . =
Vaverage
R.F . =
Vripple − rms
1 T
Vaverage = ∫ Vo (t )dt
T 0
1 T 2
Vrms =
V o (t )dt
∫
T 0
Root-mean-square
Vaverage
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Vaverage
Vrms
∫
π
∫
π
0
0
1
=
T
∫
T
0
v(t )dt =
1
π∫
π
0
1 T 2
1
v (t )dt =
=
∫
0
T
π
Vmax sin θdθ =
π
2Vmax v(t )
π
Vmax
Vmax
∫0 V max sin θdθ = 2
2
2
π
π
sin θdθ = − cos θ 0 = 2
sin θdθ = ∫
2
π
0
π
1 − cos 2θ
dθ =
2
2
Full-wave rectifier
Vaverage =
Vrms
Half-wave rectifier
2Vmax
Vaverage =
π
Vmax
=
2
Vrms
Vmax
π
Vmax
=
2
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Using ideal model
Half-wave rectifier
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Using piecewise-linear model
Transfer curve
R
R
vo =
vs −
VD 0
R + rD
R + rD
vs
PIV (peak inverse voltage) =
vs
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Using piecewise-linear model (center-tapped Rectifier)
Transfer curve
PIV (peak inverse voltage) =
2vs − VD
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Bridge rectifier
PIV (peak inverse voltage) =
vs − 2VD + VD = vs − VD
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Half-wave
Centertapped
bridge
vi
vmax sin ωt
vmax sin ωt
vmax sin ωt
v0
Vmax − VD
Vmax − VD
Vmax − 2VD
Vaverage
Vmax −VD
2 (Vmax −VD )
2 (Vmax −VD )
Vrms
Vmax −VD
2
Vmax −VD
Vmax −VD
PIV
Vmax
F .F .
R.F .
π
π
π
2
2Vmax − VD
2
Vmax − VD
π
π
π
2
2 2
2 2
( 2 ) −1
π 2
( 2π 2 ) 2 − 1
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( 2π 2 ) 2 − 1
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Filter
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iD = iC + iL = C
dvo vo
+
dt R
dvo vo
+ =0
dt R
dvo (t )
vo (t )
=−
, vo (0) = V p
dt
RC
C
vo (t ) = ke
t
− RC
vo (t ) = V p e
, k = Vp
t
− RC
RC >> T
V p − Vr = V p e
−
−
T
RC
Average diode current iD max
T
RC
T
≈ 1−
RC
Vp
T
Vr ≈ V p
=
RC fRC
e
iD max = I L (1 + 2π
2V p
Half wave
Average diode current iDav
iDav = I L (1 + π
2V p
Vr
)
next page
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Vr
/>
)
Diode on
V p − Vr = V p e
−
T
2 RC
−
T
2 RC
T
2 RC
full wave
Vp
T
Vr ≈ V p
=
RC 2 fRC
Δt
e
≈ 1−
Average diode current iDav
V p cos(ωΔt ) = V p − Vr
iDav =
1
⇒ V p [1 − (ωΔt ) 2 ] = V p − Vr
2
Vp
2Vr
(ωΔt ) 2 = Vr ⇒ ωΔt =
2
Vp
R
(1 + π
Vp
2Vr
)
Maximum diode current iD max
Qc = ic Δt = (iDav − I L )Δt
⇒ iDav = I L +
Vp
Qc
CVr
= IL +
Δt
Δt
iD max =
Vp
R
(1 + 2π
Vp
2Vr
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Limiter circuits
vs − 12
iD =
100
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Basic limiting circuits
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vo
example
zone A
D1
zone B
zone C
D2
− 5v
+ 5v
vi
zone A : D2 off, D1 on
10k
10k
(−5v)
vi +
10k + 10k
10k + 10k
1
vo = vi − 2.5v
2
vo =
zone B : D2 off, D1 off
vo = vi
zone C : D2 on, D1 off
10k
10k
(5v)
vi +
10k + 10k
10k + 10k
1
vo = vi + 2.5v
2
vo =
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example
Ideal diodes
+
Vi
R
D1
−
D2
Vy
Vx
Vo
+
Vo
−
Vx
Vy
Vx
Vi
Vy
zone A : D1 off, D2 on
Vi < V y ⇒ Vo = V y
zone C : D1 on, D2 off
Vi > Vx ⇒ Vo = Vx
zone B : D2 off, D1 off
Vx > Vi > V y ⇒ Vo = Vi
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example
+
Vi
−
Vi >> VA
15k
D1
D1on, D2 off
+
5k V
o
10V −
D2
10k
2.5V
⇒ Vo = 10V
Vi << VB
D1off , D2 on
⇒ Vo = 7.5V
VA > Vi > VB ⇒ D1on, D2 on
Vo = V
(10 // 5 )
i 15 + (10 // 5 )
=
1
11
+ 2. 5
(2Vi + 67.5)
(15 // 5 )
10 + (15 // 5 )
+ 10
(10 // 15 )
5 + (10 // 15 )
1
(2VB + 67.5)
11
1
Vo = 10V = (2VA + 67.5)
11
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Vo = 7.5V =
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DC restoration (clamping circuit)
DC restoration with load
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Voltage doubler
D1 on
D2 off
D1 off
D2 on
D1 off
D2 on
D1 off
D2 off
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Voltage doubler
+
+
Vm
Vm
−
−
+
Vm
−
+
+
−
+
Vm
Vm
−
Vm
+
Vm
−
−
+
+
Vm
−
Vm
−
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Voltage doubler
+ Vm
−
+
Vm
−
+ 2Vm
+ Vm
−
−
+ 2Vm
+ Vm
+
+
Vm
−
Vm
−
−
−
+ 2Vm
+ Vm
−
+ 2Vm
−
+ Vm
−
−
+ 2Vm
−
+
+
Vm
−
Vm
−
+ 2Vm
−
+ 2Vm
−
+ 2Vm
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−
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