Figure 3.16 Circuit symbol for a zener diode.
Zener Diodes
- Diodes operating in the breakdown region can be used in the design of voltage regulators.
Specifying and modeling the zener diode
Dynamic resistance, rZ – a few ohms to a few tens of ohms
Zener breakdown voltage, VZ – a few volts to a few hundreds of volts
In Fig. 3.17, the voltage of Q-point V r Iz
Figure 3.17 The diode i–v characteristic with the breakdown region shown in some detail.
Figure 3.18 Model for the zener diode.
V
Z V
Z0 r I
z ZV
ZV
Z0r I
z Z
Zener Diodes
Use of the zener as a shunt regulator
The change in VO corresponding to a 1-V change in VS [㎷/V] :
The change in VO corresponding to a 1-㎃ change in IL [㎷/㎃] :
Assuming R, rZ ≪ RL ,
Thus,
0 Z
( // )
O Z S L Z
Z Z
r
V V R V I R r
R r R r
O S
Line regulation V
V
O L
Load regulation V I
z z
Line regulation r
R r
( // )z Load regulation r R
Ex 3.7
VZ=6.8V at IZ=5mA , rZ=20Ω , and IZK=0.2mA.a) With no load and with V+=10V, the current through the zener is given by
b) The change in VO resulting from the ±1V change in V+
c) The change in VO resulting from connecting a load resistance RL that draws a current IL=1mA
Figure 3.19 (a) Circuit for Example 3.8. (b) The circuit with the zener diode replaced with its equivalent circuit model.
Zener Diodes
VZ0 VZ r Iz Z 6.8 20 0.005 6.7V
+
V V 0 10 - 6.7
6.35 mA , 0.5 0.02
Z Z
z
I I
R r
VO VZ0 I rZ z 6.7 6.35 0.02 6.83 V
+ 20
V V 1 38.5 mV
500 20
z O
z
r R r
Line regulation VO 38.5 mV/V V
VO r Iz Z 20 1 20 mV
Load regulation VO 20 mV/mA IL
d) The change in VO when RL=2㏀
e) When RL =0.5㏀, IL= 6.8/0.5=13.6mA > I .
→ This is impossible, therefore VO≠ 6.8V, and the zener must be cut off.
< 6.8V
f) The minimum value of RL for which the diode still operates in the breakdown region.
* Recently, zener diodes have been replaced in IC voltage-regulators.
Zener Diodes
VO r Iz Z 20 3.4 68 mV
+ 0.5
V V 10 5 V
0.5 0.5
L O
L
R R R
,min
6.7 1.5 k
L 4.4
R
6.8V / 2k 3.4mA , 3.4 mA
L Z
I I
0.2mA , 0 6.7V
Z ZK Z ZK Z
I I V V V
min
9 6.7
4.6 mA , =4.6-0.2=4.4 mA
0.5
LI I
Rectifier Circuits
The half-wave rectifier
(diode off)
(diode on) In many applications,
VD = 0.7V or 0.8V
Peak inverse voltage – the largest reverse voltage that is expected to appear across the diode.
PIV = Vs → to select a diode that has a reverse breakdown voltage at least 50% greater than the expected PIV
Figure 3.20 Block diagram of a dc power supply.
Figure 3.21 (a) Half-wave rectifier. (b) Transfer characteristic of the rectifier circuit. (c) Input and output waveforms, assuming that rD ≪ R.
Figure 3.22 Full-wave rectifier utilizing a transformer with a center-tapped secondary winding: (a) circuit; (b) transfer characteristic assuming a constant-voltage-drop model for the diodes; (c) input and output waveforms.
Rectifier Circuits
The full-wave rectifier
The primary voltage > 0 → D1 on , D2 off The primary voltage < 0 → D1 off , D2 on
During the +tive half-cycle, the voltage at the cathode of D2 is vO, and that at its anode is -vS. Thus the reverse voltage across D2 will be (vO+vS), which will reach its maximum when vO is at its peak value of (Vs - VD0), and vS is at at its peak value of Vs.
PIV = 2Vs - VD0
Rectifier Circuits
The bridge rectifier
vS > 0 → D1, D2 on D3, D4 off vS < 0 → D1, D2 off D3, D4 on
When vS > 0,
Thus
PIV = Vs 2VD VD Vs VD
3,
= +
2,= -
1,D r O D f S D f
The rectifier with a filter capacitor
Input voltage vI=Vp sin wt
Assuming that the diode is ideal,
t < T/4 → diode on , vO = vI (charging capacitor) t ≥ T/4 → diode off , vO = Vp
Figure 3.23 The bridge rectifier: (a) circuit; (b) input and output waveforms.
Figure 3.24 (a) A simple circuit used to illustrate the effect of a filter capacitor. (b) Input and output waveforms assuming an ideal diode. Note that the circuit provides a dc voltage equal to the peak of the input sine wave. The circuit is therefore known as a peak rectifier or a peak detector.
Rectifier Circuits
Figure 3.25 Voltage and current waveforms in the peak rectifier circuit with CR ≫ T.
L O / i
RI
D C L L
i i i Cd i
dt
Vp Vr VpeT CR/
V
rV
p TV
p ILCR fCR fC 0 T
The peak rectifier
Assuming that the diode is ideal,
t < 0 → diode on , vO = vI (capacitor charging) 0 < t < T -t → diode off , vO = Vp e-t/RC
(capacitor discharging) T -t < t < T → diode on , vO = vI
(capacitor charging)
Under the assumption that RC ≫ T, Load current
Diode current
Output dc voltage, Vo= Vp-Vr/2, where Vr is ripple voltage Output dc current, IL = Vo/R ≅ Vp/R
During the diode-off interval, vo = Vp e-t/CR If Δt ≪ T, at the end of the discharge interval,
Since RC ≫ T, e-t/CR ≈1-T/CR
30
Conduction interval
twhere w
= 2p
f = 2p
/ T. Sincew
tis small,
Charge that the diode supplies to the capacitor
where,
Charge that the capacitor loses
To equate two charges,
Peak value of the diode current
Rectifier Circuits
0 T
3
V 100
83.3 F V 2 60 10 10
p r
C fR
V V 2
p
r fCR
Rectifier Circuits
Figure 3.26 Waveforms in the full-wave peak rectifier.
¶ In case of the full-wave rectifier
The diode conducts for of the cycle.
Figure 3.28 General transfer characteristic for a limiter circuit.
Limiting circuits
vI ≤ L-/K , vo= L-
L-/K ≤ vI < L+/K , vo= K vI L+/K ≤ vI , vo= L+
Limiting and Clamping Circuits
Figure 3.29 Applying a sine wave to a limiter can
result in clipping off its two peaks. Figure 3.31 A variety of basic limiting circuits.
Figure 3.32 The clamped capacitor or dc restorer with a square-wave input and no load.
The clamped capacitor
Capacitor voltage
vI < 0 → diode on → vC = -vI (to charge capacitor) vI ≥ 0 → diode off → vC = minus peak of vI
Output voltage vO = vI + vC
* Reversing the diode polarity will provide an output waveform whose highest peak is clamped to 0V.
Limiting and Clamping Circuits
Figure 3.33 The clamped capacitor with a load resistance R.
Figure 3.34 Voltage doubler: (a) circuit; (b) waveform of the voltage across D1.
The voltage doubler
C1 and D1 - clamping circuit D2 and C2 - peak detector