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Chap. 3 Common Diode Application
3.1 Transformers
3.2 Half-wave rectifiers
3.3 Full-wave rectifiers
3.4 Full-wave bridge rectifiers
Basic power supply block diagram
Power supply A group of circuits used to convert ac to dc. Rectifier A circuit that converts ac to pulsating dc. Filter A circuit that reduces the variations in the output of a rectifier. Voltage regulator A circuit designed to maintain a constant power supply output Transformer A circuit connecting dc power supply to the ac input
3.1 Transformers
Transformers are made up of inductors that are in close proximity to each other, yet not electrically connected.
A transformers provides ac coupling from primary to secondary while providing physical isolation between the two circuits.
3.1 Transformers
Three type of transformer Step-up transformer (승압형) : provides a secondary voltage that is greater than the primary voltage Step-down transformer (감압형) : provides a secondary voltage that is less than the primary voltage Isolation transformer (분리형) : provides an out voltage that is equal to the input voltage
3.1 Transformers
turns ratio (권선비) : the ratio of the number of turns in the primary to the number of turns in the secondary.
turns ratio is equal to the voltage ratio of the component.
When the turns ratio an primary voltage are known, the secondary voltage can be found as :
3.1.2 calculating secondary current
Ideally, transformer transfers 100 % of it’s power to the secondary.
Since power equals the product of voltage and current ,
SP P PS S P P S P P
S P S S
VI V NV I V I I I I
I V V N
Example 3.1
Assume that the fuse limits the value of Ip to 1 A. What is the limit on the value of the secondary current ?
1 250
4
PS P
S
NI I A mA
N
If the secondary current tries to exceed the 250 mA limit, the primary current will exceed its limit and blow the fuse.
3.2 Half-wave rectifiers
Three type of rectifier circuits: Half-wave rectifier Full-wave rectifier Bridge rectifier
Half-wave rectifier circuit : a diode is connected in series with a transformer and a load resistance
3.2.1 Basic circuit operation of the positive half-wave rectifier
Positive half-wave rectifier ~ the negative half-cycle of the input to the rectifier is eliminated by the one-way conduction of the diode
During the positive half-cycle of the input, D1 is forward biased and provides a current path
When D1 is reverse biased, there is no current flow and no voltage drop through RL
Ideal model VD1 VL
Forward bias 0 V VS
Reverse bias VS 0 V
3.2.2 Negative Half-wave rectifiers
Positive & Negative Half-wave rectifiers
Basic Transformer
Small power transformer
The basic transformer is formed from two coils that are usually wound
on a common core to provide a path for the magnetic field lines.
Air core Ferrite core Iron core
Direction of windings
The direction of the windings determines the polarity of the voltage across the secondary winding
with respect to the voltage across the primary. Phase dots are sometimes used to indicate polarities.
In phase Out of phase
secondary prisec
primary pri sec
N IVn
N V I
3.2.3 Calculating load voltage & current
When we take VF into account (i.e. introducing practical model), the peak load voltage, VL(pk) is found as
where VS(pk) is the peak secondary voltage of transformer:
with VP(pk) = the peak transformer primary voltage
Example 3.2
Q. The input voltage is 120 Vac. Determine the peak load voltage
The secondary peak voltage is obtained from the primary value.
Example 3.4
Q. The input voltage is 120 Vac. Determine the peak load current.
( )
( )
( ) ( )
( ) ( )
( )
( )
120170
0.707 0.707
17056.7
3
56.7 0.7 56
565.6
10
P rms
P pk
SS pk P pk
P
L pk S pk F
L pk
L pk
L
V VV V
N VV V V
N
V V V V V V
V VI mA
R k
3.2.4 Average voltage
The value of an average voltage (Vave) is a dc voltage value equivalent to the ac waveform after an average over one period.
Vave
0
sin 2pk aveV d V
0.318ave pkV V
3.2.4 Average current
The value of an average current (Iave) is a dc current value equivalent to the ac waveform after an average over one period.
Example 3.6 Assume an input voltage of 120 Vac. Determine the value of Iave
( )
( )
( ) ( )
( ) ( )
( )
120170
0.707 0.707
17085
2
85 0.7 84.3
84.326.8
26.81.34
20
P rms
P pk
SS pk P pk
P
L pk S pk F
L pk
ave
aveave
L
V VV V
N VV V V
N
V V V V V V
V VV V
V VI mA
R k
( )
( )
( )
84.34.22
20
1.34
L pk
L pk
L
L pk
ave
V VI mA
R k
II mA
3.3 Full-wave rectifiers
Center-tapped transformer: the voltage from the center tap to each terminal of the outer winding coils is equal to one half of the secondary voltage
3.3.1 Basic circuit operation
During an positive half-cycle of the input, D1 is forward-biased and D2 is reverse- biased
When D2 is forward-biased, D1 is reverse-biased. But, the direction of the load current is the same as before.
3.3.2 Calculating load voltage & current
With assuming the practical diode model, the peak load voltage (VL(pk)) for full-wave rectifier is found as
Average load voltage (Vave) for a full-wave rectifier is found as
full-wave rectifier
half-wave rectifier
Example 3.9
Q. Determine the dc load voltage
3.3.3 Peak inverse voltage (PIV)
From practical model, D1 is forward-biased and exhibits a voltage drop of 0.7 V
Because D1 is in series with D2, the PIV across D2 is reduced by the voltage drop (VF) across D1
Note that PIV = VS(pk) from the ideal diode model
3.3.4 Negative full-wave rectifier
3.4 Full-wave bridge rectifiers
1. The bridge rectifier does not require the use of a center-tapped transformer, or the transformer itself. It can be connected directly to the AC input even without the transformer. 2. When it is connected to a transformer with the same secondary voltage, it produces nearly double the peak output voltage of the full-wave center-tapped rectifier.
3.4.1 Basic circuit operation
During the positive half-cycle of the input, D1 and D3 are switched “on”, while D2 and D4 are “off”.
During the negative half-cycle of the input, D2 and D4 are switched “on”, while D1 and D3 are “off”.
3.4.2 Load voltage & current
Since there are two conducting diodes in series with the load resistance, the peak load voltage is founds as:
Example 3.11 Determine the dc load voltage and current
( )
( ) ( )
( )
1217
0.707
1.4 15.6
2 2(15.6 )9.93
9.9382.8
120
S pk
L pk S pk
L pk
ave
aveave
L
VV V
V V V V
V VV V
V VI mA
R
3.4.3 Bridge vs. full-wave rectifier
Example 3.12 Assume a full-wave rectifier with an input voltage of 12 Vac and a load resistance of RL = 120 . Determine Vave and Iave.
12 120
( )
( )
( )
( )
1217
0.707
170.7 0.7 7.8
2 2
2 2(7.8 )4.97
4.9741.4
120
acS pk
S pk
L pk
L pk
ave
aveave
L
VV V
V VV V V V
V VV V
V VI mA
R
( )
( ) ( )
( )
1217
0.707
1.4 15.6
2 2(15.6 )9.93
9.9382.8
120
S pk
L pk S pk
L pk
ave
aveave
L
VV V
V V V V
V VV V
V VI mA
R
Bridge rectifier results Center-tapped rectifier results
Double
efficiency !
3.4.4 Peak Inverse Voltage
The PIV across each diode in the bridge rectifier is approximately equal to the peak secondary voltage, VS(pk). Counting VF results in PIV = VS(pk) – 0.7
Chap. 3 Common Diode Application
3.5 Filters
3. 6 Zener voltage regulators
Basic power supply block diagram
Power supply A group of circuits used to convert ac to dc. Rectifier A circuit that converts ac to pulsating dc. Filter A circuit that reduces the variations in the output of a rectifier. Voltage regulator A circuit designed to maintain a constant power supply output Transformer A circuit connecting dc power supply to the ac input
3.6 Filters
The effects of filtering on the output of a half-wave rectifier.
There are still voltage variations after filtering, but, the amount of variation is severely reduced.
Ripple voltage (Vr): The variation in the output voltage of a filter ~ depends on the rectifier used, the filter component value, and the load resistance.
ripple voltage
3.6.1 Basic capacitive filter
The capacitive filter is the most basic filter type and the most commonly used.
The filtering action is based on the charge/discharge action of capacitor
During the positive half-cycle, D1 conducts and capacitor charge rapidly.
When the output voltage of the rectifier becomes smaller than the capacitor voltage, the capacitor acts as the voltage source for the load resistance.
3.6.1 Basic capacitive filter : time constant
For an RC circuit, the time constant for charging or
discharging is t = RC [seconds] ~ exponential curve 100%
80%
60%
40%
20%
00 1t 2t 3t 4t 5t
99%98%
95%
86%
63%
37%
14%
5%2% 1%
Number of time constants
Perc
ent o
f final v
alu
e
C
R
C
R
It takes five time constants for a capacitor to charge or discharge fully
T = 5RC
total capacitor charge time (T)
total capacitor discharge time (T)
resistance of diode under forward bias = 5
Note that the charging time is much shorter than the discharging time!
The charging time is not determined by RL but by the diode resistance in forward bias
3.6.1 Basic capacitive filter
The amplitude of the ripple voltage at the output of a filter varies inversely with the values of filter capacitance and load resister
RLC time for discharge increases
Larger capacitance value results in highly elongated discharge time and increased charging time as well.
3.6.2 Surge current
Surge current: high initial current in a power supply
For the first instant, the discharged capacitor acts as a short circuit ~ charging of positive and negative charge on both sides of capacitor
Isurge
Example 3.14
Surge limiting resistor
The resistor Rsurge helps to limit surge current, but it also limits the output voltage on RL
The other way to limit the surge current is to use small capacitance capacitor.
Smaller results in a shorter charging time of , but it also results in larger ripple voltage
Q V VQ CV I C t C
t t I
C t
.
3.6.3 Filter Output voltage
peak output voltage of rectifier
peak-to-peak ripple vol
2
tage
rdc pk
pk
r
V
V
V V
V
where dc load current,
t = interval time between charging peaks
C = capacitance [F
]
L
L
r
I tV
I
C
3.6.3 Filter Output voltage
1 116.7 ~ half-wave rectifier
60
1 18.3 ~ full-wave rectifier
120
half wave
full wave
t msf Hz
t msf Hz
, ,2L
r r half wave r full wave
I tV V V
C
EXAMPLE 3.15
EXAMPLE 3.16
( ) ( ) ( )
( )
( )
1
2 2 2
12
16.2816.16
1 0.007381
2
dcL
L
Lr
dcr Ldc L pk L pk L pk
L
dc L pk
L
L pk
dc
L
VI
R
I tV
C
VV I t tV V V V
C C R
tV V
CR
VV
t
CR
Chap. 3 Common Diode Application
3.7 Zener Voltage Regulators
Power supply : Regulators
3.7 Zener Voltage Regulators
Connection of zener diode :
• reverse bias operation
• constant voltage for IZK < I < IZM
• parallel circuit of a zener diode and a load resistor
• VL = VZ if IZK < ID < IZM
3.7.1 Total Circuit Current
3.7.4 Zener Current
Load current
Total current
, : current through a load resistor,
,
Zener curr
, en t
L L
ZL
L
RS in ZT L Z
S S
Z T L
I R
VI
R
V V VI I I
R R
I I I
EXAMPLE 3.17 & 3.19
9.10.91
10
4.95 0.91 4.04
ZL
L
Z T L
V VI mA
R k
I I I mA mA mA
3.7.5 Load Variations
What are the practical limit on the value of RL?
If RL = 0 (the load is shorted), IT = IL and IZ = 0 < IZK ~ the regulation is lost!
There exists the minimum value of RL for the voltage regulation.
To maintain zener regulation, the minimum zener current must be equal to IZK. Therefore,
Since IT = IZ + IL and IZ(min) = IZK,
IL(max) = IT – IZ(min) = IT – IZK
Since IL(max) occurs when the load resistance is at a minimum,
(min)
(max)
Z ZL
L T ZK
V VR
I I I
regulating till IZ < IZM
EXAMPLE 3.20
Q. Assume diode current rating of IZK = 3 mA and IZM = 100 mA. Determine the minimum RL.
(max)
(min)
(max)
20 3.316.7
1
16.7 3 13.7
3.3241
13.7
in ZT
S
L T ZK
ZL
L
V V V VI mA
R k
I I I mA mA mA
V VR
I mA
3.7.7 Zener Reduction of Ripple Voltage
How does the zener diode reduce the ripple voltage?
The equivalent circuit of a zener diode consists of a zener impedance (ZZ) and VZ.
Using the voltage divider rule, the ripple output from the regulator can be found as
(
( )
)
and ( || )
where
the ripple present at the regulator output,
( || ) the parallel combiantion
( ||
of and the load resistance
= the reg
)
( || )
ulator se
Z Lr out r
Z L
Z LZ L
Z L
r out
Z
S
L Z L
S
Z RZ R
Z R
V
Z R Z R
Z RV V
Z R
R
R
ries resistance
= the peak-to-peak ripple voltage present at the regualtor inputrV
EXAMPLE 3.21
Q. The filter output has a 1.5 Vpp as a ripple voltage. Determine the ripple voltage at the load resistor.
( )
5 120( || ) 4.8
5 120
( || ) 4.81.5 0.13
( || ) 4.8 51
Z LZ L
Z L
Z Lr out r pp pp
Z L S
Z RZ R
Z R
Z RV V V V
Z R R
3.7.8 Putting It All Together :