Download UNIT 3 - VVIT EEE

VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
Rectifiers, Filters and, Regulators
1. Explain the principle of operation of Half-Wave Rectifier (HWR), Centre tap and Bridge FullWave Rectifiers (FWRs) with the help of circuits and waveforms.
Half-wave Rectifier
This is the simplest rectifier arrangement, as
shown in Fig. 1 (a).
The secondary voltage of the transformer
(being sinusoidal) can be expressed as
vs = Vm Sin ωt = Vm sin α
where α = ωt
This is shown in Fig.1 (b), where Vm is the
peak value of voltage. During the positive
half cycle, when the voltage of the point A is
positive w.r.t. point B, the diode D1 is
forward biased and will conduct. This
current, which flows through the load
resistance RL, the diode (having forward
resistance Rf) and secondary winding of the
transformer (having resistance Rs can be
expressed as
i = Im sin α for 0 ≤ α ≤ π
where Im =
Vm
Rf +Rs +RL
(This assumes that the cut-in voltage of the
diode is negligible compared to Vm )
During the negative half cycle, when the
voltage of the point A is negative w.r.t. point
B, the diode D1, is reverse-biased and will
not conduct. Therefore, the current
i = 0 for π ≤ α ≤ 2π
The waveforms of current i during both the
half cycles are shown in Fig.1(ç). The
current, though not a good d.c. is
unidirectional and has an average value
which is non-zero (actually positive in this
case).
Fig.1. Half-wave rectifier input voltage and
load current waveforms.
Centre-tap Full-Wave Rectifier
Fig.2 (a) shows a full-wave rectifier circuit. The
transformer secondary has a centre-tap and
each half give a peak voltage of Vm . In each
half (here is one diode i.e. D1 and D2 The load
resistance RL is common to both halves.
This can be seen to comprise of two half-wave
circuits. On the positive half cycle, when the
point A is positive w.r.t. B, the diode D1
conducts and current i1 flows through RL.
During this half-cycle, the point C is negative
1
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
w.r.t point B and hence the diode D2 does not
conduct. Therefore i2 = 0.
On the negative half-cycle, the point C is
positive w.r.t. point B. Hence the diode D2
conducts and current i2 flows through RL.
During this half-cycle, the point A is negative
w.r.t. point B and hence the dode D1 does not
conduct. Therefore i1 = 0. Fig.2
(b) and (c) shows the waveforms of currents i1
and i2 since both i1 and i2 flow through the load
RL. The total current i through RL is i = i1 +i2,
which is obtained by adding the two waveform
and is shown fig.2 (d).
Fig. 2 Full-Wave Rectifier and the current
waveforms
Bridge Full-Wave Rectifier
Fig. 3 shows the bridge rectifier circuit, there
are four diodes D1, D2, D3 and D4 which form
the four arms of the “bridge”. The a.c. from
the transformer secondary, in fed to the two
corners and the load resistance RL is
connected to the other two corners. Please
note the directions of all the diodes and also
the polarity of d.c. output.
Fig. 3 Bridge rectifier circuit
During positive half-cycle, the point A is more
positive than the point B. The current path is
point A, through D1, RL, D3 and back to point
B. During this time, the diodes D2 and D4 are
reverse-biased. In the negative half-cycle, the
point B is more positive than point A. The
current path is point B, through D2, RL, D4 and
back to point A. During this time, the diodes
D1 and D2 are reverse biased.
It is interesting to note that in both the halfcycles, the direction of current flow through RL
is the same. Thus we obtain full-wave
rectification, the waveform being same as
shown Fig. 2(d).
2. Derive the following HWR and FWR
a) Average value
b) RMS value
c) PIV
d) Efficiency
f) TUF
Half-wave Rectifier
a) Average value:
If we put a d.c. ammeter in series with RL, it
cannot show a reading the readings as shown
in the Fig.1 (c), because a meter cannot
(i) Average Current (Idc)
2
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
respond to such fast changes in wave form.
(However an oscilloscope can show the exact
waveform). Instead, the d.c. ammeter will
show an average reading. For finding out the
average value of such an waveform, we have
to determine the area under the curve and
divide by 2π.
Since the value of the current i=0 for π ≤ α ≤
2π , therefore the second term above is equal
to zero.
Idc =
Im π
∫ sin α
2π 0
2
1 2π
1 2π
Idc = 2π ∫0 i dα = 2π ∫0 Im sin α
π
2π
Im
(∫ sin α dα + ∫π sin α dα)
2π 0
Im
[cos α]π0
2π
I
= πm
dα =
= Idc =
Im
2π
dα =
(ii) D.C. Output Voltage (Vdc)
The d.c. (i.e. average) output voltage
appearing across RL is
Vdc = Idc R L
Vdc =
Im
R
π L
V
= π(R +Rm +R
f
s
L)
RL
Normally Rf and Rs are quite small in value.
If
Substiting the value of Idc from equation 2 in
the above equation,
Rf +Rs
RL
≪1
Vdc =
Then
Vm
π
b) RMS Value
Because i = Im sin α
Root-means-square (RMS) current can be
obtained by carrying out the following
operations in the sequence s-m-r i.e.
(i) Square Since current is given by i ,
therefore its square is i2;
Irms =
1
2
1
2
Im 2 π 1
Im 2 π 1
Irms = [
∫
dα −
∫
cos2αdα ]
2π 0 2
2π 0 2
Im
Irms =
2
1
Irms
Im 2 π
[ 2π
∫0 sin2 α
dα] (as i =0 for π ≤
α ≤ 2π
1
2
As sin α = 2 (1 − cos2α)
(ii) Mean By calculating the area under the
curve (by integrating) and then taking the
mean over the period of 2π;
(iii) Root: take the square-root.
2
1 2π
= [ ∫ i2 dα]
2π 0
1
2
Im 2 2π 2
Irms = [
∫ sin α dα]
2π 0
1
2
1 2π
Irms = [ ∫ Im 2 sin2 α dα]
2π 0
C) PIV
In the reverse bias, each diode has
maximum voltage across it as it is open.
Peak Inverse Voltage is defined as the
maximum voltage to which a diode is
subjected in the reverse bias.
Referring to Fig.1 (a) in the negative halfcycle, the point B is positive Vm volts w.r.t.
the point A. Therefore, in half-wave circuit,
PIV of the diode Vm volts.
D) Efficiency
3
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
4
1
× 100%
2
π Rf + Rs + 1
RL
If we assume (Rf + Rs) ≪ RL then
4
η = 2 × 100% = 40.6%
π
This means, even under ideal conditions (i.e.
RL and Rs equal to zero), only 40.6% of the
a.c. input is converted into d.c. power. In
general it can be understood that higher the
rectification efficiency, lower would be the
ripple content.
Rectification efficiency is defined as
η=
d. c. power delivered to the load
a. c. input power
Pdc
η=
× 100%
Pac
2
Pdc = Idc
RL
Pac is the rms power delivered into the circuit,
comprising of RL, RS, and Rf
2
Pac = Irms
(R L + R s + R f )
2
Idc
RL
η= 2
× 100%
Irms (R L + R s + R f )
I 2
( πm ) R L
η=
× 100%
Im 2
( 2 ) (R L + R s + R f )
e) Ripple factor
η=
r. m. s. value of the alternating
component of the wave
r=
average value of the wave
The purpose of a rectifier is to convert a.c.
into d.c. But the simple circuit as used in halfwave rectifier does not fully achieve this
purpose.
r=
We redraw the Fig.1 (c), but this time Idc is
drawn in firm line and the rectified output in
dotted line. From this, we can consider that
an alternating component (of course nonsinusoidal) has been superimposed on Idc.
r=
2
2
√Irms
− Idc
Idc
=
′
′
Irms
Vrms
=
Idc
Vdc
2 − V2
√Vrms
dc
Vdc
2 − V2
Vrms
dc
= √
2
Vdc
Im
r = √( 2 ) − 1 = 1.21
Im
π
Idc = Average value of the waveform
i’ = The value of alternating component of the
current wave form
I’rms = r.m,s value of the alternating
component, i.e. of i’.
If the above expression is expressed as a
percentage, it indicates that the amount of the
alternating component present in the output is
121% of the d.c. value. In other words, it
means the undesired ac. component is in fact
more the desired half wave rectifier is not a
good
rectifier.
Ripple factor, r. is defined as
f) TUF (Transformer Utilization Factor)
Most of rectifier circuits make use of a
transformer, whose secondary feeds the ac.
power. It is necessary to determine the
rating of transformer while designing a
power supply. TUF is useful for this
purpose. TUF is defined as
4
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
TUF
d. c. power delivered to the load
a. c. power rating of transformer secondary
Pdc
=
Pac (rated)
=
I 2
( πm ) R L
TUF =
Im (R f + R s + R L )
Pac (rated) = Vac(rms) Irms
Im
√22
If (R f + R s ) ≪ R L , then
I 2
( πm ) R L
TUF =
Vm Im
√2 2
TUF =
√22
= 0.287
π2
Full-wave Rectifier
1
a) Average value: (i) Average Current (Idc)
2π
Idc = 2π ∫0 i dα
If we put a d.c. ammeter in series with RL, it
cannot show a reading the readings as shown
in the Fig.2, because a meter cannot respond
to such fast changes in wave form. (However
an oscilloscope can show the exact
waveform). Instead, the d.c. ammeter will
show an average reading.
1 2π
∫ I sin α dα
2π 0 m
π
2π
Im
=
(∫ sin α dα + ∫ sin α dα)
2π 0
π
=
Since the value of the current i for π ≤ α ≤ 2π
is the same as the period 0 ≤ α ≤ π, therefore
the curve will be twice that of the half wave
case.
For finding out the average value of such an
waveform, we have to determine the area
under the curve and divide by 2π.
Idc =
2 Im
π
(ii) D.C. Output Voltage (Vdc)
The d.c. (i.e. average) output voltage
appearing across RL is
2 Im
RL
π
Vdc =
2V
Vdc = π(R +Rm+R
Vdc = Idc R L
f
s
L)
RL
Normally Rf and Rs are quite small in value.
Rf +Rs
If
≪1
R
substiting the value of Idc from equation 2 in
the above equation,
L
Then
2 Vm
π
Vdc =
b) RMS Value
Root-means-square (RMS) current can be
obtained by carrying out the following
operations in the sequence s-m-r i.e.
(i) Square Since current is given by i ,
therefore its square is i2;
(ii) Mean By calculating the area under the
curve (by integrating) and then taking the
mean over the period of 2π;
(iii) Root: take the square-root.
5
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
1
(1 − cos2α)
2
2 2π
Im
1
1
Irms
As sin2 α =
2
1 2π
= [ ∫ i2 dα]
2π 0
Irms = [
2π
1
1
2
Im 2 2π 2
Irms = [
∫ sin α dα]
2π 0
Irms =
1
2
dα]
α ≤ 2π
0
2
dα
1
2
Im 2 2π 1
−
∫
cos2αdα ]
2π 0 2
Im
Irms =
√2
2
1 2π
Irms = [ ∫ Im 2 sin2 α dα]
2π 0
Because i = Im sin α
Im 2 2π
[ 2π
∫0 sin2 α
∫
(as i =0 for π ≤
C) PIV
In the reverse bias, each diode has maximum
voltage across it as it is open. Peak Inverse
Voltage is defined as the maximum voltage to
which a diode is subjected in the reverse bias.
D) Efficiency
FWR the non conducting period, the inverse
voltage across the non conducting dode is
2Vm
2I 2
( m) R L
π
Rectification efficiency is defined as
η=
d. c. power delivered to the load
η=
a. c. input power
η=
Pdc
× 100%
Pac
η=
2
Pdc = Idc
RL
2 (R + R + 2R )
Irms
L
s
f
8
1
× 100%
2
π Rf + Rs + 1
RL
η=
2
Pac = Irms
(R L + R s + 2R f )
η=
× 100%
If we assume (2Rf + Rs) ≪ RL then
Pac is the rms power delivered into the
circuit, comprising of RL, RS, and Rf
2
Idc
RL
I 2
( m ) (R L + R s + 2R f )
√2
8
× 100% = 81.2%
π2
This means, even under ideal conditions
(i.e. RL and Rs equal to zero), only 81.2% of
the a.c. input is converted into d.c. power. In
general it can be understood that higher the
rectification efficiency, lower would be the
ripple content.
× 100%
e) Ripple factor
i’ = The value of alternating component of
the current wave form
I’rms = r.m,s value of the alternating
component, i.e. of i’.
The purpose of a rectifier is to convert a.c.
into d.c..
Idc = Average value of the waveform
6
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
Im
r = √( √2 ) − 1 = 0.483
2 Im
π
Ripple factor, r. is defined as
r. m. s. value of the alternating
component of the wave
r=
average value of the wave
If the above expression is expressed as a
percentage, it indicates that the amount of
the alternating component present in the
output is 121% of the d.c. value. In other
words, it means the undesired ac.
component is in fact more the desired half
wave rectifier is not a good rectifier.
′
′
Irms
Vrms
r=
=
Idc
Vdc
r=
2
2
√Irms
− Idc
Idc
=
2 − V2
√Vrms
dc
Vdc
2 − V2
Vrms
dc
= √
2
Vdc
f) TUF (Transformer Utilization Factor)
In FWR, the average TUF is obtained by
considering yhe primary and secondary
windings separately. In each half of the
secondary winding, the current flows through
1800, whre in the primary winding the current
flows through complete 3600
(ii) TUF (primary)
Since the primary winding carries a full wave
current, its rated capacity is utilized to the
same extent as the rectification.
TUF = 0.812
(i) TUF (secondary)
(iii) Average TUF
Pac(rated secondary) = Pac(rated) one half + Pac(rated) other half
=
Vm
√2
=
X
Im
2
Vm Im
√2
+
Vm
√2
X
Average TUF =
Im
2
=
= 0.574
TUF (secondary)+TUF (primary)
2
0.574+0.812
2
= 0.693
3. Explain why a bridge rectifier is preferred over a centre tap rectifier
(i) The current flow through both the primary
and the secondary windings are sinusoidal
(i.e. full cycle). Due to this, the TUF of the
both the primary and the secondary are 0.812
and the overall TUF is 0.812. This is higher
than the TUF of full-wave rectifier (0.693) and
hence for the same d.c. output power, a
smaller transformer can be used in the bridge
circuit.
(iii) The PIV of each diode = Vm. Let us
consider the positive half cycle when diodes
D1 and D3 are conducting and therefore the
voltage drops across each of them is
negligible. Therefore the voltage appearing
across D2 is equal to Vm. Also the voltage
appearing across D4 is equal to Vm In a similar
way considering the negative half-cycle, it can
be understood that PIV of D1 and D3 are also
equal to Vm. Thus, for a given voltage rating of
the diodes, these can be used for highvoltage applications.
(ii) A centre-tap is not required in the
transformer secondary.
7
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
(d) Bridge circuit requires four diodes. At any
given time two of them conduct in series,
causing voltage drop due to two diodes. This
additional voltage drop cannot be ignored in
low voltage d.c. power supplies.
4. Derive the expression for ripple factor for FWR with L-section filter. Explain the necessity a
bleeder resistor.
2 Vm
This is also known as the choke-input filter or L
Vdc =
− 𝐼𝑑𝑐 (R c + R s + R f )
C filter. Figure shows the filter circuit, in which
π
an inductor L is used at the input of the filter
and a capacitor in parallel with the load where R f , R c and, R s are the resistances of
resistance RL. Also shown in the dotted line is diode, choke and secondary winding. Both the
Rb the bleeder resistance.
above equations are similar to those of only
inductor filter. However the ripple is much
lower in L-section filter than inductor filter.
Ripple Factor:
The choke L is series with the parallel
combination of the capacitor C and the load
resistance RL. In order the choke can suppress
the ripple components, the reactance of the
choke must be very large compared to the
parallel impedance of the capacitor and the
resistance. This parallel impedance is made
small by making the reactance of C much
Figure : L-section filter
smaller than resistance RL. So we can
The inductor L offers series impedance to the reasonably assume that the entire a.c. ripple
ripple currents and the capacitor C offers low passes through the capacitor and none
impedance in shunt with the load. A corn through the resistance.
hin4tion of these two effects result in a load
current which smoothed much more effectively Due to the above consideration, the total
impedance between point A and B is
than either L or C alone.
practically that of the choke alone and is
If L is small, then the capacitor C will get approximately,
charged to peak voltage Vm and diodes will
𝑋𝐿 = 2𝜔𝐿, since ripple frequency = 2𝜔
conduct only for short periods in each half
cycle. If the L is large, the diodes will conduct
for longer period due to smoothening effect of The ripple voltage has the magnitude 4𝑣𝑚 and
3𝜋
the inductor if the value of L is more then
therefore the ripple current,
certain minimum value (called critical
inductance), each diode will conduct for
4𝑣𝑚 1
complete half-cycle. The output d.c. voltage is
𝛤=
3𝜋 𝑥𝐿
given by
Vdc =
2 Vm
π
The r.m.s. value of ripple current is
𝛤𝑟𝑚𝑠 =
The output d.c. voltage on load is
8
4𝑣𝑚 1
√23𝜋 𝑥𝐿
𝛤
√2
so,
VASIREDDY VENKATADRI INSTITUTE OF TECNOLOGY::NAMBUR
ELECTRONIC DEVICES & CIRCUITS (UNIT-3)
The a.c. ripple voltage the load RL is the same
as that across the capacitor and this is equal to
𝐿𝑐 ≥
′
′
𝑉𝑟𝑚𝑠
= 𝐼𝑟𝑚𝑠
𝑋𝑐
=
A practical value (taking into account the
approximations made in the derivation) would
be more than the above and a figure of
√2 𝑉𝑑𝑐
𝑋
3 𝑋𝐿 𝑐
𝐿𝑐 ≥
Where Xc = reactance of the capacitor at the
ripple frequency i.e. 2𝜔
By defining ripple factor
√2 𝑉𝑑𝑐
3 𝑋𝐿
𝑉𝑑𝑐
𝑋𝑐
=
√2 𝑋𝑐
3 𝑋𝐿
=
r=
′
𝑉𝑟𝑚𝑠
𝑉𝑑𝑐
𝑅𝐿
3𝜔
𝑅𝐿
800
Bleeder Resistance:
The output voltage of L-section filter at load is
Vm whereas when some load current flows the
2𝑣
value is 𝜋𝑚 . This means in the region of very
low current, the output voltage drops sharply,
as shown in figure
=
√2
12 𝜔2 𝐿𝐶
We note the ripple factor is dependent on the
values of L and C, but is independent of load
Critical Inductance :
All the foregoing calculations are based on the
assumption that value of the inductance is
more than the critical value so that the current i
does not fill to zero at any time. For this to
happen, the value of Idc must be more than the
peak value of the a.c. (ripple) component.
Figure: (a) D.C. Output Voltage Vs. Load
Current.
′
𝐼𝑑𝑐 > √2 𝐼𝑟𝑚𝑠
In order to avoid this sharp change in output
voltage, a resistance Rb (called bleeder
resistor) is connected in parallel to RL so that a
current of Imin, bleeds (i.e. flows) through Rb.
Making use of equation Lc, the value of the
bleeder resistance is
𝑉𝑑𝑐 2𝑉𝑑𝑐
≥
𝑅𝐿
3
Let the critical value of inductance be Lc.
therefore
2𝜔𝐿𝑐 ≥
𝑅𝑏 ≤ 800 𝐿𝑐
2 𝑅𝐿
3
9