Diode rectifiers, power supplies,
Zener Diodes and Voltage Regulators
EGB120 – Lecture 12
Associate Professor Geoff Walker
Practice Exams!
• 4 practice exams are available for you on Blackboard!
• Week 13 Lecture
– First half will be a review of content from this semester
– Second half we will go through some tips for solving the practice exams
• Week 13 Tutorials
– You will be able to work through the practice exams, and receive feedback from
your tutors in these sessions (as well as through email of course!)
– You can attend multiple sessions if you wish
2
Exam Prep Tips – Week 12
Increase your
productivity with the
Pomodoro
Technique. Set your
timer for 25 min and
focus on ONE thing
 5 min
Facilitator
Namebreak
 do another 25 min
Your brain
works best with
balanced
activity. For
example:
study  take a walk
study  cook dinner
study  dance!
Convert class work
into flow charts,
mind maps or
diagrams.
Explain the topic to
friends or family
members to
memorise subject
knowledge.
↓
GET AHEAD: bit.ly/Problem-BasedExams
CRICOS No. 00213J
Prac 6 Exam
• Prac 6 exam in weeks 12 or 13,
• Similar to Prac 5, but easier: Build a simple op-amp circuit, all circuit values
are given. Then measure and calculate gain & phase with the DSO.
• Example exam is available on Canvas
• Exam is open book, but no phones / laptops etc.
• Group Exam (unless you have already nominated for an individual exam)
– Consists of a multiple element circuit of operational amplifier and resistors.
– 3 students per group. Each student must demonstrate their skills and
understanding.
– 1 hour to complete exam with 5 mins perusal.
– Discuss within your group to build and debug your circuit.
Prac Drop-in Sessions
• Additional tutor supervised drop in sessions have been
organised
• These session are an opportunity for you to practice for your
assessed practical (prac 6 in weeks 12/13)
• All sessions are in S919 – your practical room
– Monday Week 12
– Monday Week 13
2pm-4pm
2pm-4pm
• If you cannot make it at that time, you can also practice in
S905-C (no tutor supervision). Access details on the door.
5
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·
↳
-
Rectifiers
• Rectifiers covert AC to DC.
s U
• Historically, rectifiers have been intricate mechanical devices,
mercury arc rectifiers and vacuum tubes.
• Recently, diodes have been the dominant technology.
• Modern devices use computer controlled “switch mode”
rectification
.
– This creates greater efficiencies and can provide voltage control too.
13
AC Plugpacks
• Steps the mains AC voltage (240 V RMS)
down to a safe AC voltage level.
• The underlying technology is a
transformer (which we won’t cover).
(Not examinable
->
I
Yz
L
...
)
⑤115
=
.
D
I
mor
12Nac
0
N
=
240N
=
12
-
14
Revision: Diode Characteristic
• The diode has non-linear characteristic.
• This is the ideal version.
-
Be
A
0
A
K
A
i
v
5- 1
-
02
Forward voltage
-
.
0
i
+
.
.
7-0
.
8W
v
-
Diode
15
Half wave Rectifier Circuit
A single diode makes a half-wave rectifier.
• Conducts on the positive half cycle, (output follows input),
– Output is forward voltage drop of diode less than the input,
• Blocks on the negative half cycle (output is zero).
[16
20
15
AC plugpack
12 VAC +
RMS
Um E
=
+
12
=17 IV
.
-
50
170
-
-
.
10
5
0
-5
0
0.02
0.04
0.06
0.08
0.1
-10
-15
ov
.
*
-20
>V
AC
16
Ripple voltage
• To smooth the ripples, we add a capacitor.
• Diode conducts and capacitor charges just at the peaks.
• Capacitors supplies the output current for the rest of the
I
period.
I
-
van
20
I
15
AC plugpack
12 VAC +
RMS
10
5
0
A
0
need
0.02
0.04
0.06
0.08
0.1
-5
-10
ac
-15
-20
17
Calculating Ripple
• Assume capacitor is instantly charged
when diode is forward biased.
• Assume constant current discharge of
capacitor.
dv
i=C
dt
dv i
=
dt C
where
i
=
VDC
E
17
16.5
AC plugpack
12 VAC +
RMS
A
16
15.5
15
14.5
14
E0
0.005
0.01
0.015
0.02
0.025
18
Calculating Ripple
• If we approximate Δt as the period of the AC signal, we can
work out the total voltage ripple:
∆v i
=
∆t C
i
∆v = ∆t
C
i
=
fC
17
16.5
16
15.5
Δv
15
14.5
14
0
0.005
0.01
0.015
Δt
0.02
=
0.025
20ms
since f
=
50H2
19
i
∆v =
fC
Ripple Example
I
• Choose a capacitor to keep the ripple voltage less than 0.1 V.
12
DV
6 12
--
-
=
.
0
.
↑ 1
-
6W
1
AC plugpack
+
5 VAC
RMS
50 Hz
-
I Foo
1=GomA ROms
-
C
100 Ω
(f
=
--
Uk
=
7012
-
C
ist
=
-
AV
50Hz
.
60m
=
.
20m
-
0
·
1
12000pF
↑
=
-
20
Ripple Example
i
∆v =
fC
• Choose a capacitor to keep the ripple voltage less than 0.1 V.
𝑖𝑖
Δ𝑣𝑣 =
𝑓𝑓𝑓𝑓
AC plugpack
+
5 VAC
RMS
50 Hz
𝑖𝑖
𝑣𝑣/𝑅𝑅
𝑓𝑓 =
=
Δ𝑣𝑣𝑓𝑓
Δ𝑣𝑣𝑓𝑓
C
100 Ω
5 × 2/100
=
0.1 × 50
= 14100 μF
2x 6800 μF 10 V
Dia 16mm x 32mm high
21
Full Wave Rectifier
• Four diodes can make a full-wave rectifier.
• Two diodes conduct every half cycle – both positive and negative.
• Now have forward voltage drop of two diodes.
AC plugpack
12 VAC +
RMS
50 Hz
22
Waveforms
AC plugpack
12 VAC
RMS
50 Hz
20
15
+
&
D
10
Voltage (V)
5
Input
0
0
0.02
0.04
0.06
0.08
0.1
Output
Capacitor
-5
-10
+
-15
-20
⑤
Time (s)
23
Ripple Voltage
• Because the capacitor is recharged every half cycle, the ripple
voltage is halved. Sinc Dt=10ms , 27 100H1
• But there are now two diode voltage drops
=
.
17
i
∆v =
2 fC
-
16.5
16
15.5
15
14.5
14
0
0.005
0.01
0.015
0.02
0.025
24
Dual Half-Wave Rectifier
• We can create both positive and negative supplies by using two
half-wave rectifiers.
• This is very useful for powering op-amps.
⑧
AC plugpack
12 VAC
RMS
+17 V
+
b
- 17 V
25
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2.
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All survey responses are reported after grades
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so teaching staff will not be able to identify
individual students).
Voltage Regulation
• Most circuits require a “clean” DC power supply to operate
correctly
– Battery voltage changes due to discharge.
– Rectified AC voltage still has ripples.
– Elements switching on and off can change current and hence affect
voltage.
• A voltage regulator ensures a DC power supply acts like an
ideal voltage source.
– Stable voltage no matter what the current.
Zener Diode
• A Zener Diode has a specific reverse breakdown
voltage.
• Allows current at a particular reverse voltage.
Vahisneg
Note the different symbol
,
iA
I
LA
+
A
-
v
Ar
-
.
-
K
va
isweg
.
vAR
:
->
Reverse breakdown
voltage: vZ
Forward voltage
=0
↓
iA
-
Clarence Zener
Zener Diode
>U
Zener Diode Regulator
• Find the voltage v across the Zener diode for vS = 6 V.
SV
10 Ω
6U
O
k
-
-
vS = 6 – 12 V
E
+
50
vZ = 5 V
v
+
*
0
&i
A
i
Voc
-bu
-
v
=
(5V, ..)
0
vZ
=
5V
=>
-
USC
boomA=
3
soch
Zener Diode Regulator
• Now find the voltage v across the Zener diode for vS = 12 V.
lartV10 Ω
au
vS = 6 – 12 V
-
+
5V
50
vZ = 5 V
v
+
↑i
i
Voc
=
-12U
:
-
Sw
I
-
- ise:
-
0-7A
*
v
vZ
0
=
1-2A
Zener Diode Regulator
• Find the voltage v across the Zener diode for vS = 6 V, and then
for 12 V.
– With the ideal model, v is always –5 V.
i
10 Ω
vS = 6 – 12 V
+
vZ = 5 V
v
+
v
i
vZ
Zener Diode Regulator
• Find the current i through the diode for each case (vS = 6 V and
6-5V
vS = 12 V).
bU
When Vs=
IOM
,
Leaver
• Find the power dissipated
=0 1A
by the diode for each case.
12
Vs=
12-5W
10 Ω
,
-
--
-
--
~
zener
-
-
1O
vS = 6 – 12 V
+
vZ = 5 V
v
+
i
=-0
.
TA
Zener Diode Regulator
• Find the current i through the diode for each case (vS = 6 V and
vS = 12 V).
• Find the power dissipated
by the diode for each case.
v s + iR + v = 0
10 Ω
->
-
=
vS = 6 – 12 V
+
vZ = 5 V
v
+
#
↑i
• For vS = 6 V
6 + 10i − 5 = 0 ⇒ i = −0.1 A
P = vi = 0.5 W
• For vS = 12 V
12 + 10i − 5 = 0 ⇒ i = −0.7 A
P = vi = 3.5 W
-
Effect of a Load
• Consider the voltage regulator with a 6 V input and a 1 Ω load.
Will the voltage across the load resistor be 5 V? If not, what
rute
divider
will it be?
Voltage
bu Dion*
i
O
vS = 6 V
+
~75
12
10 Ω
v
+
zi
=
0
z
vZ = 5 V
V25V
X i
at
1Ω
Ole
Effect of a Load
• If there was 5 V across the load, there would be 5 A through
the load. That would mean a 50 V drop across the 10 Ω
resistor.
– Voltage across resistor ≠ 5 V.
• If the diode is not
conducting, the circuit
becomes:
10 Ω
vS = 6 V
+
10 Ω
v
+
vZ = 5 V
i
1Ω
vS = 6 V
+
1Ω
v = 6/11 V
Maximum Current
• Find the maximum current iL that the ideal diode can regulate.
For U
in
12 70 mA
50
=
2
is
ic
.
6-5V
=-
IOR
->
0
.
/A
is 10 Ω
Rs
vS = 6 V
+
iL
-
bi>0
v
+
I
:A
.
vZ = 5 V
i
0
issum
I
RL
?
in
0
.
1A
Maximum Current
• Find the maximum current iL that the ideal diode can regulate.
i s Rs < (v s − v z )
For a small negative i, is ≈ i L
is 10 Ω
Rs
vS = 6 V
+
iL
∴ i L Rs < (v s − v z )
v
+
vZ = 5 V
i
RL
?
iL <
(v s − v z )
Rs
i L < 100 mA
Moral of the Story …
• Be careful of power dissipation in a Zener voltage regulator,
particularly if you want to drop large voltages.
• You can reduce power dissipation by increasing the series
resistance, BUT, increasing the series resistance reduces the
current you can regulate.
• A zener shunt regulator always draws power from the source,
regardless of load requirement.
• In practice, always allow some tolerance in these values, as
Zener diodes are not ideal.
Real Zener Diodes
• Maximum power rating, and voltage range available:
– BZX79 series = 500mW, 2.4V to 75V.
– 1N5333 series = 5W, 3.3V to 200V.
• Available voltages follow preferred series with some special values also
available. Specified at a particular test current (eg 1mA) or power (eg
25% of rating).
• Voltage tolerances typically 5%, but 2% or even better are available.
• Voltage variation with temperature (Temp-co) and impedance (series
resistance) best at 5V – 6V.
NXP BZX79 series
I
I
-
Stop
I
riff
.
Shunt vs. Series regulators
Shunt regulator
Series regulator
RS
RS
vS
vL
+
RSh
RL
• Regulates the output voltage by
shunting excess current away from the
load.
• Simple, rugged, but
• Not generally efficient.
• Zener regulators are shunt regulators.
vS
+
vL
RL
• Regulates the output voltage by limiting
the load current by varying the series
element.
• More efficient, but
• Series element sees input to output
voltage difference.
• Most IC regulators (next slide) are series
regulators.
Linear Regulators
• Three-terminal linear regulators are available to regulate fixed
voltages of 0.5 V to 24V.
• Adjustable regulators also available (eg LM317)
• Current ratings typically 100mA, 1A, 3A.
• Include current limit & over-temp protection
LM7805
vi = 7 – 35 V
0.33 μF
0.1 μF
vo = 5 V
Linear Regulators
• Need input voltage at least 2V greater in magnitude than the
regulated output voltage.
• When input-output voltage difference may be less than 2V,
use a Low Drop-Out (LDO) regulator.
• Need capacitors as shown on input and output.
• Need to consider power dissipation.
Power Dissipation
• How much power will be dissipated by the regulator? P VI
• Hint 1: assume 7805 gnd current ≈ 0A (= 3mA).
(12 5) 5
• Hint 2: find the voltage drop and the current.
3 Ste
=
=
-
0
.
-
ou
->
0
.
5A
.
+
0.33 μF
-
-
TV
-
sw
-
LM7805
12
vi = 12 V
I aut
joct
in
·
↓
=
3mA
0.1 μF
vo = 5 V
20
10 Ω
-
5A
Power Dissipation
• How much power will be dissipated by the regulator?
• Hint 1: assume 7805 gnd current ≈ 0A (= 3mA).
• Hint 2: find the voltage drop and the current.
LM7805
vi = 12 V
0.33 μF
0.1 μF
vo = 5 V
10 Ω
P = vi = 7 × 0.5 = 3.5 W
Numbering Convention
• The "78xx" series (7805, 7812, etc.) regulate positive voltages
while the "79xx" series (7905, 7912, etc.) regulate negative
voltages.
• The last two digits are the output voltage; e.g., a 7805 is a +5 V
regulator, while a 7915 is a -15 V regulator.
vi = –17 to –35 V
0.33 μF
0.1 μF
LM7915
vo = – 15 V
Ripple Voltage
Power Supply Design
∆v =
i
fC
--
• Design a power supply to provide 5V @ 100mA from a 12 VAC
0-1A
i
RMS 50 Hz plugpack.
C =Dom
Tupk->
->
fav
100m
1TW
1&
IV
↓
55V
.......
-
↓
-
On
-
-
-
IOU
.
50
-
I
inI #
↓
-
⑧
-
->
D..
Iph
=
-
0 002x0
.
.
200pF
-
1
Ripple Voltage
Power Supply Design
∆v =
i
fC
• Design a power supply to provide 5V @ 100mA from a 12 VAC
RMS 50 Hz plugpack.
LM7805
AC plugpack
12 VAC +
RMS
50 Hz
C
0.33 μF
Peak voltage is 17V.
Regulator needs minimum 7V.
Maximum ripple voltage is 10 V.
0.1 μF
vo = 5 V
i
0.1
C>
=
= 200μF
f∆v 50 × 10
Choose C = 330 μF.
Op-amp as a regulator
• First … Three circuits to remember (Lecture 8)
– Inverting amplifier:
𝑣𝑣𝑜𝑜𝑜𝑜𝑜𝑜
𝑅𝑅2
= − 𝑣𝑣𝑖𝑖𝑖𝑖
𝑅𝑅1
– Non-inverting amplifier:
𝑣𝑣𝑜𝑜𝑜𝑜𝑜𝑜
R2
vin R1
vout
vin
vout
𝑅𝑅1 + 𝑅𝑅2
𝑅𝑅2
=
𝑣𝑣𝑖𝑖𝑖𝑖 = 1 +
𝑣𝑣
𝑅𝑅1
𝑅𝑅1 𝑖𝑖𝑖𝑖
R1
– Voltage follower:
𝑣𝑣𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑣𝑣𝑖𝑖𝑖𝑖
vin
R2
vout
49
Op-amp as a regulator
• Simplest version using a follower after a zener regulator.
SV
• What is the voltage across RL?
Shut regulator
qu
1 kΩ
vS = 9 V
+
=
4mAf
vZ = 5 V
Usage
O
=
vin
E
lower
vout
Vottage
regulator
Vi i 50
=
-
L
RL
Op-amp as a regulator
• Simplest version using a follower after a zener regulator.
• What is the voltage across RL? vL = vout = 5V
1 kΩ
vS = 9 V
vin
vout
iL
+
vZ = 5 V
RL
Op-amp as a regulator
• Add gain to get adjustable output voltage
• What resistor value will set vout = 5V?
Nominating
Amp
-
1 kΩ
vS = 9 V
+
lissing
↓
vin
-ins
-
R =30k
vout -5 i
L
To
1252
vZ = 1.25 V
.
Ri
18282
Rz
?? kΩ
10 kΩ
,
RL
Op-amp as a regulator
• Add gain to get adjustable output voltage
• What resistor value will set vout = 5V?
1 kΩ
vS = 9 V
+
vZ = 1.25 V
• Non-inverting amplifier
vin
vout
iL
?? kΩ
RL
10 kΩ
 R 
R1 + R2
vin = 1 + 2 vin
R1
 R1 
R 

5 = 1 + 2 1.25
 10k 

 5
− 1 = 30kΩ
R2 = 10k 
 1.25 
vout =
Vont
Op-amp as a regulator
=
Vadj +102s
• LM317 adjustable regulator “floats”…
5V
• NB: This circuit is not examinable content.
-
vin
in
-
470x3
.
~1500
-
-
5V
V out
.
vout
iL
vS = 9 V
+
vZ = 1.25 V
↳
adj
RL
Op-amp as a regulator
for
Vout
Ret
• LT3080 family set by one resistor
Fion
=
in
out
vS = 9 V
Rset
5
500k
-
iset = 10 uA
+
.
.5x10
-
adj
50
YI
=
=
vin
=
LOMA
vout
iL
RL
Next Steps…
•
•
•
•
•
Final Quiz, Quiz 6 (due Sunday week 12 at 11:59pm)
Attend your Prac 6 Prac exam (week 12/13)
Attempt Practice exams
Attend next weeks review lectures
Attend next weeks practice exam tutorials