Chapter 2
DIODE part 2
MENJANA MINDA KREATIF DAN INOVATIF
objectives
¾Diode with DC supply – circuit analysis serial & parallel
¾Di d
¾Diode applications ‐
li ti
th DC
the DC power supply & Clipper
l & Cli
¾Analysis & Design of rectifier with capacitor filter : ¾Analysis
& Design of rectifier with capacitor filter :
clamper
Diode circuit analysis
Diode circuit analysis
Series configuration
¾Voltage shift between input and output voltages in transfer characteristics.
¾The diode only conducts when v1 > Vγ.
Diode circuit analysis
Diode circuit analysis
Series configuration
1.
2.
3
3.
4.
Determine the state of the diode whether it is in ON or OFF state.
•The applied Voltage is matched with the arrow of the diode symbol or not.
y
•The VD > Vknee for the diode to be operated.
Substitue the equivalent circuit for ON or OFF diode.
Perform KVL to find node voltages
Perform KVL to find node voltages Perform Ohm’s Law to find current
Diode circuit analysis
Diode circuit analysis
Series configuration
F
Forward
d bias
bi di
diode:
d
¾Using practical model diode, assume rd=0Ω:
¾VK for Si = 0.7V and Ge = 0.3V
¾If voltage source (E) > knee voltage(Vk), diode is assume replaced by a battery 0.7V for Si or 0.3V for Ge. ( E = Vs )
¾hence, using
•Kirchoff's voltage law (KVL): E = VR + VD •IID=IIR ,, Ohm’s law :V
Ohm s law :VR = IIRR
R = IIDR
Diode circuit analysis
Diode circuit analysis
Series configuration
Reverse bias diode:
- VD +
- VD +
Current flow is approximately 0A (ID=0A)
Using
Kirchoff's voltage law (KVL): E = -V
VD + VR
E = -VD + ID R where ID= 0 A
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 1
Find VD, VR and IR in the circuit below.
Si
0.7V
VR
2k
3V
2k
IR
3V
+
VR
IR
-
Using KVL,
3 V = 0.7 V + VR
VR = 2.3 V
23V
VD = 0.7 V
Using Ohm’s Law,
2.3V
V
= 1.15mA
=
IR =
2kΩ
R
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 2
Find VD, VR and IR in the circuit below.
Si
+
3V
VR
3V
2k
2k
3V
IR
Since it is an open circuit, IR = 0
Using Ohm’s Law, VR = IRR = 0
U i KVL VD = 3V
Using KVL, V
3V
+
VR
IR
-
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 3
Find VR and I
Find V
and IR in the circuit below.
in the circuit below
Si
Ge
+12V
VR
5.6k
IR
0.7V
0.3V
+12V
5 6k
5.6k
+
VR
IR
-
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 3 cont..
Using KVL,
12 V = 0.7 V + 0.3 V + VR
VR = 11 V
Using Ohm’s
Law I = V
Law,
R
R
11V
=
5.6kΩ
= 1.96mA
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 4
+ V1 -
Si
Vo
+10V
Find V1, V2 and
Vo in the circuit.
4.7k
+
V2
2.2k
-
+ V1 -
-5V
Vo
0.7V
+10V
4.7k
+
V2
2.2k
-
-5V
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 4 cont..
Using KVL,
10 V = V1 + 0.7 V + V2 – 5
V1 + V2 = 10 – 0.7 + 5
= 14.3 V
Using Ohm’s Law to get current
V +V
I= 1 2
R1 + R2
14.3V
=
4 .7 k + 2 .2 k
14.3V
=
6.9kΩ
= 2.07 mA
Therefore,
V1 = IR1
= 2.07 mA x 4.7 kΩ
= 9.73 V
V2 = IR2
= 2.07 mA x 2.2 kΩ
2 07 mA x 2 2 kΩ
= 4.55 V
V2 = Vo + 5 = 4.55 V
Vo = 4.55 – 5
= ‐0.45 V
0 45 V
Diode circuit analysis
Diode circuit analysis
Series configuration
Example 5
-- Find Vo1 and Vo2 in the figure below
Ge
-10V
10V
Si
Vo1
Vo2
1.2k
3 3k
3.3k
Solution:
V 1 = -9
Vo1
9V
Vo2 = - 6.6 V
Diode circuit analysis
Diode circuit analysis
Parallel configuration
Diode circuit analysis
Diode circuit analysis
Parallel configuration
Example 1
Calculate the value of current when the diode is ideal.
Diode circuit analysis
Diode circuit analysis
Combination Series & Parallel configuration
The
The DC power supply
DC power supply
The basic function of a DC power supply is to convert an AC voltage to a smooth DC voltage.
The
The DC power supply
DC power supply
.
¾The diode only conducts when it is in forward bias, hence only half of the AC cycle ¾Th
di d
l
d t h it i i f
d bi h
l h lf f th AC
l
passes through the diode.
¾The diode is OFF during the negative cycle. ¾Diode is reverse biased, and is therefore open circuit. No output during this cycle
Half wave rectifier
Half wave rectifier
•A half wave rectifier(ideal) allows conduction for only 180° or half of a complete cycle
a complete cycle.
• The output frequency is the same as the input. •The average voltage V
Th
lt
VDC or V
VAVG
V AVG = V d . c =
Vp
π
= 31 . 8 % V p
The rms voltage
V rms =
Vp
2
Half‐wave rectifier operation. The diode is considered to be ideal.
Half wave rectifier
Half wave rectifier
PIV
•Peak inverse voltage = is the maximum voltage when it is in reverse bias.
•The diode must be capable of withstanding this amount of voltage.
PIV = Vp
Full wave rectifier
Full wave rectifier
¾A full‐wave rectifier allows current to flow during both the positive and negative half cycles or the full 360º.
negative half cycles or the full 360
¾Note that the output frequency is twice the input frequency
¾The average voltage VDC or VAVG
¾The rms voltage
V rms =
V AVG = V d . c =
Vp
2 2
2V p
π
= 63 . 7 % V p
Full wave rectifier
Full wave rectifier Center tap
This method of rectification employs two diodes connected to a center‐tapped transformer.
The peak output is only half of the transformer’s peak secondary voltage.
Full wave rectifier
Full wave rectifier Note the current flow direction Note
the current flow direction
during both alternations. Being that it is center tapped, the peak output i b
is about half of the secondary h lf f h
d
windings total voltage. Each diode is subjected to a PIV of the full secondary winding output minus one diode voltage drop.
g
p
PIV = V p ( in ) − 0 . 7V
or
PIV = 2V p ( out ) + 0 . 7V
V p ( out ) =
V p ( in )
2
− 0 . 7V
Center tap
Full wave bridge rectifier
Full wave bridge rectifier The full‐wave bridge rectifier ta es ad a tage o t e u
takes advantage of the full output of the secondary winding.
It employs four diodes arranged such that current flows in the same direction
flows in the same direction through the load during each half of the cycle. Full wave bridge rectifier
Full wave bridge rectifier PIV
V p ( out ) = V p ( in ) − 1 . 4V
PIV = V p ( out ) + 0 . 7V
The PIV for a bridge rectifier is approximately half the PIV for a center
The
PIV for a bridge rectifier is approximately half the PIV for a center‐tapped
tapped rectifier.
rectifier
Note that in most cases we take the diode drop into account.
Summary of rectifier circuit
Summary of rectifier circuit
rectifier
Ideal(VDC)
Practical(VDC)
Half wave rectifier
VDC=0.318Vp
VDC=0.318Vp - 0.7
Full wave bridge
rectifier
ifi
VDC=0
0.636V
636Vp
VDC=0
0.636V
636Vp - 2(0.7)
2(0 7)
Full wave center tap
rectifier
VDC=0.636Vp
VDC=0.636Vp - 0.7
Vp = peak of the AC voltage.
Power supply filter & regulator
Power supply filter & regulator
The output of a rectifier is a pulsating DC. With filtration and regulation this pulsating voltage can be smoothed out and kept to a steady value.
Power supply filter & regulator
Power supply filter & regulator
A capacitor‐input filter will charge and p
p
g
discharge such that it fills in the “gaps” between each peak.
This reduces variations of voltage. The remaining voltage variation is called The
remaining voltage variation is called
ripple voltage.
Power supply filter & regulator
Power supply filter & regulator
The advantage of a full‐wave rectifier over a half‐wave is quite clear.
The capacitor can more effectively reduce the ripple when the time between peaks is shorter.
Power supply filter & regulator
Power supply filter & regulator
Vr(p-p)
Vr( p− p) =
V pT
RLC
=
Vp
fR L C
Th
The rms voltage,
lt
V r ( rms ) =
The average voltage, V DC
Ripple factor,
Vr( p− p)
2 3
1
= V p − V r ( p − p ) ≈ V p ( since V r ( p − p ) << V p )
2
V rms
1
r =
=
V DC
2 3 fR L C
P ki
Peak inverse voltage , PIV =V
l
PIV Vp
Power supply filter & regulator
Power supply filter & regulator
Vr(p-p)
Vr( p− p) =
V pT
RLC
=
Vp
fR L C
Th
The rms voltage,
lt
V r ( rms ) =
Vr( p− p)
2 3
1
The average voltage, V DC = V p − V r ( p − p ) ≈ V p ( since V r ( p − p ) << V p )
2
V rms
1
r =
=
Ripple factor,
l f
V DC
4 3 fR L C
PID =V
PID
=Vp (center tap); (center tap);
PID = 2Vp (bridge)
The
The DC power supply
DC power supply
Diode conducts current for only small portion of the period
Δt
1
=
T
π
2V r
VM
Power supply filter & regulator
Power supply filter & regulator
Being that the capacitor appears as a short during the
appears as a short during the initial charging, the current through the diodes can momentarily be quite high.
t il b
it hi h
To reduce risk of damaging the diodes, a surge current limiting resistor is placed in series with the filter and load.
Power supply filter & regulator
Power supply filter & regulator
Regulation is the last step in eliminating the remaining ripple and maintaining the output voltage to a specific value Typically this
maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current y
yp
g
requirements.
The
The DC power supply
DC power supply
How well the regulation is performed by a regulator is
measured by it’s regulation percentage. There are two
db i ’
l i
Th
types of regulation, line and load. Line and load
regulation percentage is simply a ratio of change in
regulation percentage is simply a ratio of change in
voltage (line) or current (load) stated as a percentage.
Line Regulation = (VOUT/VIN)100%
Load Regulation = (VNL – VFL)/VFL)100%
Summary
Diode Limiters
Diode Limiters
Limiting circuits limit the positive or negative amount of an input voltage to a specific value
input voltage to a specific value.
This positive limiter will limit the output to VBIAS + 0.7V
p
p
Diode Limiters
Diode Limiters
The desired amount of limitation can be attained by a power
pp y
g
pp
j
supply or voltage divider. The amount clipped can be adjusted
with different levels of VBIAS.
This positive limiter will limit
the output to VBIAS + 0.7V
The voltage divider provides the
VBIAS . VBIAS =(R3/R2+R3)VSUPPLY
Summary – clipper(series)
Summary –
Summary Summary – clipper(parallel)
Summary –
Summary Diode clampers
Diode clampers
A diode clamper adds a DC level to an AC voltage. The capac to c a ges to t e
capacitor charges to the peak of the supply minus the diode drop. Once charged the capacitor acts
charged, the capacitor acts like a battery in series with the input voltage. The AC voltage will “ride” along
voltage will “ride” along with the DC voltage. The polarity arrangement of the diode determines h d d d
whether the DC voltage is negative or positive.
Clampers
Clampers Clampers
Example 1
Diode Forward Biased
–ON
•C Charges to peak
value V in direction
shown
•V0 = 0 [V‐V = 0]
•V0 = 0 [V‐V = 0]
The
The DC power supply
DC power supply
Example 1
•• Diode reverse biased –
Diode reverse biased – OFF
• V0 equals voltage across C
and battery –2V
Clamper circuit
Clamper circuit
A diode
di d and
d
capacitor can be
combined to
“clamp” an AC
signal
g
to a specific
p
DC level
Bias clamper circuit
Bias clamper circuit
Bias clamper circuit
Bias clamper circuit
Example 2
•Diode forward biased – ON state
•C charges to peak value equals (20+25) volts
•Vo = 5 Volts
Bias clamper circuit
Bias clamper circuit
Example 2
•Diode reverse biased – OFF state
•Vo = ((25+10)) volts
Bias clamper circuit
Bias clamper circuit
Example 2
Bias clamper circuit
Bias clamper circuit
The input signal can be any type of waveform such as sine, square, triangle. ti l
The DC source adjusts the DC The
DC source adjusts the DC
clamping level
Voltage multiplier
Voltage multiplier
Clamping action can be used to increase peak rectified voltage. Once C1 and C2 charges to the peak voltage they act like two batteries in series, effectively doubling the voltage output. The current capacity for voltage multipliers is low.
Voltage multiplier
Voltage multiplier
The ffull-wave
Th
ll
voltage
lt
doubler
d bl arrangementt off di
diodes
d and
d
capacitors takes advantage of both positive and negative
peaks to charge the capacitors giving it more current
capacity. Voltage triplers and quadruplers utilize three and
four diode-capacitor arrangements respectively.
Summary
Summary clamper circuit
clamper circuit
Summary
•Filtering and Regulating the output of a rectifier helps keep the DC voltage smooth and accurate
the DC voltage smooth and accurate.
• Limiters are used to set the output peak(s) to a given value.
p p ()
g
• Clampers are used to add a DC voltage to an AC voltage.
• Voltage Multipliers allow a doubling, tripling, or quadrupling of rectified DC voltage or low current
quadrupling of rectified DC voltage or low current applications.