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Contents Diode circuit model
Diode in Series
Diode in Parallel
Half-wave rectifiers
Full-wave rectifiers
Power Supply
Line regulation
Limiter/clipper
Clamper
Diode Circuit Models
The Ideal Diode
Model
The diode is designed to allow current to flow in only one
direction. The perfect diode would be a perfect conductor in
one direction (forward bias) and a perfect insulator in the
other direction (reverse bias). In many situations, using the
ideal diode approximation is acceptable.
Example: Assume the diode in the circuit below is ideal. Determine the value
of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias)
+
_VA
ID
RS = 50 a) With VA > 0 the diode is in forward bias and
is acting like a perfect conductor so:
ID = VA/RS = 5 V / 50 = 100 mA
b) With VA < 0 the diode is in reverse bias and
is acting like a perfect insulator, therefore no
current can flow and ID = 0.
Series Diode Configurations
Constants
Silicon Diode: VD = 0.7 V
Germanium Diode: VD = 0.3 V
Analysis (for silicon)
VD = 0.7 V (or VD = E if E < 0.7 V)
VR = E – VD
ID = IR = IT = VR / R
4
Forward Bias
Load-Line Analysis
5
The load line plots all possible
combinations of diode current (ID)
and voltage (VD) for a given circuit.
The maximum ID equals E/R, and
the maximum VD equals E.
The point where the load line and
the characteristic curve intersect is
the Q-point, which identifies ID and
VD for a particular diode in a given
circuit.
Diode Circuit Models
The Ideal Diode with
Barrier Potential
This model is more accurate than the simple ideal diode
model because it includes the approximate barrier
potential voltage. Remember the barrier potential
voltage is the voltage at which appreciable current starts
to flow.
Example: To be more accurate than just using the ideal diode model include
the barrier potential. Assume V = 0.3 volts (typical for a germanium diode)
Determine the value of ID if VA = 5 volts (forward bias).
+
_VA
ID
RS = 50 With VA > 0 the diode is in forward bias and
is acting like a perfect conductor so write a
KVL equation to find ID:
0 = VA – IDRS - V
ID = VA - V = 4.7 V = 94 mA
RS 50
V+
V+
Diode Circuit Models
The Ideal Diode
with Barrier
Potential and
Linear Forward
Resistance
This model is the most accurate of the three. It includes a linear
forward resistance that is calculated from the slope of the linear
portion of the transconductance curve. However, this is usually
not necessary since the RF (forward resistance) value is pretty
constant. For low-power germanium and silicon diodes the RF
value is usually in the 2 to 5 ohms range, while higher power
diodes have a RF value closer to 1 ohm.
Linear Portion of
transconductance
curve
VD
ID
VD
IDRF = VD
ID
+V RF
Diode Circuit Models
The Ideal Diode
with Barrier
Potential and
Linear Forward
Resistance
Example: Assume the diode is a low-power diode with a
forward resistance value of 5 ohms. The barrier potential
voltage is still: V = 0.3 volts (typical for a germanium diode)
Determine the value of ID if VA = 5 volts.
+
_VA
ID
RS = 50
V+
RF
Once again, write a KVL equation for the circuit:
0 = VA – IDRS - V - IDRF
ID = VA - V = 5 – 0.3 = 85.5 mA
RS + RF 50 + 5
Diode Circuit Models
Values of ID for the Three Different Diode Circuit Models
Ideal Diode
Model
Ideal Diode
Model with
Barrier
Potential
Voltage
Ideal Diode
Model with
Barrier
Potential and
Linear Forward
Resistance
ID 100 mA 94 mA 85.5 mA
These are the values found in the examples on previous slides where
the applied voltage was 5 volts, the barrier potential was 0.3 volts and
the linear forward resistance value was assumed to be 5 ohms.
The Q Point
The operating point or Q point of the diode is the quiescent or no-signal
condition. The Q point is obtained graphically and is really only needed
when the applied voltage is very close to the diode’s barrier potential voltage.
The example 3 below that is continued on the next slide, shows how the Q
point is determined using the transconductance curve and the load line.
+
_VA
= 6V
ID
RS = 1000
V+
First the load line is found by substituting in different
values of V into the equation for ID using the ideal
diode with barrier potential model for the diode. With
RS at 1000 ohms the value of RF wouldn’t have much
impact on the results.
ID = VA – V
RS
Using V values of 0 volts and 1.4 volts we obtain IDvalues of 6 mA and 4.6 mA respectively. Next we will
draw the line connecting these two points on the
graph with the transconductance curve. This line is
the load line.
The Q Point
ID (mA)
VD (Volts)
2
4
6
8
10
12
0.2 0.4 0.6 0.8 1.0 1.2 1.4
The
transconductance
curve below is for a
Silicon diode. The Q
point in this example
is located at 0.7 V
and 5.3 mA.
4.6
0.7
5.3
Q Point: The intersection of the load
line and the
transconductance curve.
Series Diode Configurations
Diodes ideally behave as open circuits
Analysis
VD = E
VR = 0 V
ID = 0 A
12
Reverse Bias
Look example in monograph pages 23, 24
Parallel Configurations
13
mA 142
mA 28
D2I
D1I
mA 28.33kΩ
V .7V 10
R
DVE
RI
V 9.3R
V
V 0.7O
VD2
VD1
V
V 0.7D
V
Look example in monograph page 25
Sine Wave
The sine wave is a common type of alternating current (ac) and alternating voltage.
The time required for a sine wave to complete one full cycle is called the period (T).
Frequency ( f ) is the number of cycles that a sine wave completes in one second. The more cycles completed in
one second. The higher the frequency.
Frequency is measured in hertz(Hz)
Relationship between frequency ( f) and period (T) is:
f = 1/T
Peak-to-Peak / Average / RMS
The peak-to-peak value of a sine wave is the voltage or current from the positive peak to the negative peak.
The peak-to-peak values are represented as:
Vpp and Ipp
Where: Vpp = 2Vp and Ipp = 2Ip
The rms (root mean square) value of a sinusoidal voltage is equal to the dc voltage that produces the same amount of heat in a resistance as does the sinusoidal voltage.
Vrms = 0.707Vp
Irms = 0.707Ip
Half-Wave Rectification
16
The diode only
conducts when it is
forward biased,
therefore only half
of the AC cycle
passes through the
diode to the
output.
The DC output voltage is 0.318Vm, where Vm = the peak AC voltage.
Half-wave Rectifier
Note that the frequency stays the same
Strength of the signal is reduced
Vavg = Vp(out)/ = 0.318 x Vp(out) [31.8 % of Vp]
Vp(out) = Vp(in) – VBar
For Silicon VBar = 0.7 V
Half-wave
Rectifier
Vp(out)Vp(in)
Vavg=0.318Vm
2
PIV (PRV)
18
Because the diode is only forward biased for one-half of the AC cycle, it is
also reverse biased for one-half cycle.
It is important that the reverse breakdown voltage rating of the diode be
high enough to withstand the peak, reverse-biasing AC voltage.
PIV (or PRV) > Vm
• PIV = Peak inverse voltage
• PRV = Peak reverse voltage
• Vm = Peak AC voltage
Half-wave Rectifier - Example
• Draw the output signal– Vp(out) = Vp(in) – 0.7– Vavg = 99.3/– What happens to thefrequency?
Output:
Peak Inverse Voltage (PIV)
– The peak voltage at which the
diode is reverse biased
– In this example PIV = Vp(in)-
– Hence, the diode must be rated
for PIV = 100 V
Transformers (Review)
Transformer: Two inductors coupled together – separated by a dielectric
When the input magnetic field is changing voltage is induced on the
second inductor
The dot represents the + (voltage direction)
Applications:
Step-up/down
Isolate sources
Turns ratio (n)
n = Sec. turns / Pri. turns = Nsec/Npri
Vsec = n. Vpri
depending on value of n : step-up or step-down
Center-tapped transformer
Voltage on each side is Vsec/2
Half-wave Rectifier - Example Example:
– Assume that the input is a sinusoidal signal with Vp=156 V & T = 2
msec; assume Nsec:Npri = 1:2
– Draw the signal
– Find turns ratio;
– Find Vsec;
– Find Vout.
n = ½ = 0.5
Vsec = n.Vpri = 78 V
Vout = Vsec – 0.7 = 77.3 V
78-0.7
Full-wave Rectifier Note that the frequency is doubled
Vavg = 2Vp(out)/ = 0.636 x Vp(out)
Full-Wave Rectification
23
• Half-wave: Vdc = 0.318Vm
• Full-wave: Vdc = 0.636Vm
The rectification process can be improved by
using a full-wave rectifier circuit.
Full-wave rectification produces a greater
DC output:
Full-Wave Rectification
24
There are 2 types of full-wave rectifiers circuit connection as below:
Full-Wave Rectification
25
Center-Tapped Transformer Rectifier
Requires
• Two diodes
• Center-tapped transformer
VDC = 0.636Vm
Full-wave Rectifier Circuit
Center-tapped full-wave rectifier
Each half has a voltage = Vsec/2
Only one diode is forward biased at a time
The voltages at different halves are opposite of each other
Full-wave Rectifier Circuit
Center-tapped full-wave rectifier
Each half has a voltage = Vsec/2
Only one diode is forward biased at a time
The voltages at different halves are opposite of each other
Full-wave Rectifier Circuit
• Vout = Vsec /2 – 0.7
• Peak Inverse Voltage (PIV)
– PIV = (Vsec/2 – 0.7)- (-Vsec/2) = Vsec – 0.7
• Vout = Vsec/2 – 0.7
Assuming D2 is
reverse-biased
No current through D2
Full-wave Rectifier - Example
Assuming a center-tapped transformer
Find the turns ratio
Find Vsec
Find Vout
Find PIV
Draw the Vsec and Vout
What is the output freq? Vsec
n=1:2=0.5
Vsec=n*Vpri=25
Vout = Vsec/2 – 0.7
PIV = Vsec-0.7=24.3 V
Bridge Full-wave Rectifier Uses an untapped transformer larger Vsec
Four diodes connected creating a bridge When positive voltage
D1 and D2 are forward biased
When negative voltage D3 and D4 are forward biased
Two diodes are always in series with the load Vp(out) = Vp(sec) – 1.4V
The negative voltage is inverted
The Peak Inverse Voltage (PIV) PIV=Vp(out)+0.7
Full-Wave Rectification
31
Bridge Rectifier
Four diodes are connected in a bridge configuration
VDC = 0.636Vm
Summary of Rectifier Circuits
32
Vm = peak of the AC voltage.
In the center tapped transformer rectifier circuit, the peak AC voltage
is the transformer secondary voltage to the tap.
Rectifier Ideal VDC Realistic VDC
Half Wave Rectifier VDC = 0.318Vm VDC = 0.318Vm – 0.7
Bridge Rectifier VDC = 0.636Vm VDC = 0.636Vm – 2(0.7 V)
Center-Tapped Transformer
RectifierVDC = 0.636Vm VDC = 0.636Vm – 0.7 V
Bridge Full-wave Rectifier - Comparison
12
0Vr
ms
Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc
load current
Make sure you understand this!
Simulation using PSPICE You can download the Orcad Pspice software
at http://www.cadence.com/products/orcad/pages/downloads.aspx#pspice.
OrCAD PCB Designer Lite DVD (All Products)
The file size is about 1GB and need to fill the form.
Example on how to draw a simple circuit on PSPICESimulation on PSPICE involves 4 steps
1. Creating a new project
2. Drawing schematic diagram
3. Creating simulation profile
4. Running simulation
1. Creating a new project On the Windows start menu , All programs >>
CADENCE >> ORCAD >> ORCAD CAPTURE CIS
In the program, File >> New >> Project
A new window will appear
1. Creating a new project
In New Project window,
Enter the Name and Location
Choose “Analog or Mixed A/D”
Click “OK”
A new dialog will appear
Choose “Create a blank project”
Click “OK”
A new schematic window will appear as follows.
2. Drawing schematic diagram (Half wave rectifier)
• Place the components
• Press Place Part icon located in the upper right side of the tool palette. (shortcut key for Place Part is P)
• Choose all libraries (if you can’t find the libraries, click
Add Library and choose all libraries files in Pspice folder
• Find the parts listed below and add to the windows by left clicking.
Press “Esc” to quit the place part mode.
Parts for half wave rectifier circuit
• VSIN (voltage source)
• 1n4148 (diode)
• R (load resistor)
2. Drawing schematic diagram (Continued) Place the ground by clicking “Place Ground” on the
right tool palette (shortcut: G) and choose “0/CAPSYM”.
Arrange the parts as follows
Connect the parts with wire. (Place wire or shortcut: w)
2. Drawing schematic diagram (Continued) Change the properties of
voltage source by double clicking on each properties
VOFF=0
VAMPL = 10V
FREQ = 50Hz
3. Creating simulation profile• From the top toolbar
PSPICE>>New simulation profile
• Type in the Name, inherited from “None” and click “Create”
• Simulation Settings windows will appear• Choose “Time Domain” in
Analysis type• Set Run to time: 50ms and
Max step size 0.0001• Click OK
4. Runing Simulation Place voltage probe from
the top probe toolbar as shown in the figure.
Click on the Run PSpice icon from the toolbar.
A new simulation window will open showing the plots of two probes.
Bridge Rectifier Simulation To create a new schematic page
to simulate bridge rectifier , click on New Document button at the toolbar.
Draw the Bridge rectifier circuit as shown in the diagram as in step 2
Repeat steps 3 and 4 above guidelines to simulate the circuit.
In the simulation, use Voltage differential markers across Voltage source instead of single Voltage marker .
Bridge Rectifier Simulation (Continued) Full wave rectifier simulation results
Bridge Full-wave Rectifier - Example Assume 12 Vrms secondary voltage for the standard 120 Vrms across the
primary Find the turns ratio
Find Vp(sec)
Show the signal direction when Vin is positive
Find PIV rating
Solution:
n=Vsec/Vpri = 0.110:1
Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V
Vp(out) = V(sec) – (0.7 + 0.7) = 15.6 V through D1&D2
PIV = Vp(out) + 0.7 = 16.3 V
120Vr
ms
Note: Vp-Vbr ; hence, always
convert from rms to Vp
Filters and Regulators
Filters
Filters
Filters
-Ripple voltage depends on voltage variation across the capacitor
- Large ripple means less effective filter
Filters
peak-to-peak
ripple voltage
Too much ripple is bad!
Ripple factor = Vr (pp) / VDC
Vr (pp) = (1/ fRLC) x Vp(unfiltered)
VDC = (1 – 1/ fRLC) x Vp(unfiltered)
Filters
FiltersPlease refer your filter notes in the Electronics Monograph
Diode Clippers
55
The diode in a series clipper “clips”
any voltage that does not forward
bias it:•A reverse-biasing polarity
•A forward-biasing polarity less than 0.7 V (for a silicon diode)
Biased Clippers
56
Adding a DC source in
series with the clipping
diode changes the
effective forward bias of
the diode.
Parallel Clippers
57
The diode in a parallel clipper
circuit “clips” any voltage that
forward bias it.
DC biasing can be added in
series with the diode to change
the clipping level.
Summary of Clipper Circuits
58
Summary of Clipper Circuits
59
more…
Diode Limiting What is Vout?
Vout+ = Vin (RL)/(RL+R1) = 9.09
Vout- = -0.7
Forward biased when positive
Reverse biased when negative,
hence voltage drop is only -0.7
So how can we change the offset?
Diode Limiting – Changing the offset
What if we mix these together?
Positive limiter
Negative limiter
Remember:
When positive voltage reverse biased No current no clipping!
Diode Limiter When the input signal is positive D1 is reversed biased;
acting as positive limiter
Pos. Limiter
-VBIAS-0.7+VBIAS+0.7
Clampers
63
A diode and capacitor can be
combined to “clamp” an AC
signal to a specific DC level.
Biased Clamper Circuits
64
The input signal can be any type
of waveform such as sine, square,
and triangle waves.
The DC source lets you adjust
the DC camping level.
Diode Clamper
It adds a dc level
When the input voltage is negative, the capacitor is charged Initially, this will establish a positive dc
offset
Note that the frequency of the signal stays the same
RC time constant is typically much larger than 10*(Period)
Note that if the diode and capacitor are flipped, the dc level will be negative
Output:
Summary of Clamper Circuits
66
Zener Diodes
When Vi VZ
The Zener is on
Voltage across the Zener is VZ
Zener current: IZ = IR – IRL
The Zener Power: PZ = VZIZ
When Vi < VZ
The Zener is off
The Zener acts as an open circuit
67
The Zener is a diode operated
in reverse bias at the Zener
Voltage (Vz).
Zener Resistor Values
68
ZKRL I II min
min
max L
ZL
I
VR
min
max
L
Z
L
LL
R
V
R
V I
Zi
ZL
VV
RVR min
If R is too large, the Zener diode cannot conduct because the available amount of
current is less than the minimum current rating, IZK. The minimum current is
given by:
The maximum value of resistance is:
If R is too small, the Zener current exceeds the maximum current
rating, IZM . The maximum current for the circuit is given by:
The minimum value of resistance is:
Voltage-Multiplier Circuits
Voltage Doubler
Voltage Tripler
Voltage Quadrupler
69
Voltage multiplier circuits use a combination of diodes and
capacitors to step up the output voltage of rectifier circuits.
Voltage Doubler
70
This half-wave voltage doubler’s output can be calculated by:
Vout = VC2 = 2Vm
where Vm = peak secondary voltage of the transformer
Voltage Doubler
71
• Positive Half-Cycle
o D1 conducts
o D2 is switched off
o Capacitor C1 charges to Vm
• Negative Half-Cycle
o D1 is switched off
o D2 conducts
o Capacitor C2 charges to Vm
Vout = VC2 = 2Vm
Voltage Tripler and Quadrupler
72
Practical Applications
Rectifier Circuits
Conversions of AC to DC for DC operated circuits
Battery Charging Circuits
Simple Diode Circuits
Protective Circuits against
Overcurrent
Polarity Reversal
Currents caused by an inductive kick in a relay circuit
Zener Circuits
Overvoltage Protection
Setting Reference Voltages
73
DiodeCharacteristics (VRRM & IF(AV) )
References
Robert Boylestad and Louis Nashelsky. Electronic Devices and
Circuit Theory, 10e Pearson 2009
www.ccsu.edu/.../ccsu/courses/cet323/lectures/Chapter2.ppt