Diode Applications Chapter 2

```Diode Applications
Chapter 2
Overview
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Power Supply
Half-wave rectifiers
Full-wave rectifiers
Line regulation
Limiter
Clamper
DC Power Supply
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Basic components
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Transformer (not
shown)
Rectifier
Filter
Regulator
Or Full-wave rectifier
Sine Wave
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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.
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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
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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
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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 Rectifiers
Forward biased
Half-wave Rectifiers
Reverse biased
Output result
Half-wave Rectifier
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Note that the frequency stays the same
Strength of the signal is reduced
Vavg = Vp(out)/p = 0.318 x Vp(out) [31.8 % of Vp]
Vp(out) = Vp(in) – VBar
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For Silicon VBar = 0.7 V
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Vp(in)
Vp(out)
Half-wave
Rectifier
Vavg
2p
Half-wave Rectifier - Example
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Draw the output signal
Vp(out) = Vp(in) – 0.7
– Vavg = 99.3/p
– What happens to the
frequency?
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Peak Inverse Voltage (PIV)
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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
Output:
Transformers (Review)
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Transformer: Two inductors coupled together – separated by a dielectric
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When the input magnetic field is changing voltage is induced on the
second inductor
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The dot represents the + (voltage direction)
Applications:
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Step-up/down
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Isolate sources
Turns ratio (n)
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n = Sec. turns / Pri. turns = Nsec/Npri
Vsec = n. Vpri
depending on value of n : step-up or step-down
Center-tapped transformer
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Voltage on each side is Vsec/2
Half-wave Rectifier - Example
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Example:
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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.
78-0.7
n = ½ = 0.5
Vsec = n.Vpri = 78 V
Vout = Vsec – 0.7 = 77.3 V
Full-wave Rectifier
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Note that the frequency is doubled
Vavg = 2Vp(out)/p = 0.637 x Vp(out)
Full-wave Rectifier Circuit
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Center-tapped full-wave rectifier
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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
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Center-tapped full-wave rectifier
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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
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Vout = Vsec /2 – 0.7
Peak Inverse Voltage (PIV)
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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
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Assuming a center-tapped transformer
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Find the turns ratio
Find Vsec
Find Vout
Find PIV
Draw the Vsec and Vout
What is the output freq?
n=1:2=0.5
Vsec=n*Vpri=25
Vout = Vsec/2 – 0.7
PIV = Vsec-0.7=24.3 V
Vsec
Full-wave Rectifier - Multisim
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XFrmr can be virtual or real
Use View Grapher to see the details of your
results
The wire-color can determine the waveform
color
Make sure the ground is connected to the
scope.
Bridge Full-wave Rectifier
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Uses an untapped transformer  larger Vsec
Four diodes connected creating a bridge
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When positive voltage 
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When negative voltage 
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D3 and D4 are forward biased
Two diodes are always in series with the load
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D1 and D2 are forward biased
Vp(out) = Vp(sec) – 1.4V
The negative voltage is inverted
The Peak Inverse Voltage (PIV)
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PIV=Vp(out)+0.7
Bridge Full-wave Rectifier - Example
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Assume 12 Vrms secondary voltage for the standard 120 Vrms
across the primary
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Find the turns ratio
Find Vp(sec)
Show the signal direction when Vin is positive
Find PIV rating
120Vr
ms
n=Vsec/Vpri = 0.110:1
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
Note: Vp-Vbr ; hence, always
convert from rms to Vp
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!
Filters and Regulators
Filters
Filters
Filters
Filters
-Ripple voltage depends on voltage variation across the capacitor
- Large ripple means less effective filter
peak-to-peak
ripple voltage
Filters
Too much ripple is bad!
Ripple factor = Vr (pp) / VDC
Vr (pp) = (1/ fRLC) x Vp(unfiltered)
VDC = (1 – 1/ fRLC) x Vp(unfiltered)
Diode Limiting
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What is Vout?
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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
Negative limiter
Remember:
When positive voltage  reverse biased  No current  no clipping!
Positive limiter
What if we mix these together?
Diode Limiter
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When the input signal is positive D1 is
reversed biased; acting as positive limiter
Pos. Limiter
+VBIAS+0.7
-VBIAS-0.7
Diode Clamper
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It adds a dc level
When the input voltage is negative, the
capacitor is charged
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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:
Diode
Characteristics (VRRM & IF(AV) )
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