1. Diode Applications
Chapter 2

2. Overview Power Supply Half-wave rectifiers Full-wave rectifiers
Line regulation
Limiter
Clamper

3. DC Power Supply Basic components Transformer (not shown) Rectifier
Filter
Regulator
Or Full-wave rectifier

4. 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

5. Peak-to-Peak / Average / RMS
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
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

6. Half-wave Rectifiers
Forward biased

7. Half-wave Rectifiers
Reverse biased
Output result

8. Half-wave Rectifier For Silicon VBar = 0.7 V
Note that the frequency stays the same
Strength of the signal is reduced
Vavg = Vp(out)/p = 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 2p

9. Half-wave Rectifier - Example
Draw the output signal
Vp(out) = Vp(in) – 0.7
Vavg = 99.3/p
What happens to the
frequency?
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
Output:

10. 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

11. Half-wave Rectifier - 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

12. Full-wave Rectifier Note that the frequency is doubled
Vavg = 2Vp(out)/p = x Vp(out)

13. 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

14. 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

15. 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

16. 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

17. Full-wave Rectifier - Multisim
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.

18. 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

19. 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
120Vrms
n=Vsec/Vpri = 0.110:1
Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 V
Vp(out) = V(sec) – ( ) = 15.6 V through D1&D2
PIV = Vp(out) = 16.3 V
Note: Vp-Vbr ; hence, always convert from rms to Vp

20. Bridge Full-wave Rectifier - Comparison
120Vrms
Vp(2)=Peak secondary voltage ; Vp(out) Peak output voltage ; Idc = dc load current
Make sure you understand this!

21. Filters and Regulators

22. Filters

23. Filters

24. Filters

25. Filters
Ripple voltage depends on voltage variation across the capacitor
Large ripple means less effective filter
peak-to-peak ripple voltage

26. 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)

27. Diode Limiting What is Vout? Vout+ = Vin (RL)/(RL+R1) = 9.09
Forward biased when positive
Reverse biased when negative,
hence voltage drop is only -0.7
So how can we change the offset?

28. 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?

29. Diode Limiter
When the input signal is positive D1 is reversed biased; acting as positive limiter
Pos. Limiter
+VBIAS+0.7
-VBIAS-0.7

30. 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:

31. Diode Characteristics (VRRM & IF(AV) )