Diode Applications
Chapter 2
Overview
Power Supply Half-wave rectifiers Full-wave rectifiers Line regulation Limiter Clamper
Basic components– Transformer (not
shown)– Rectifier– Filter– Regulator
DC Power Supply
Or Full-wave rectifier
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 Rectifiers
Forward biased
Half-wave Rectifiers
Reverse biased
Output result
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-waveRectifier
Vp(out)Vp(in)
Vavg
2
Half-wave Rectifier - Example
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
Draw the output signal– Vp(out) = Vp(in) – 0.7– Vavg = 99.3/– What happens to the
frequency?
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 = 0.5Vsec = 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)/= 0.637 x Vp(out)
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
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.
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
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
n=Vsec/Vpri = 0.110:1Vp(sec) = (0.707)-1 x Vrms = 1.414(12)=17 VVp(out) = V(sec) – (0.7 + 0.7) = 15.6 V through D1&D2 PIV = Vp(out) + 0.7 = 16.3 V
120Vrms
Note: Vp-Vbr ; hence, always convert from rms to Vp
Bridge Full-wave Rectifier - Comparison
120Vr
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
-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) / VDCVr (pp) = (1/ fRLC) x Vp(unfiltered)VDC = (1 – 1/ fRLC) x Vp(unfiltered)
Filters
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
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:
DiodeCharacteristics (VRRM & IF(AV) )