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Recall Lecture 7
• Rectification – transforming AC signal into a
signal with one polarity
– Half wave rectifier
• Full Wave Rectifier
– Center tapped
– Bridge
Rectifier Parameters
Relationship between the number of turns of a
step-down transformer and the input/output
voltages
𝑉𝑃
𝑉𝑆
=
𝑁1
𝑁2
The peak inverse voltage (PIV) of the diode is the peak value of the voltage
that a diode can withstand when it is reversed biased
Duty Cycle: The fraction of the wave cycle over which the diode is
conducting.
Duty Cycle: The fraction of the wave cycle over
which the diode is conducting.
EXAMPLE 1 – Half Wave Rectifier
Determine the currents and voltages of the half-wave rectifier circuit. Consider the
half-wave rectifier circuit shown in Figure.
Assume VB = 6V, R = 120 Ω , V = 0.6 V and vs(t) = 18.6 sin t.
Determine the peak diode current, maximum reverse-bias diode voltage, the fraction
of the wave cycle over which the diode is conducting.
-VR + VB + 18.6 = 0
VR = 24.6 V
- VR +
+
A simple half-wave battery charger circuit
This node must be
at least 6.6V
6V
The peak inverse voltage (PIV) of the diode is
the peak value of the voltage that a diode can
withstand when it is reversed biased
Type of Rectifier
PIV
Half Wave
Peak value of the input secondary voltage, vs (peak)
Full Wave :
Center-Tapped
2vs (peak)- V
Full Wave:
Bridge
vs (peak) - V
Example: Half Wave Rectifier
Given a half wave rectifier with input primary voltage, Vp = 80 sin t and the
transformer turns ratio, N1/N2 = 6. If the diode is ideal diode, (V = 0V),
determine the value of the peak inverse voltage.
1. Get the input of the secondary voltage:
𝑉𝑃
𝑉𝑆
=
𝑁1
𝑁2
80 / 6 = 13.33 V
2. PIV for half-wave = Peak value of the input voltage = 13.33 V
Example: Full Wave Rectifiers
Calculate the transformer turns ratio and the PIV voltages for each type of the full wave
rectifier
a) center-tapped
b) bridge
Assume the input voltage of the transformer is 220 V (rms), 50 Hz from AC main line
source. The desired peak output voltage is 9 volt; also assume diodes cut-in voltage =
0.6 V.
Solution: For the centre-tapped transformer circuit the peak voltage of the
transformer secondary is required
The peak output voltage = 9V
Output voltage, vo = vs - V
Hence, vs = 9 + 0.6 = 9.6V  this is peak value! Must change to rms value
Peak value = Vrms x 2
So, vs (rms) = 9.6 / 2 = 6.79 V
The turns ratio of the primary to each secondary winding is
The PIV of each diode: 2vs (peak) - V = 2(9.6) - 0.6 = 19.6 - 0.6 = 18.6 V
Solution: For the bridge transformer circuit the peak voltage of the transformer
secondary is required
The peak output voltage = 9V
Output voltage, vo= vs - 2V
Hence, vs = 9 + 1.2 = 10.2 V  this is peak value! Must change to rms value
Peak value = Vrms x 2
So, vs (rms) = 10.2 / 2 = 7.21 V
The turns ratio of the primary to each secondary winding is
The PIV of each diode: vs (peak) - V = 10.2 - 0.6 = 9.6 V
Filters

A capacitor is added in parallel with the
load resistor of a half-wave rectifier to
form a simple filter circuit. At first there
is no charge across the capacitor

During the 1st quarter positive
cycle, diode is forward biased, and C
charges up.

VC = VO = VS - V.

As VS falls back towards zero, and
into the negative cycle, the
capacitor discharges through the
resistor R. The diode is reversed
biased ( turned off)

If the RC time constant is large, the
voltage across the capacitor
discharges exponentially.
Filters

During the next positive cycle of the input
voltage, there is a point at which the input
voltage is greater than the capacitor
voltage, diode turns back on.

The diode remains on until the input
reaches its peak value and the
capacitor voltage is completely
recharged.
Vp
Quarter cycle;
capacitor
charges up
Capacitor discharges
through R since diode
becomes off
VC = Vme – t / RC
Input voltage is greater
than the capacitor
voltage; recharge before
discharging again
NOTE: Vm is the peak value of the capacitor voltage = VP - V
Since the capacitor filters out a large portion of the sinusoidal signal, it is called a
filter capacitor.
Ripple Voltage, and Diode Current
Vr = ripple voltage
Tp
Vr = VM – VMe -T’/RC
T’
where T’ = time of the
capacitor to discharge to its
lowest value
Vr = VM ( 1 – e -T’/RC )
Expand the exponential in
series,
Vr= ( VMT’) / RC
Figure: Half-wave rectifier with smoothing capacitor.
• If the ripple is very small, we can approximate T’ = Tp
• Hence for half wave rectifier
Vr = ( VMTp) / RC
Vr = VM / ( f RC)

For full wave rectifier
Vr = ( VM 0.5Tp) / RC
Vr = VM / ( 2 f RC)
MULTIPLE DIODE CIRCUITS
Example:
Cut-in voltage of each diode in the circuit shown in Figure is 0.65 V. If the input voltage VI = 5 V,
determine the value of R1 when the value of ID2 = 2ID1. Also find the values of VI , ID1 and ID2.
Assume that all diodes are forward-biased.
End of Chapter 3