Download Lecture 9 - PIV, Filters and Multiple Diodes

Recall Lecture 8
• Full Wave Rectifier
• Center tapped
• Bridge
• Rectifier Parameters
• PIV
• Duty Cycle
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 = 19.2 - 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
Vm
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
which is the period of the input signal
• Hence for half wave rectifier
Vr = ( VMTp) / RC

For full wave rectifier
Vr = ( VM 0.5Tp) / RC
Example
Consider a full wave center-tapped rectifier. The capacitor is connected in
parallel to a resistor, R = 2.5 k. The input voltage has a peak value of 120 V
with a frequency of 60 Hz. The output voltage cannot be lower than 100 V.
Assume the diode turn-on voltage, V = 0.7 V. Calculate the value of the
capacitor.
VM = 120 – 0.7 = 119.3 V
Vr = 119.3 – 100 = 19.3 V
19.3 = 119.3 / (2*60*2500*C)
C = 20.6 F
Example
Consider a full wave bridge rectifier. The capacitor C = 20.3 F is connected in
parallel to a resistor, R = 10 k. The input voltage, vs = 50 sin (2(60)t).
Assume the diode turn-on voltage, V = 0.7 V. Calculate the value of the ripple
voltage.
Frequency = 60 Hz
VM = 50 – 1.4 = 48.6 V
Vr = 48.6 / (2*60*10x103*20.3x10-6)
Vr = 2 V
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 VO, ID1 and ID2 . Assume that all diodes are forwardbiased.