09/03/2014
DC Voltage Component
(Average voltage)
• v(t) = VDC + vac(t)
Properties of electrical signals
• VDC is the voltage value displayed on a DC
voltmeter
Triangular waveform – DC component
Half-wave rectifier output
Full-wave rectifier output
Triangular waveform – AC component
vac(t) = v(t) – VDC = v(t) – Vm / 2
Effective Current
• Average power over a load resistor
• Effective current is given by
• Irms is the current value displayed on a AC
ammeter
1
09/03/2014
Effective Voltage
Example:
v(t) = A + B cos(wt)
Vrms=?
• Average power over a load resistor
• Effective voltage is given by
• Vrms is the voltage value displayed on a AC
voltmeter
Triangular waveform – RMS value
Half-wave rectifier output
RMS value
Full-wave rectifier output
RMS value
AC Triangular Waveform – RMS value
Half-wave rectifier output
AC component RMS value
Full-wave rectifier output
AC component RMS value
2
09/03/2014
Half Wave Rectifier
Half Wave Rectifier Output
Diodes convert AC to DC in a process called rectification.
The diode only conducts for one-half of the AC cycle. The remaining half is either all
positive or all negative. This is a crude AC to DC conversion.
The DC Voltage out of the diode :
VDC = 0.318Vm
where Vm = the peak voltage
PIV (PRV)
Because the diode is only forward biased for one-half of the AC
cycle, it is then also off for one-half of the AC cycle. It is important
that the reverse breakdown voltage rating of the diode be high
enough to withstand the peak AC voltage.
Full Wave Rectification
The rectification process can be improved by using more diodes in a Full Wave
Rectifier circuit.
Full Wave rectification produces a greater DC output.
PIV (PRV) > Vm
PIV = Peak Inverse Voltage
PRV = Peak Reverse Voltage
Vm = Peak AC Voltage
Half Wave Rectifier
PIV(half-wave rectifier) = Vm
Center–Tapped Transformer Rectifier Circuit
Full Wave Rectifier Circuits
There are two Full Wave Rectifier circuits:
• Center –Taped Transformer Rectifier
• Bridge Rectifier
Two diodes and a center-tapped transformer are required.
VDC = 0.636(Vm)
Note that Vm here is the transformer secondary voltage to the tap.
3
09/03/2014
Operation of the Center–Tapped Transformer
Rectifier Circuit
Bridge Rectifier Circuit
For the positive half of the AC cycle:
PIV(center-tapped) = 2Vm
For the negative half of the AC cycle:
Four diodes are required.
VDC = 0.636 Vm
Operation of the Bridge Rectifier Circuit
Rectifier Circuit Summary
For the positive half of the AC cycle:
PIV(bridge) = Vm
For the negative half of the AC cycle:
Note: Vm = peak of the AC voltage.
Practical Applications of Diode Circuits
Rectifier Circuits
Conversions of AC to DC for DC operated circuits
Battery Charging Circuits
Power Supply Diagram
Simple Diode Circuits
Protective Circuits against
Overcurrent
Polarity Reversal
Currents caused by an inductive kick in a relay circuit
Zener Circuits
Overvoltage Protection
Setting Reference Voltages
4
09/03/2014
Filter Circuits
Voltage Regulation
The amount of variation in DC output voltage due to varying loads from no-load to full-load
is called voltage regulation.
Voltage Regulation:
VNL = no-load voltage
VFL = full-load voltage
The output from the rectifier section is a pulsating DC. The filter circuit reduces the peak-topeak pulses to a small ripple voltage.
Ripple Factor
Half-wave Rectifier Ripple Factor
DC output:
AC Ripple output:
Ripple Factor:
(Note Vm is the peak rectifier output voltage.)
Full-wave Rectifier Ripple Factor
DC output:
After the filter circuit a small amount of AC is still remaining. The amount of ripple voltage
can be rated in terms of Ripple Factor (r).
AC Ripple output:
Ripple Factor:
Vr(rms) = RMS value of the AC ripple voltage
Types of Filter Circuits
The full-wave rectifier has a significantly lower ripple factor.
Capacitor Filter
• Capacitor Filter
• RC Filter
Capacitor Filter
5
09/03/2014
Capacitor Filter Output
Ripple Voltage with a Capacitor Filter
A capacitor significantly reduces the AC content of the rectified signal.
For light load (r < 6.5%), ripple voltage:
The larger the capacitor the smaller the ripple voltage.
DC Output with a Capacitor Filter
Diode Ratings with Capacitor Filter
A capacitor increases the DC output.
DC Output:
Note: Vn= peak rectified voltage
IDC is the load current in mA.
Ripple Factor with a Capacitor Filter
The size of the capacitor increases the current drawn through the diodes. The larger the
capacitance, the greater the amount of current.
Peak Current vs. Capacitance:
C = capacitance
V = change in capacitor voltage during charge/discharge
 t = the charge/discharge time
The capacitor reduces the ripple factor.
A smaller capacitor will reduce the peak current through the diodes.
Ripple factor for light load (r < 6.5%):
Exercise (Midterm 1, 2004-2005)
Output voltage: (a) small C (b) large C
Since the average current drawn from the supply must equal to the average diode
current during the charging period, the following relation can be used
6
09/03/2014
RC Filter Circuit
Additional RC Filter
Adding an RC section will further reduce the ripple voltage and decrease the surge current
through the diodes.
DC Operation
Since both capacitors are open-circuit
for DC operation, the resultant output
voltage is
AC Operation
Due to voltage divider action of the
capacitor AC impedance and the load
resistor, the AC component voltage over
the load is given by
where VDC is the voltage on C1
SIMPLIFICATION
Example: fripple = 50Hz, C2=10F, RL=2K
If RL >> XC e.g. RL ≥ 5XC then |Z′|  XC
Consequently, V'r(rms) could be written as
If RL >> XC and R >> XC (generally this won’t hold)
e.g. RL > 5XC and R > 6XC
then V'r(rms) may be written as
314  318, thus the assumption |Z| = XC holds as RL > 5XC
7
09/03/2014
Ripple Voltage in an RC Filter Circuit
Relation between r' and r
The ripple voltage is significantly reduced by the addition of an RC circuit.
Ripple Voltage:
Vr(rms) = ripple voltage after the RC filter
Vr(rms) = ripple voltage before the RC filter
R = resistor in the added RC filter
XC = reactance of the capacitor in the added RC filter
This RC filter gives better results for light loads as the capacitor filter. For heavy loads, it
shows high ripples and low voltage regulation
Example: C1=15F, R=500, C2= 10F, RL=5K, VDC=150V,
VAC (rms) =15V. (fmains=50Hz)
Ripples are 3 times smaller
– Find the DC and AC voltages over the load
– Find the ripple factors, %r and %r' values
– Find the voltage regulation factor %VR.
Ripple factor dropped by 3 times
13.6V voltage drop
 - filters
Resistor R in the RC filter is replaced by inductor L, as the DC
resistance Rl of the coil is small and but the AC reactance XL is
high.
Example: For the -filter shown in the figure, the output DC voltage and current
are given as 200V and 50mA. VDC(C1) = 220V, Vr(C1)= 12V (rms) and
the frequency of the ripple voltage fripple = 100Hz. In order to satisfy
r  0.02 (%2) find RL, Rl, L and C2.
NOTE: Rl denotes the DC resistance of the coil.
8