Diode Applications Diode Switching
As we have already seen the diode can act as a switch – Forward biased or reverse biased - On or Off. Voltage Rectifier
A voltage rectifier is a circuit that converts an AC voltage to a DC voltage. AC Voltage
An AC voltage, as represented by a sine wave, is one where the polarity of the voltage changes (and the current flow reverses) in a periodic manner. This waveform has RMS, Peak, and Peak-Peak values. It also has an average value of 0 volts - VAvg = 0 V. Note: Average value is what a voltmeter reads when set to DC volts.
DC Voltage
A DC voltage is one where the polarity of the voltage does not change (and the current flow does not reverse) in a periodic manner. The diagram below left shows a constant DC voltage while at right there is a non-constant DC voltage.
Half Wave Rectifier
A Half Wave rectifier can be constructed using a single diode as shown in the circuit.
An AC voltage is connected to a transformer and the transformer output voltage is rectified by a single diode. The secondary voltage from the transformer has a voltage that can be described as VS,Pk. During the positive half cycle of the transformer secondary voltage, the diode is forward biased and turns ON. During the negative half cycle of the transformer secondary voltage, the diode is reverse biased and turns OFF. The waveform shown below is called the Half Wave Rectified output waveform for this circuit. The peak load voltage VL,Pk is the voltage across the load resistor.
Question
Does this Half Wave Rectified output waveform represent a DC or an AC voltage? Average Value
As the waveform is always positive, the Half Wave Rectified output load voltage waveform has a non-zero average value. This DC or average value can be calculated as
The peak load voltage V
L,Pk= V
S,Pk- 0.7 V
Example
1. A transformer with a turns ratio of 1:12 has a 120 VAC connected to the primary. The secondary voltage is connected to a Half Wave rectifier circuit with a 1 kΩ load resistor.
What is the RMS secondary voltage? What is the peak secondary voltage? What is the peak load voltage? What is the peak load current? What is the average load voltage?
Answers
RMS secondary voltage VS,RMS = VP x 1/12 = 120 V x 1/12 = 10 V RMS
Peak secondary voltage VS,Pk = 10 V x 1.41 = 14.1 V Pk √2 = 1.41 Peak load voltage VL,Pk = VS,Pk - 0.7 = 14.1 - 0.7 = 13.4 V Pk
Peak load current IL,Pk = VL,Pk/1 kΩ = 13.4 mA Pk
Average load voltage VAvg = VL,Pk/π = 13.4 V Pk/π = 4.3 V DC
Frequency of Load Voltage waveform
If the voltage connected to the primary of the transformer is 60 Hz then the frequency of Load Voltage waveform will also be 60 Hz.
Capacitive Filtering
The rectifier circuit has created a pulsating DC voltage at the load. This can be improved to a more constant DC voltage by the use of a filter capacitor. The capacitor charges up during the time when the diode is conducting and discharges during the time the diode is off. The amount of discharge depends on the value of the load resistor and the capacitor – the larger the values of R and C the smaller the amount of discharge will be and the filtered waveform will be closer to a constant value.
Ripple Voltage
The filtering action of the capacitor produces an output (load) voltage waveform that is called a Ripple Voltage. The ripple voltage is measured as a Peak-Peak value – Vripple,P-P. The ripple voltage is defined in terms of the maximum and minimum values of the ripple voltage waveform.
V
ripple,P-P= V
max- V
min The RMS value of the ripple voltage isV
ripple,RMS= V
ripple,P-P/ 2√3
The Average or DC value of the load voltage is
V
L,DC= V
L,Avg= (V
max+ V
min)/2
The frequency of the ripple voltage waveform is 60 Hz.
Example
2. A filtered output load voltage has a ripple voltage with Vmax = 13.4 V and Vmin = 11.2 V. What are the values of
V
ripple,P-P, V
ripple,RMSand V
L,Avg(V
L,DC).
V
ripple,P-P= V
max- V
min= 13.4 V - 11.2 V = 2.2 V P-P
V
ripple,RMS= V
ripple,P-P/ 2√3 = 2.2 V P-P/ 2√3 = 0.64 V RMS
V
L,DC= V
L,Avg= (V
max+ V
min)/2 = (13.4 V + 11.2 V)/2 = 12.3 V DC
Capacitor Size
If the size of the capacitor is increased the ripple voltage will decrease. At some point the capacitor will be sufficiently large that the average DC load voltage will increase until it is approximated by the peak load voltage. In practice capacitor size is not increased excessively as there are other ways to help reduce the size of the ripple voltage.
Percent Ripple
The value of the ripple voltage alone is not sufficient by itself to accurately describe the circuit
performance. A better descriptor is percent ripple which relates the ripple voltage to the average or DC load voltage.
Percent Ripple (% ripple) =
V
ripple,P-P/
V
L,DCx 100%
Consider these two examples.1.
V
L,DC= 5.0 V
V
ripple,P-P= 1.0 V P-P
% ripple = 1.0 V/
5.0 V x 100% = 20 %
2.
V
L,DC= 25 V
V
ripple,P-P= 1.0 V P-P
% ripple = 1.0 V/
25 V x 100% = 4 %
Which example has the better ripple performance?
Full Wave Rectifier – 2 diode circuit
The two drawings below show a full wave rectifier. You should satisfy yourself that they are equivalent representations. Note: The transformer for this rectifier circuit requires a center tap connection.
Full Wave Rectified Waveform
The full wave rectified load voltage waveform is shown. Each diode in the circuit contributes one half of the output waveform. The center tap of the transformer divides the secondary voltage in half so the voltage connected to each diode is VS,Pk/2.
The frequency of this waveform is double that of the half wave rectifier. If the primary frequency is 60 Hz then the frequency of the load waveform is 120 Hz.
Example
3. A transformer with a 1:4 turns ratio is connected to 120 VAC and a full wave rectifier (2 diode) circuit.
What is the RMS and Peak secondary voltage?
What is the peak load voltage? What is the average load voltage? Answers
VS = VP x 1/4 = 120 VAC x 1/4 = 30 V RMS = 42.2 V Pk VL,Pk = VS,Pk/2 - 0.7 = 42.4 V/2 - 0.7 = 21.2 - 0.7 = 20.5 V Pk VL,Avg = VL,DC = 2 VL,Pk/π = 2 x (20.5)/π = 13.1 V DC
The peak load voltage is now VL,Pk = VS,Pk/2 - 0.7 V
Full Wave Rectifier – Bridge Rectifier circuit
As with the 2 diode full wave rectifier the circuit for the bridge rectifier circuit can be represented in several but equivalent ways. Each circuit however requires 4 diodes connected in a “bridge circuit” arrangement.
Circuit Operation
On the positive half cycle of the secondary voltage two of the bridge diodes (D3 and D1) are forward biased and conduct as shown below left. On the negative half cycle of the secondary voltage two of the bridge diodes (D4 and D2) are forward biased and conduct as shown below right.
The frequency of this waveform is also double that of the half wave rectifier. If the primary frequency is 60 Hz then the frequency of the load waveform is 120 Hz.
The voltage connected to each diode is VS,Pk.
Example
4. A transformer with a 1:4 turns ratio is connected to 120 VAC and a full wave Bridge rectifier circuit.
What is the RMS and Peak secondary voltage? What is the peak load voltage?
What is the average load voltage? Answers
VS = VP x 1/4 = 120 VAC x 1/4 = 30 V RMS = 42.2 V Pk VL,Pk = VS,Pk - 1.4 = 42.4 V - 1.4 V = 41.0 V Pk
VL,Avg = VL,DC = 2 VL,Pk/π = 2 x (41.0)/π = 27.3 V DC
Bridge Rectifier Component
The bridge rectifier circuit configuration is very popular and rather than using 4 discrete diodes to build it, manufacturers make a device that incorporates the 4 diodes in a single component. These devices can often be attached to a heat sink and can handle larger
currents. Two examples are shown below.
Capacitive Filtering
As with the half wave rectifier a capacitor can be added (for either of the full wave rectifier circuits) in parallel with the load resistor to help “smooth” the rectified waveform.
The ideas and definitions developed earlier about the ripple voltage still apply to this waveform.
V
ripple,P-P= V
max- V
min The RMS value of the ripple voltage isV
ripple,RMS= V
ripple,P-P/ 2√3
The Average or DC value of the load voltage is
