Upon completion of the chapter the student should be able to
State the operation and characteristics of diode rectifier.
Discuss the performance parameters and use different technique for analyzing and design of diode rectifier circuits.
Simulate different arrangement of diode rectifiers by using PSpice.
3 Overview Single-phase full wave rectifier R load R-L load Controlled R R-L Load continuous and discontinuous current mode Three-phase rectifier uncontrolled controlled
Single-phase half wave rectifier
Free wheeling diode
DEFINITION Converting AC (from mains or other AC source) to DC power by using power diodes or by controlling the firing angles of thyristors/controllable switches.
Basic block diagram
Input can be single or multi-phase (e.g. 3-phase).
Output can be made fixed or variable
DC welder DC motor drive Battery charger DC power supply HVDC
6 Root-Mean-Squares (RMS) 7 Root Mean Squares of f 8 Concept of RMS t Average of v0 9 Ideal RectifierSingle-Phase Half-Wave R-Load
Considering the diode is ideal the voltage at R-load during forward biased is the positive cycle of voltage source while for negative biased the voltage is zero.
10 Ideal RectifierSingle-Phase Half-Wave R-Load
We observe that
DC voltage is fixed at 0.318 or 31.8 of the peak value
RMS voltage is reduced from 0.707 (normal sinusoidal RMS) to 0.5 or 50 of peak value.
Half wave is not practical because of high distortion supply current. The supply current contains DC component that may saturate the input transformer
11 Example 1
Consider the half-wave rectifier circuit with a resistive load of 25 and a 60 Hz ac source of 110Vrms.
Calculate the average values of Vo and Io. Justify the significant value of Vo and Io.
Calculate the rms values of Vo and Io.
Calculate the average power delivered to the load.
12 Example 1 (Cont)
(i) The average values of Vo and Io are given by
In this case for the particular circuit possible dc output voltage obtained from the circuit is 49.52V and dc output current is 1.98A. That means for any dc application within this value this circuit can be used. 13 Example 1 (Cont) (iii) average power delivered to the load over one cycle
(ii) The rms value of the of Vo and Io
14 Example 2
For the half-wave rectifier the source is a sinusoid of 120Vrms at a frequency of 60Hz. The load resistor is 5. Determine
(i) the average load current
(ii) the average power absorbed by the load and
(iii) the power factor of the circuit.
15 Example 2 (Cont) (ii) The average power absorbed by the load
(i) The average load current
16 Example 2 (Cont) (iii) Power Factor Note The power factor at the input of the rectifier circuit is poor even for resistive load and decreases as triggering angle for controlled rectifier is delayed. 17 Half-wave with R-L load
Industrial load typically contain inductance as well as resistance.
By adding an inductor in series with the load resistance causes an increase in the conduction period of the load current hence resulting the half-wave rectifier circuit working under an inductive load.
That means the load current flows not only during Vs 0 but also for a portion of Vs This is due to
18 Half-wave with R-L load
Until certain time (VR (hence VL Vs-VR is positive) the current builds up and inductor stored energy increases.
At maximum of VR VsVR hence VL 0V.
Beyond this point VL becomes negative (means releasing stored energy) and current begins to decrease.
After T the input Vs becomes negative but current still positive and diode is still conducts due to inductor stored energy. The load current is present at certain period but never for the entire period regardless of the inductor size.
This will results on reducing the average output voltage due to the negative segment. The larger the Inductance the larger negative segment
19 Half-wave with R-L load
The point when the current reaches zero is when the diode turns off given by
20 Example 3
For half-wave rectifier with R-L load R100 L0.1H 377rad/s and Vs100V. Determine
An expression for the current in this circuit
The point where diode turns off
The average current
The rms current
The power absorbed by the R-L load and
The power factor
21 Example 3 (cont)
For parameter given
(ii) (diode stop) Using numerical root finding is found to be 3.50 rads or 201o (i) Current Equation 22 Example 3 (cont) v) Power absorbed by resistor
iii) Average current
iv) rms current vi) Power factor 23 Half-wave with R-C load
In some applications in which a constant output is desirable a series inductor is replaced by a parallel capacitor.
The purpose of capacitor is to reduce the variation in the output voltage making it more like dc.
The resistance may represent an external load while the capacitor is a filter of rectifier circuit.
24 Half-wave with R-C load
Assume the capacitor is
unchargedand as source
diode is forward biased
As diode is on the output voltage is the same as source voltage and capacitor charges.
Capacitor is charged to Vm as input voltage reaches its positive peak at t /2.
As source decreases after t /2 the capacitor discharges into load resistor. As diode is reversed biased the load is isolated from source and the output voltage (capacitive charge) decaying exponentially with time constant RC.
25 Half-wave with R-C load
The angle t is the point when diode turns off.
The diode will stay off until the capacitor and input voltages become equal again.
The effectiveness of capacitor filter is determined by the variation in output voltage or expressed as maximum and minimum output voltage which is peak-to-peak ripple voltage.
26 Half-wave with R-C load (Ripple Voltage)
if VVm and /2 then ripple can be approximated as
The output voltage ripple is reduced by increasing the filter capacitor C. Anyhow this results in a larger peak diode current.
27 Example 4
The half-wave rectifier has 120Vrms source at 60Hz R500 C100F and delay when diode turns on is given 48. Determine
The expression of output voltage
Peak diode current
Sketch and label the output waveform
Value of C as ripple voltage is 1 of Vm and hence find new under this condition.
28 Example 4 (cont)
For parameter given
(ii) Ripple Voltage (iii) Peak diode current (i) Output Voltage 29 Example 4 (cont) (iv) Waveform must be properly labeled according to data (v) Capacitor value 30 RL Source Load
To supply a dc source from an ac source
The diode will remain off as long as the voltage of ac source is less than dc voltage.
Diode starts to conduct at t. Given by
31 RL Source Load 32 Example 5
The RL half-wave rectifier has 120Vrms source at 60Hz R2 L20mH Vdc 100V with extinction angle given by 193o. Determine
The expression of current in the circuit
Power absorbed by resistor
Power absorbed by dc source
Power supplied by ac source
Draw the waveform
33 Example 5 (cont)
For parameter given
(i) Current Equation ii) Power absorbed by resistor 34 Example 5 (cont)
iii) Power absorbed by dc source
v) Power factor iv) Power supplied v) Waveform - Refer notes 35 Freewheeling Diode (FWD)
Note that previously discussed uncontrolled half-wave RL load rectifier allows load current to present at certain period (current decreasing by time since opposing negative cycle of input) hence reducing the average output voltage due to the negative segment.
In other word for single-phase half wave rectifier with R-L load the load (output) current is NOT CONTINUOUS.
A FWD (sometimes known as commutation diode) can be placed in parallel to RL load to make the load (output) current continuous.
36 Freewheeling Diode (FWD)
Note that both D1and D2 cannot be turned on at the same time.
For a positive cycle voltage source
D1 is on D2 is off
The voltage across the R-L load is the same as the source voltage.
For a negative cycle voltage source
D1 is off D2 is on
The voltage across the R-L load is zero.
However the inductor contains energy from positive cycle. The load current still circulates through the R-L path.
37 Freewheeling Diode (FWD)
negative cycle voltage source (cont)
But in contrast with the normal half wave rectifier negative cycle of FWD does not consist of supply voltage in its loop.
Hence the negative part of Vo as shown in the normal half-wave disappear.
Irms is determined from Fourier component of current
- same as uncontrolled RLoad Rectifier 38 Example 6
Uncontrolled R-L load rectifier has a problem of discontinuous load current. Suggest a solution to the problem by justifying your answer through its principles of operation and waveform.
Solution Operation of FWD and its waveform (refer notes) 39 Example 7
Determine the average load voltage and current and determine the power absorbed by the resistor in the FWD circuit where R2 and L25mH Vm100V 60Hz.
The average load voltage and current
40 Example 7 (cont) Fourier Impedance
The ac voltage amplitudes
Note angle note included in calculation 41 Example 7 (cont) Resulting Fourier Terms are as follows
42 The Controlled Half-wave Rectifier
Previously discussed are classified as uncontrolled rectifiers.
Once the source and load parameters are established the dc level of the output and power transferred to the load are fixed quantities.
A way to control the output is to use SCR instead of diode. Two condition must be met before SCR can conduct
The SCR must be forward biased (VSCR0)
Current must be applied to the gate of SCR
43 Controlled Half-wave R load
A gate signal is applied at t where is the delay/firing angle.
44 Example 8
Design a circuit to produce an average voltage of 40V across 100 load resistor from a 120Vrms 60 Hz ac source. Determine the power absorbed by the resistor and the power factor.
Briefly describe what happen if the circuit is replaced by diode to produce the same average output.
45 Example 8 (Cont)
In such that to achieved 40V average voltage the delay angle must be
If an uncontrolled diode is used the average voltage would be
That means some reducing average resistor to the design must be made. A series resistor or inductor could be added to an uncontrolled rectifier while controlled rectifier has advantage of not altering the load or introducing the losses
46 Controlled Half-wave R-L load
The analysis of the circuit is very much similar to that of uncontrolled rectifier.
47 Controlled Half-wave R-L load 48 Example 9
For controlled RL rectifier the source is 120Vrms at 60Hz R20 L0.04H delay angle is 45o and extinction angle is 217o. Determine
An expression for i(t)
Average current and voltage
Power absorbed by load
49 Example 9 (cont)
For parameter given
(i) Current Equation 50 Example 9 (cont) iv) Power absorbed by resistor
ii) Average current and voltage
v) Power factor iii) rms current
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