ECE 121 - Electronics (1) Lecture 3 - BU
Transcript of ECE 121 - Electronics (1) Lecture 3 - BU
ECE 121 - Electronics (1) Lecture 3
Dr. Said Emamhttp://bu.edu.eg/staff/saidemam3
CHAPTER OUTLINE
2โ1 Diode Operation2โ2 Voltage-Current (V-I) Characteristics of a Diode2โ3 Diode Models2โ4 Half-Wave Rectifiers2โ5 Full-Wave Rectifiers2โ6 Power Supply Filters and Regulators2โ7 Diode Limiters and Clampers2โ8 Voltage Multipliers2โ9 The Diode Datasheet2โ10 Troubleshooting
CHAPTER OBJECTIVES
โ Use a diode in common applicationsโ Analyze the voltage-current (V-I) characteristic of a diodeโ Explain how the three diode models differโ Explain and analyze the operation of half-wave rectifiersโ Explain and analyze the operation of full-wave rectifiersโ Explain and analyze power supply filters and regulatorsโ Explain and analyze the operation of diode limiters and clampersโ Explain and analyze the operation of diode voltage multipliersโ Interpret and use diode datasheetsโ Troubleshoot diodes and power supply circuits
OR GATE Determine ๐๐ for the network of Fig. 2.39.
10-V level is assigned a โ1โ
0-V level is assigned a โ0.โ
AND GATE Determine the output level for the positive logic AND gate of Fig. 2.42 .
10-V level is assigned a โ1โ
0-V level is assigned a โ0.โ
Half-Wave Rectifier Operationโข The process of removing one-half the input signal to establish a dc level is called half-wave
rectification.
Using the ideal model for the diode:
During the positive half cycle:
โข ๐ฃ๐๐ is positive the diode is forward-biased (ON - closed switch).
โข The output voltage has the same shape as the positive half-cycle of the input
voltage. ๐ฝ๐๐๐ = ๐ฝ๐๐
During the negative half cycle:
โข ๐ฃ๐๐ is negative the diode is reverse-biased.
โข the output voltage is 0 ๐ (๐ฝ๐๐๐ = ๐๐ฝ)
โข Since the output does not change polarity it is a pulsating dc voltage with a frequency equal the input frequency
HWR: (๐๐๐๐ = ๐๐๐)
Average Value of the Half-Wave Output Voltageโข The average value of the half-wave rectified output voltage is the value you
would measure on a dc voltmeter.
โข Mathematically, it is determined by finding the area under the curve over a full cycle, and then dividing by 2๐ (the number of radians in a full cycle).
๐ฝ๐จ๐ฝ๐ฎ ๐๐ ๐ฝ๐ ๐ =๐ฝ๐ท๐
= ๐. ๐๐๐ ๐ฝ๐ท
Notice that ๐๐ด๐๐บ is 31.8% of ๐๐.
Effect of the Barrier Potential on the Half-Wave Rectifier Output(Using the practical diode model)
โข When ๐ฝ๐๐ < ๐. ๐ the diode is OFF (O.C.) and (๐ฝ๐๐๐ = ๐๐ฝ)
โข When ๐ฝ๐๐ > ๐. ๐ the diode is ON and (๐ฝ๐๐๐ = ๐ฝ๐๐ โ ๐. ๐)
โข The peak output ๐ฝ๐ท ๐๐๐ = ๐ฝ๐ท(๐๐) โ ๐. ๐
๐๐ด๐๐บ โ๐๐ โ 0.7
๐When ๐๐ โซ 0.7 ๐๐พ = 0.7 ๐
EXAMPLE1 Draw the output voltages of each rectifier for the indicated input voltages. The 1N4001 and 1N4003 are specific rectifier diodes.
Example2 a. Sketch the output ๐๐ and determine the dc level of the output.b. Repeat part (a) if the ideal diode is replaced by a silicon diode.c. Repeat parts (a) and (b) if ๐ฝ๐ is increased to 200 V.
Solution (a)
โข During the positive half cycle the diode is R.B.(O.C.) and ๐ฃ๐ = 0
โข During the negative half cycle the diode is F.B.(S.C.) and ๐ฃ๐ = ๐ฃ๐๐
โข ๐๐๐ = โ๐๐
๐= โ0.318 ร 20 = โ6.36๐
(b)For a silicon diode, the output has the appearance of Fig. below , and๐๐๐ โ โ0.318 ๐๐ โ 0.7 ๐ = โ0.318 19.3 ๐ = โ๐. ๐๐ ๐ฝ
The resulting drop in dc level is 0.22 V, or about 3.5%.
(c) For ๐๐ = 200๐
Ideal: ๐๐๐ = โ๐๐
๐= โ0.318 ร 200 = โ๐๐. ๐ ๐
Practical: ๐๐๐ โ โ0.318 ๐๐ โ 0.7 ๐ = โ0.318 199.3 ๐ = โ๐๐. ๐๐ ๐ฝ
which is a difference that can certainly be ignored for most applications
Peak Inverse Voltage (PIV)
โข The peak inverse voltage (PIV) equals the peak value of the input voltage, and the diode must be capable of withstanding this amount of repetitive reverse voltage.
โข The peak inverse voltage (PIV) rating of the diode is the voltage rating that must not be exceeded in the reverse-bias region or the diode will enter the breakdown region.
โข A diode should be rated at least 20% higher than the PIV for better performance.
๐๐ผ๐ ๐๐๐ก๐๐๐ โฅ ๐๐(๐๐)
๐๐ผ๐ = ๐๐(๐๐)
Half-wave rectifier with transformer coupled input voltage.
๐ ๐๐ข๐๐๐ ๐๐๐ก๐๐ =๐๐ ๐๐๐๐๐๐
๐๐ ๐๐ = ๐๐๐๐๐
If ๐ < 1 step down transformer and ๐๐ ๐๐ < ๐๐๐๐
If ๐ > 1 step up transformer and ๐๐ ๐๐ > ๐๐๐๐
The peak output of the rectifier is: ๐ฝ๐(๐๐๐) = ๐ฝ๐(๐๐๐) โ ๐. ๐
and ๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐)
EXAMPLE 3 Determine the peak value of the output voltage.
โข ๐ =1
2= 0.5
โข ๐๐(๐ ๐๐) = ๐๐๐(๐๐๐) = 0.5 ร 170 = 85 ๐
โข ๐ฝ๐(๐๐๐) = ๐ฝ๐(๐๐๐) โ ๐. ๐ = ๐๐. ๐ ๐ฝ
โข ๐ท๐ฐ๐ฝ = ๐ฝ๐ ๐๐๐ = ๐๐ ๐ฝ
2โ5 FULL-WAVE RECTIFIERSโข A full-wave rectifier allows unidirectional (one-way) current through the load
during the entire cycle of the input, whereas a half-wave rectifier allows current through the load only during one-half of the cycle.
โข FWR : ๐๐๐๐ = ๐ ๐๐๐๐๐๐ [twice the input frequency]
โข๐๐ด๐๐บ =2๐๐
๐= 0.637 ๐๐ [twice that of HWR]
โข During the +๐๐ half cycle D1 is ON and D2 is OFF
โข During the โ๐๐ half cycle D1 is OFF and D2 is ON
โข The current passes through ๐ in the same direction in both half cycles
Center-Tapped Full-Wave Rectifier (CT-FWR)
๐ฝ๐๐๐ =๐ฝ๐๐๐๐
โ ๐. ๐
Effect of the Turns Ratio on the Output Voltageโข ๐ = 1
โข The total secondary voltage is: ๐๐(๐ ๐๐) = ๐๐(๐๐๐)
โข ๐ฝ๐(๐๐๐) =๐ฝ๐(๐๐๐)
๐โ ๐.๐
=๐ฝ๐(๐๐๐)
๐โ ๐. ๐
=๐ฝ๐(๐๐๐๐๐)
๐โ ๐. ๐
โข ๐ = 2
โข The total secondary voltage is: ๐๐(๐ ๐๐) = 2 ๐๐(๐๐๐)
โข ๐ฝ๐(๐๐๐) =๐ฝ๐(๐๐๐)
๐โ ๐. ๐
= ๐ฝ๐(๐๐๐) โ ๐. ๐= ๐ฝ๐(๐๐๐๐๐) โ ๐. ๐
๐๐๐ข๐ก =๐๐ ๐๐2
โ 0.7
Center-Tapped Full-Wave Rectifier (CT-FWR) (Similar configuration)
Network conditions for the positive region of ๐ฃ๐ .
Network conditions for the negative region of ๐ฃ๐.
โข During the +๐๐ half cycle D1 is ON and D2 is OFF
โข During the โ๐๐ half cycle D1 is OFF and D2 is ON
โข The current passes through ๐ in the same direction in both half cycles
Ideal : ๐ฝ๐๐๐ =๐ฝ๐๐๐
๐Practical :๐ฝ๐๐๐ =
๐ฝ๐๐๐
๐โ ๐. ๐
Peak Inverse Voltage
๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐) โ ๐. ๐
Example 4
(a) Show the voltage waveforms across each half of the secondary winding and across ๐ ๐ฟ
(b) Find the DC value of the output.
(c) What minimum PIV rating must the diodes have?
Solution (a) ๐ =1
2= 0.5
โข The total peak secondary voltage is:๐๐(๐ ๐๐) = ๐๐๐(๐๐๐) = 0.5 ร 100 = 50 ๐
โข Hence There is a ๐๐(๐ ๐๐)
2= 25 V peak across each
half of the secondary with respect to ground.
โข The load voltage has a peak value:
๐๐(๐๐ข๐ก) =๐๐(๐ ๐๐)
2โ 0.7 = 24.3 ๐
(b) ๐๐ด๐๐บ =2๐๐(๐๐ข๐ก)
๐= 0.637 ๐๐(๐๐ข๐ก) = 15.48 ๐
(c) Each diode must have a minimum PIV rating of
๐๐ผ๐ = ๐๐ ๐ ๐๐ โ 0.7 = 49.3 ๐
Bridge Full-Wave RectifierDuring the positive half-cycle
โข ๐ซ๐ and ๐ซ๐are forward-
biased and conduct current.
โข ๐ซ๐ and ๐ซ๐ are reverse-biased.
During the negative half-cycle
โข ๐๐ and ๐๐ are forward-
biased and conduct current.
โข ๐ซ๐ and ๐ซ๐are reverse-biased.
๐ฝ๐(๐๐๐) = ๐ฝ๐(๐๐๐) โ ๐. ๐
Practical:
๐ฝ๐(๐๐๐) = ๐ฝ๐(๐๐๐)
๐ฝ๐(๐๐๐) = ๐ฝ๐(๐๐๐) โ ๐. ๐
Practical:
Ideal:
Peak Inverse Voltage
๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐)
๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐) + ๐. ๐
Practical:
Ideal:
Solution
โข The peak secondary voltage is : ๐ฝ๐(๐๐๐) = ๐๐ฝ๐๐๐(๐๐๐) = ๐ ร ๐๐ = ๐๐ ๐ฝ
โข The peak output voltage is: ๐ฝ๐(๐๐๐) = ๐ฝ๐(๐๐๐) โ ๐.๐ = ๐๐. ๐ ๐ฝ
โข The ๐๐ผ๐ rating for each diode is : ๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐) + ๐. ๐ = ๐๐. ๐ ๐ฝ
Example 5
Determine the peak output voltage for the bridge rectifier in Figure. Assuming the practical model, what ๐๐ผ๐ rating is required for the diodes? The transformer is specified to have a 12 ๐ ๐๐๐ secondary voltage for the standard 120 V across the primary.
Bridge FWR (similar configuration)
During the positive half cycle
๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐)
๐ท๐ฐ๐ฝ = ๐ฝ๐(๐๐๐) + ๐. ๐Practical:
Ideal:
๐ฝ๐(๐๐๐๐) = ๐ฝ๐ โ ๐. ๐
The Basic DC Power Supply
โข The dc power supply converts the standard 220 V, 60 Hz ac voltage available at wall outlets into a constant dc voltage.
โข The voltage produced is used to power all types of electronic circuits including consumer electronics (televisions, DVDs, etc.), computers, industrial controllers, and most laboratory instrumentation systems and equipment.
โข The dc voltage level required depends on the application, but most applications require relatively low voltages.
โข The ac input line voltage is stepped down to a lower ac voltage with a transformer
โข A transformer changes ac voltages based on the turns ratio between the primary and secondary. If the secondary has more turns than the primary, the output voltage across the secondary will be higher and the current will be smaller.
โข If the secondary has fewer turns than the primary, the output voltage across the secondary will be lower and the current will be higher.
โข The rectifier can be either a half-wave rectifier or a full-wave rectifier.
โข The rectifier converts the ac input voltage to a pulsating dc voltage.
โข The filter eliminates the fluctuations in the rectified voltage and produces a relatively smooth dc voltage.
โข The regulator is a circuit that maintains a constant dc voltage for variations in the input line voltage or in the load.
The Basic DC Power Supply
Multisim
โข Open the Multisim file E02-03 in the Examples folder. For the inputs specified in the example, measure the resulting output voltage waveforms.
โข Open the Multisim file E02-01 in the Examples folder. Measure the voltages across the diode and the resistor in both circuits.
โข Open the Multisim file E02-07 in the Examples folder. Measure the output voltage and compare to the calculated value.
โข Open the Multisim file E02-04 in the Examples folder. For the specified input, measure the peak output voltage.
โข Open the Multisim file E02-06 in the Examples folder. For the specified input voltage, measure the voltage waveforms across each half of the secondary and across the load resistor.
Homework
โข Determine the output waveform for the network of Figure below and calculate the output dc level and the required PIV of each diode.
โข Determine ๐๐, calculate the output dc level and the required PIV rating of each diode for the configuration of Fig. below . In addition, determine the maximum current through each diode. (solve for ideal and practical diodes)
Homework