Chapter 2 MENJANA MINDA KREATIF DAN INOVATIF

Transcription

1 Chapter 2 DIODE part 2 MENJANA MINDA KREATIF DAN INOATIF

2 objectives Diode with DC supply circuit analysis serial & parallel Diode d applications the DC power supply & Clipper Analysis & Design of rectifier with capacitor filter : Analysis & Design of rectifier with capacitor filter : clamper

3 Diode circuit analysis Series configuration oltage shift between input and output voltages in transfer characteristics. The diode only conducts when v1 > γ.

4 Diode circuit analysis Series configuration 1. Determine the state of the diode whether it is in ON or OFF state. The applied oltage is matched with the arrow of the diode symbol or not. The D > knee for the diode to be operated. 2. Substitue the equivalent circuit for ON or OFF diode. 3. Perform KL to find node voltages 4. Perform Ohm s Law to find current

5 Diode circuit analysis Series configuration Forward bias diode: d Using practical model diode, assume r d =0Ω: K for Si = 0.7 and Ge = 0.3 If voltage source (E) > knee voltage(k), diode is assume replaced by a battery 0.7 for Si or 0.3 for Ge. ( E = s ) hence, using Kirchoff's voltage law (KL): E = R + D II D =I R, Ohm s law : R = I R R = I D R

6 Diode circuit analysis Series configuration Reverse bias diode: - D + - D + Current flow is approximately 0A (I D =0A) Using Kirchoff's voltage law (KL): E = - D + R E = - D + I D R where I D = 0 A

7 Diode circuit analysis Example 1 Find D, R and I R in the circuit below. Si 0.7 Series configuration 3 IR R 2k 3 IR 2k + R - Using KL, 3 = R R = D = 0.7 I R = R Using Ohm s Law, 2.3 = =1.15mA R 2kΩ k Ω

8 Diode circuit analysis Example 2 Series configuration Find D, R and I R in the circuit below. Si IR R 2k 3 IR 2k + R - Since it is an open circuit, I R = 0 Using Ohm s Law, R = I R R = 0 Ui Using KL, D = 3

9 Example 3 Find R and I R in the circuit below. Diode circuit analysis Series configuration +12 Si Ge R IR 5.6k IR 56k 5.6k + R -

10 Diode circuit analysis Example 3 cont.. Using KL, 12 = R R = 11 Using Ohm s Law, I R = R 11 = 5.6kΩ =1.96mA Series configuration

11 Diode circuit analysis Example 4 Series configuration Find 1, 2 and o in the circuit k Si 2.2k o o 4.7k 2.2k

12 Diode circuit analysis Example 4 cont.. Using KL, 10 = = = 14.3 Using Ohm s Law to get current I = R 1 + R = 4.7k + 2.2k Series configuration Therefore, 1 = IR 1 = 2.07 ma x 4.7 kω = = IR 2 = 207mA 2.07 x 22kΩ 2.2 = = o + 5 = 4.55 = 6.9kΩ o = = 2.07mA =

13 Diode circuit analysis Series configuration Example 5 -- Find o1 and o2 in the figure below -10 Ge Si o1 o2 1.2k 33k 3.3k Solution: o1 = -9 o2 = - 6.6

14 Diode circuit analysis Parallel configuration

15 Diode circuit analysis Example 1 Parallel configuration Calculate the value of current when the diode is ideal.

16 Diode circuit analysis Combination Series & Parallel configuration

17 The DC power supply The basic function of a DC power supply is to convert an AC voltage to a smooth DC voltage.

18 The DC power supply. The diode d only conducts when it is in forward bias, hence only half of the AC cycle passes through the diode. The diode is OFF during the negative cycle. Diode is reverse biased, and is therefore open circuit. No output during this cycle

19 Half wave rectifier A half wave rectifier(ideal) allows conduction for only 180 or half of a complete cycle. The output frequency is the same as the input. The average voltage DC or AG p AG = d. c = = 31.8% π The rms voltage = rms p 2 p Half wave rectifier operation. The diode is considered to be ideal.

20 Half wave rectifier PI Peak inverse voltage = is the maximum voltage when it is in reverse bias. The diode must be capable of withstanding this amount of voltage. PI = p

21 Full wave rectifier A full wave rectifier allows current to flow during both the positive and negativehalf cycles orthefull 360º. Note that the output frequency is twice the input frequency The average voltage DC or AG The rms voltage = rms 2 p AG = d. c = = 63.7% π p 2 2 p

22 Full wave rectifier Center tap This method of rectification employs two diodes connected to a center tapped transformer. The peak output is only half of the transformer s peak secondary voltage.

23 Full wave rectifier Center tap Note the current flow direction during both alternations. Being that it is center tapped, the peak output is about hlf half of the secondary windings total voltage. Each diode is subjected to a PI of the full secondary winding output minus one diode voltage drop. PI PI = = p ( in ) 0. 7 or 2 p ( out ) p ( in ) = p ( out ) 7

24 Full wave bridge rectifier The full wave bridge rectifier takes advantage a of the full output of the secondary winding. It employs four diodes arranged such that current flows in the same direction through the load during each half of the cycle.

25 PI Full wave bridge rectifier p ( out ) = p ( in ) 1. 4 PI = p ( out ) The PI for a bridge rectifier is approximately half the PI for a center tapped rectifier The PI for a bridge rectifier is approximately half the PI for a center tapped rectifier. Note that in most cases we take the diode drop into account.

26 Summary of rectifier circuit rectifier Ideal(DC) Practical(DC) Half wave rectifier DC=0.318p DC=0.318p -0.7 Full wave bridge rectifier Full wave center tap rectifier DC= p DC= p - 2(0.7) DC=0.636p DC=0.636p -0.7 p = peak of the AC voltage.

27 Power supply filter & regulator The output of a rectifier is a pulsating DC. With filtration and regulation this pulsating voltage can be smoothed out and kept to a steady value.

28 Power supply filter & regulator A capacitor input p filter will charge and discharge such that it fills in the gaps between each peak. This reduces variations of voltage. The remaining voltage variation is called The remaining voltage variation is called ripple voltage.

29 Power supply filter & regulator The advantage of a full wave rectifier over a half wave is quite clear. The capacitor can more effectively reduce the ripple when the time between peaks is shorter.

30 Power supply filter & regulator r(p-p) T p r ( p p ) = = R LC fr The rms voltage, r ( p p ) The average voltage, Ripple factor, p L r ( rms r = C ) = r DC = p r p p ) p (since r ( p p ) << 2 rms 1 = 2 3 fr DC Peak kinverse voltage, PI =p L ( p C )

31 Power supply filter & regulator T p r ( p p ) = = R LC fr p L C r(p-p) r The rms voltage, r ( p p ) The average voltage, Ripple factor, r ( rms ) = DC = p r p p ) p (since r ( p p ) << 2 rms 1 r = = 4 3 fr C DC L ( p ) PID = (center tap); PID =p (center tap); PID = 2p (bridge)

32 The DC power supply Diode conducts current for only small portion of the period Δt 1 2 = T π r M

33 Power supply filter & regulator Being that the capacitor appears as a short during the initial charging, the current through the diodes can momentarily be quite high. h To reduce risk of damaging the diodes, a surge current limiting resistor is placed in series with the filter and load.

34 Power supply filter & regulator Regulation is the last step in eliminating the remaining ripple and maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current requirements.

35 The DC power supply How well the regulation is performed by a regulator is measured by i it s regulation percentage. There are two types of regulation, line and load. Line and load regulation percentage is simply a ratio of change in voltage (line) or current (load) stated as a percentage. Line Regulation = (OUT/IN)100% Load Regulation = (NL FL)/FL)100%

36 Summary

37 Diode Limiters Limiting circuits limit the positive or negative amount of an input voltage to a specific value. This positive limiter will limit the output to BIAS + 0.7

38 Diode Limiters The desired amount of limitation can be attained by a power supply or voltage divider. The amount clipped can be adjusted with different levels of BIAS. This positive limiter will limit the output to BIAS The voltage divider provides the BIAS. BIAS =(R3/R2+R3)SUPPLY

39 Summary clipper(series)

40 Summary clipper(parallel)

41 Diode clampers A diode clamper adds a DC level to an AC voltage. The capacitor charges gesto the peak of the supply minus the diode drop. Once charged, thecapacitor acts like a battery in series with the input voltage. The AC voltage will ride along with the DC voltage. The polarity arrangement of the diode d determines whether the DC voltage is negative or positive.

42 Clampers

43 Clampers Example 1 Diode Forward Biased ON C Charges to peak value in direction shown 0 = 0 [ = 0]

44 Example 1 The DC power supply Diode reverse biased OFF Diode reverse biased OFF 0 equals voltage across C and battery 2

45 Clamper circuit Adi diode d and capacitor can be combined to clamp an AC signal to a specific DC level

46 Bias clamper circuit

47 Bias clamper circuit Example 2 Diode forward biased ON state C charges to peak value equals (20+25) volts o = 5 olts

48 Example 2 Diode reverse biased OFF state o = (25+10) volts Bias clamper circuit

49 Example 2 Bias clamper circuit

50 Bias clamper circuit The input signal can be any type of waveform such as sine, square, ti triangle. The DC source adjusts the DC The DC source adjusts the DC clamping level

51 oltage multiplier Clamping action can be used to increase peak rectified voltage. Once C1 and C2 charges to the peak voltage they act like two batteries in series, effectively doubling the voltage output. The current capacity for voltage multipliers is low.

52 oltage multiplier The full-wave voltage doubler arrangement of diodes d and capacitors takes advantage of both positive and negative peaks to charge the capacitors giving it more current capacity. oltage triplers and quadruplers utilize three and four diode-capacitor arrangements respectively.

53 Summary clamper circuit

54 Summary Filtering and Regulating the output of a rectifier helps keep the DCvoltage smooth and accurate. Limiters are used to set the output peak(s) to a given value. Clampers are used to add a DC voltage to an AC voltage. oltage Multipliers allow a doubling, tripling, or quadrupling of rectified DCvoltage or low current applications.