Electronic Circuits. Diode Applications. Dr. Manar Mohaisen Office: F208 Department of EECE

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1 Electronic Circuits Diode Applications Dr. Manar Mohaisen Office: F208 Department of EECE

2 Review of the Precedent Lecture Doping It is a controlled addition of impurities to the pure semiconductive material to increase its conductivity. The impurities addition is for target to increase either the holes or the free electrons. Si Free (conduction) electron from Sb atom Si Hole fro m B atom Si Sb Si Si B Si Si Si n-type semiconductor. The extra electron from the antimony atom becomes a free electron. The impurity element is a pentavalent. Impurities: Arsenic (As), antimony (Sb) p-type semiconductor An extra hole is created because the impurity element is a trivalent (i.e., it has 3 valence electrons) Impurities: Boron (B), indium (In), gallium (Ga)

3 Diode Approximations Review of the Precedent Lecture contd. Reverse Current = 0 I F After BIAS exceeds the barrier potential, the diode acts like a short circuit I F B = 0.7 Reverse Current 0 I F R F R 0.7 F R 0.7 F I R I R I R Ideal Diode Model -Reverse current, barrier potential, and dynamic resistance are neglected. Not accurate model. - Useful for troubleshooting to check whether the diode is worki ng properly or not. - Forward current is given by: I F = R BIAS LIMIT Practical Diode Model -Reverse current and dynamic resistance are neglected. -Forward current is given by: I F = R BIAS LIMIT B Complete Diode Model - Nothing is neglected. - The most accurate model. - Forward current is given by I F = R BIAS LIMIT r B ' d

4 Class Objectives Explain the Operation of the Half-wave Rectifier Explain the Operation of the Full-wave Rectifier Discussions Explain the Use of Power Supply Filters and Regulators Explain Diode Limiting and Clamping Circuits Explain the Operation of the oltage Multipliers Discussions

5 Introduction DC Power Supply can be half-wave or full-wave rectifier smoothing filter (Low-pass filter)

6 Half-Wave Rectifier Connection 1 Forward Bias In the first half cycle, the diode is forward-biased. If the diode is ideal, the output voltage equals the input voltage. Reverse Bias In the second half cycle, the diode is reverse-biased. There is no output ( out = 0). Connection 2 Reverse biasing in the first half period Forward bias in the second half period. in 0 R L 0 t0 t1 2 t 0 t1 I out t t 2 I = 0 A out in 0 R t L 0 0 t1 t 2 t0 t1 t 2 in 0 Connection 1 R L 0 t0 t1 2 t 0 t1 I t t 2 Connection 2 out

7 Half-Wave Rectifier contd. Average alue AG 1 T = x() t dt T 0 1 T /2 = sin( ), 2 0 p ωt dt ω= π T T = 1 T p cos( ωt) ω T /2 0 cos 2π T cos(0) p T 2 p = = T 2π π T Peak Inverse oltage Equivalent to the peak value of the input voltage. Diode should be capable of withstanding this repetitive reverse voltage.

8 Half-Wave Rectifier contd. Example 2-2 Ideal Diode The peak output voltage = p (in) = 5 and 100, respectively. AG = out / π = 1.59 and Practical Diode The peak output voltage = p (in) 0.7 = 4.3 and 99.3, respectively. Find the average output voltage in this case [Homework: 2points].

9 Full-Wave Rectifier Center-Tapped Full-Wave Rectifier in 0 F D 1 I R L out 0 During the first half cycle, the upper diode is forward-biased and the lower diode is reverse -biased. D 2 F D 1 in 0 I R L out 0 During the second half cycle, the lower diode is forwardbiased and the upper diode is reverse-biased. D 2

10 Full-Wave Rectifier contd. Turn Ratio The ratio between the number of turns in the secondary winding and those in the primary winding. Designated n. Affects the output voltage. Peak Inverse oltage PI = 0.7 = 0.7 p(sec) p(sec) 2 2 = pout ( ) p(sec)

11 Full-Wave Rectifier contd. Example 2-5 p = n = 0.5(100) = 50 (sec) p( pri) p(sec) = 0.7 = 24.3 p( out) 2 PI = = 2(24.3) 0.7 = 49.3 p ( out ) Is this figure correct?

12 Full-Wave Rectifier contd. The Bridge Full-Wave Rectifier Uses four diodes connected as shown. F I D 3 D 1 p( out) = p(sec) 1.4 in D R L out 2 D 4 0 PI = 0.7 p(sec) = 0.7 p( out) F I D 3 D 1 in D R L out 2 D 4 0

13 Full-Wave Rectifier contd. The Bridge Full-Wave Rectifier p(sec) = 12 rms Input is sinusoidal, then = 2 = 17 p(sec) p(sec)( rms) = 1.4= = 15.6 ( ) p(sec) p out PI = 0.7 = = 16.3 p ( out ) Root Mean Square sin( ) 1 1 rms = t dt = cos(2 ωt) dt T T 2 T p 2 p ω T T T p dt 0 0 = 1 cos(2 ωt ) dt 2T 2 2 p p = T = 2T 2

14 Class Objectives Explain the Operation of the Half-wave Rectifier Explain the Operation of the Full-wave Rectifier Discussions Explain the Use of Power Supply Filters and Regulators Explain Diode Limiting and Clamping Circuits Explain the Operation of the oltage Multipliers Discussions

15 Power Supply Filters and Regulators Objective Convert the AC power line voltage into a DC voltage. The fluctuation in the filter output voltage is called ripple. in OUT Full-wave Rectified 0 Filter 0 rectif ier output Capacitor-Input Filter Consists of a capacitor connected from the output of the rectifier to the ground. in C R L

16 Capacitor-Input Filter contd. Positive 1st quarter cycle: Diode is forward-biased. Capacitor charges to within 0.7 of the input voltage. Load and capacitor have the same voltage. Remaining part of the cycle Diode is reverse-biased. Capacitor discharges in the load at a rate determined by RC. Power Supply Filters and Regulators contd. 1st quarter of next cycle Diode becomes forward-biased.

17 Capacitor-Input Filter contd. Ripple voltage Power Supply Filters and Regulators contd. without filter The variation in the capacitor voltage due to charging and discharging The smaller the ripple, the better the filtering performance. Ripple factor It is an indication of the effectiveness of the filter. It is defined as ( ) r = r pp DC 1 r( pp) fr C p( rect) L DC voltage of the filter s output voltage = 1 1 DC 2 fr C prect ( ) L

18 Capacitor-Input Filter contd. Half-wave vs. Full-wave rectifiers Power Supply Filters and Regulators contd.

19 Power Supply Filters and Regulators contd. Example 2-7 p(pri) = rms = 1.414(120 ) = 170 p(sec) = n p(pri) = 0.1(170 ) = 17 p(rect) = p(sec) 1.4 = = r( pp) p( rect) = 15.6 = fr C (120Hz)(220 Ω)(1000μF) DC = ( ) prect = = frlc 2(120Hz)(220 )(1000μF) Ω ( ) r = r pp = = L DC

Surge Current in the Capacitor-Input Filter Power Supply Filters and Regulators contd. Before closing the switch, the capacitor is uncharged Uncharged capacitor acts as a short circuit.

20 Surge Current in the Capacitor-Input Filter Power Supply Filters and Regulators contd. Before closing the switch, the capacitor is uncharged Uncharged capacitor acts as a short circuit. When the switch is closed An initial current, called surge current, passes through the diodes D1 and D2. To avoid defecting the diodes, a fuse is used. A fuse is a protection device. A slow-blow fuse is generally used. Slow-blow indicates a slow melting of the wire inside the fuse. Fuse (source: wikipedia.org]

21 Power Supply Filters and Regulators contd. oltage Regulators A regulator is connected to the output of the capacitor-input filter. It maintains a constant output despite changes in the input. The input of the regulator should have a ripple < 10%. Percent Regulation Δ Line regulation = OUT 100%. Δ IN NL Load regulation = FL 100%, where NL and FL refer to no load FL and full load (i.e., maximum load) Capacitor-input filter SW1 F 1 T 1 D 3 D 1 oltage D 2 D 4 regulator Output capacitor To improve the transient response C 1 C 2

22 Diode Limiter (Clipper) Clips either positive or negative part Example 2-9 R 1 = 10 kω R L = 100 kω p(in) = 10 Diode Limiting and Clamping Circuits R L p( out) = R p( in) 1 RL R L p( out) = R p( in) 1 RL 100k = Ω = 110k Ω

23 Diode Limiting and Clamping Circuits contd. Biased Limiters A limiter that has a bias voltage in series with the diode. R 1 0 in R L BIAS Positive limiter BIAS The diode is forward-biased when p(in) exceeds the voltage bias by 0.7 (i.e., p(in) = 0.7 BIAS ). Since the load is connected in parallel with the diode and bias voltage (out) = BIAS 0.7. Otherwise, the diode is reverse-biased. The peak output voltage is then given by: R = L p( out) R p( in) 1 RL

24 Diode Limiting and Clamping Circuits contd. Biased Limiters contd. What is the input at this point?

25 Biased Limiters contd. Example 2-10 When A exceeds 5.7 D 1 becomes forward-biased. (out) equals 5.7. D 2 is still reverse-biased. When A goes below -5.7 D 2 becomes forward-biased. (out) equals D 1 is still reverse-biased For values between 5.7 and -5.7 Diode Limiting and Clamping Circuits contd. Both diodes are reverse-biased. Depending on the load value, the input voltage is partitioned between R 1 and the load.

26 oltage-divider Bias Diode Limiting and Clamping Circuits contd. BIAS is the voltage applied to R 3 BIAS is simply given by BIAS = R R 3 R 3 2 SUPPLY

27 Example 2-11 Diode Limiting and Clamping Circuits contd. Note that there is a power supply of 12 and R2 parallel to the bias voltage. These two elements are not shown in the figure. BIAS = R R 3 R 3 2 SUPPLY = = R 1 in BIAS 0.7 = R L BIAS 0

28 Diode Limiting and Clamping Circuits contd. Diode Clampers A clamper is a circuit that adds a DC level to an AC signal. Clampers are also referred to as DC restorers. The capacitor charges and acts as a DC voltage source. RC time constant should be large enough to avoid discharging of the capacitor. RC = 100 for excellent clamping action.

29 Diode Clampers Example 2-12 Diode Limiting and Clamping Circuits contd. When the input reaches about 0.7, the diode is forward-biased. The capacitor is charged up to (-(24 0.7) = ) Consider the polarity of the capacitor! The diode is then reverse-biased. The capacitor acts as a DC voltage source (consider a negligible discharging) The circuit can be analyzed as if it includes an AC voltage source and a DC voltage source (The later one replaces the capacitor).

30 oltage Multipliers oltage Doubler Half-wave Doubler During the positive half cycle D1 is forward-biased, and C1 is charged to p less the diode drop. D2 is reverse-biased. During the negative half cycle D2 is forward-biased and C2 is charged by both the input and C1 which acts as a DC voltage source. C2 is charged to (2 p 1.4 ) D1 is reverse-biased. The output is a half-wave filtered voltage.

31 oltage Multipliers oltage Doubler Full-wave Doubler Class discussion Groups of students analyze the full-wave doubler based on the following Figures. 0 p D 1 I C 1 p 0 Reverse-biased p I D 1 C 1 p 2 p D 2 Reverse-biased C 2 D 2 C 2 p

32 oltage Tripler & Quadrupler oltage Tripler 1st positive half cycle D1 is forward-biased. C1 charges to p st negative half cycle D2 is forward-biased. C1 acts as a DC voltage source. C2 charges to 2(p 0.7). 2nd positive half cycle D3 is forward-biased. C1 and C2 acts as DC voltage sources. C3 charges to 2(p 0.7). oltage Tripler oltage Quadrupler

33 Summary and Discussion Explain the Operation of the Half-wave Rectifier Explain the Operation of the Full-wave Rectifier Discussions Explain the Use of Power Supply Filters and Regulators Explain Diode Limiting and Clamping Circuits Explain the Operation of the oltage Multipliers Discussions

34 Discussion & Notes K K A K A A A A K K K K A K A K K A K A K A

CHAPTER 2. Diode Applications

CHAPTER 2. Diode Applications CHAPTER 2 Diode Applications 1 Objectives Explain and analyze the operation of both half and full wave rectifiers Explain and analyze filters and regulators and their characteristics Explain and analyze