Part I Lectures 1-7 Diode Circuit Applications

Transcription

1 Part Lectures -7 iode Circuit Applications

2 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page of 7 Second Year, Electronics, 9 - The PN Junction iode Basic Construction: When acceptor impurities are introduced into one side and donors into the other side of a single crystal of a semiconductor, a p-n junction is formed. n general, the acceptor ion is indicated by a minus sign because, after this atom "accepts" an electron, it becomes a negative ion. The donor ion is represented by a plus sign because, after this impurity atom "donates" an electron, it becomes a positive ion. Now, if a junction is formed between a sample of p-type and one of an n-type semiconductor, this combination possesses the properties of a rectifier (permits the flow of charge in one direction). Such a two-terminal device is called a p-n junction diode. The two single crystal semiconductors (having four valence electrons) used most frequently in the construction of p-n junction diodes are silicon () and germanium (Ge). The p-type is created by introducing those impurity elements (acceptors) that have three valence electrons (trivalent), such as boron, gallium, and indium. The n-type is created by introducing those impurity elements (donors) that have five valence electrons (pentavalent), such as antimony, arsenic, and phosphorus. n a p-type material the hole is the majority carrier and the electron is the minority carrier. n an n-type material the electron is called the majority carrier and the hole the minority carrier. The electrons and holes in the region of the junction will combine, resulting in a lack of carriers in the region near the junction. This region of uncovered positive and negative ions is called the "depletion region" due to the depletion of carriers in this region. Essential Characteristics: The essential electrical characteristic of a p-n junction is that it constitutes a rectifier which permits the easy flow of charge in one direction but restrains the flow in the opposite direction. We consider now how this diode rectifier action comes above. No Applied Bias ( = ): n the absence of an applied bias voltage, the net flow of charge in any one direction for a semiconductor diode is zero (see Fig. -). Fig. -

3 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page of 7 Second Year, Electronics, 9 - everse Bias ( < ): The current that exists under reverse-bias conditions is called the reverse saturation current and is represented by s (see Fig. -). Fig. - Forward Bias ( > ): A semiconductor diode is forward-biased when the association p-type and positive and n-type and negative has been established (see Fig. -3). Fig Characteristic Carve and Current Equation: N P Cath ode An ode Forward-bias region everse-bias region everse-breakdown region Fig. -4 k / T K = ( e ) [.] S Where k = 6/η with η = for Ge and η = for for relatively low levels of diode current and η = for Ge and for higher levels of diode current. T K = T C 73 o.

4 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page 3 of 7 Second Year, Electronics, 9 - esistance Levels:. C or Static esistance: The application of a dc voltage to a circuit containing a p-n junction diode will result in an operating point on the characteristic carve that will not change with time. The resistance of the diode at the operating point can found simply by finding the corresponding levels of and as shown in Fig. -5 and applying the following equation: = [.] Fig. -5. Ac or ynamic esistance: f a sinusoidal rather than dc input is applied, the varying input will move the instantaneous operating point up and down a region of the characteristics and thus defines a specific change in current and voltage as shown in Fig. -6. With no applied varying signal, the point of operation would be the Q-point determined by the applied dc levels. A straight line drawn tangent to the curve through the Q-point will define a particular change in voltage and current that can be used to determine the ac or dynamic resistance for this region of the diode characteristics. n equation form, r d Δ Δ d = [.3] d Fig. -6

5 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page 4 of 7 Second Year, Electronics, 9 - n differential calculus, the derivative of a function at a point is equal to the slope of the tangent line drawn at that point. Eq. [.3], as defined by Fig. -6, is, therefore, essentially finding the derivative of the function at the Q-point of operation. f we find the derivative of the general Eq. [.] for the p-n junction diode with respect to the applied forward bias and then invert the result, we will have an equation for the dynamic or ac resistance in that region. That is; d d / K ( ) = [ ( k T S e )] d d d k = ( S ) d TK d k ( Generally, >> S ) d Tk d = ( η = & T K = 98 o K 6 => = ) d T K 98 d.6 r = v / i = d 6m rd = [.4] All the resistance levels determined thus far have been defined by the p-n junction and do not include the resistance of the semiconductor material itself (called body resistance) and the resistance introduce by the connection between the semiconductor material and the external metallic conductor (called contact resistance). These additional resistance levels can be included in Eq. [.4] by adding resistance denoted by r B appearing in Eq. [.5]. 6m r d = r B [.5] 3. Average AC esistance: f the input signal is sufficiently large to produce a board swing such as indicated in Fig. -7, the resistance associated with the device for this region is called the average ac resistance. The average ac resistance is, by definition, the resistance determined by a straight line drawn between the two intersection establish by the maximum and minimum value of input voltage. n equation form, r Δ d av = [.6] Δ d pt. to pt.

6 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page 5 of 7 Second Year, Electronics, 9 - Fig. -7 Equivalent Circuits (Models):. Piecewise-Linear Model: (see Fig.-8); Forward-bias; T r av(f ) > T > T < T T everse-bias; Fig. -8 < T r av(). mplified Model: (see Fig. -9); Forward-bias & network >> r av(f) ; T everse-bias, r av() = Ω & = A; Fig. -9 > r av() 3. deal Model: (see Fig. -9); Forward-bias, E network >> T, network >> r av(f) & = ; T everse-bias, r av() = Ω & = A; Fig. - r av(f ) <

7 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page 6 of 7 Second Year, Electronics, 9 - Load-Line Analysis: From Fig. -: E = E = E = [.7] Eq. [.7] is a linear equation; y = mx c, where m = / & c = E /. = = E E [.8] Fig. - = = E Example -: etermine the currents,, and for the network of Fig kΩ E 5.6kΩ Fig. - Solution:.7 = = 3.3k Appling KL yields: E and =.ma. = = E =.7.7 = 8., with = = = 3.3mA. 5.6k Finally, = = 3.3m.m = 3. 8mA.

8 The PN Junction iode Electrical and Electronic Engineering epartment Lecture One - Page 7 of 7 Second Year, Electronics, 9 - Exercises: Find the values of and o in the circuits shown in Fig. -3. Ge 6 kω kω o Ge.kΩ Ge o o 3.3kΩ 3kΩ o kω (a) (b) (c) kω o.47kω o nωt (t) d r r =. kω = MΩ av( F ) av( ) Ge kω o (t) (d) (e) Fig. -3

9 iode Switching Circuits Electrical and Electronic Engineering epartment Lecture Two - Page of 3 Second Year, Electronics, 9 - iode Switching Circuits Basic Concepts: iode switching circuits typically contain two or more diodes, each of which is connected to an independent voltage source. Understanding the operation of a diode switching circuit depends on determining which diodes, if any, are forward biased and which, if any, are reverse biased. The key to this determination is remembering that a diode is forward biased only if its anode is positive with respect to its cathode (see Fig. -). One of the very import applications of diode switching circuits is logic gates Fig. - kω kω kω kω kω 3 8 Logic Gates: iodes can be used to form logic gates, which perform some of the logic operations required in digital computers. O Gate: t has output when there a signal in any input channels (see Fig. -). A B o nput voltages State of diodes Output voltage A B o off off off on on off on on Fig. - AN Gate: t has output only when all inputs are present (see Fig. -3). A B o nput voltages State of diodes Output voltage A B o on on on off off on off off Fig. -3

10 iode Switching Circuits Electrical and Electronic Engineering epartment Lecture Two - Page of 3 Second Year, Electronics, 9 - Example -: etermine which diodes are forward biased and which are reverse biased in the circuits shown in Fig. -4. Assuming a.7- drop across each forward-biased diode, determine the output voltage o o 5 o 5 o (a) (b) (c) Fig. -4 Solution: n (a) the net forward-biasing voltage between supply and input for each diode is & 3 : 5 - (5) =, & 4 : 5 - (-5) =. Therefore, and 4 are forward biased and and 3 are reverse biased. o = -5.7 = While in (b) the net forward-biasing voltage between supply and input for each diode is : 5 - (5) =, : 5 - = 5, 3 : 5 - (-) = 5. Therefore, 3 is forward biased and and are reverse biased. o = -.7 = Finally, in (c) the net forward-biasing voltage between supply and input for each diode is : -5 - (-) = 5, : 5 - (-) = 5. Therefore, is forward biased and is reverse biased. o = = 4.3.

11 iode Switching Circuits Electrical and Electronic Engineering epartment Lecture Two - Page 3 of 3 Second Year, Electronics, 9 - Exercises: etermine o and for each circuit in Fig. -5. Assume that each of the diodes in these circuits has a forward voltage drop of kω o 8 5 kω o 4 kω o (a) (b) (c) A B.. 3. kω kω (d) o 5.kΩ = B, = B 5,and = & 5. A = A = A B = (e) kω C A kω 3 5 o o C kω B kω 5. No pulses at either A or B,. A 3 positive pulse at A or B, and 3. Positive pulses (3 ) at both A and B. Fig. -5

12 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page of 8 Second Year, Electronics, 9 - iode Clipping Circuits Basic efinition: There are a variety of diode circuits called clippers (limiters or selectors) that have the ability to "clip" off a portion of the input signal above (positive) or below (negative) certain level without distorting the remaining part of the alternating waveform. epending on the orientation of the diode, the positive or negative region of the input signal is "clipped" off. There are two general categories of clippers: series and parallel. The series configuration is dined as one where the diode is in series with the load. While the parallel variety has the diode in a branch parallel to the load (see Fig. 3-). T/ T/ T T t v o mple Series (Positive) Clipper t mple Parallel (Negative) Clipper Fig. 3- T/ T t T/ T t Example 3-: Biased Series (Negative) Clipper, see Fig E T/ T t Fig. 3-

13 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page of 8 Second Year, Electronics, 9 - E T i i 4.5 deal 3.5 For t = t and t T; ON, and = 4.5. For t = t t ; OFF, and = T/ t t T t i v t Fig. 3- (cont.) Example 3-: Biased Parallel (Positive) Clipper, see Fig T/ T t E Fig. 3-3

14 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page 3 of 8 Second Year, Electronics, 9 - transition i d = E 5.7 transition i d d E = ; transition = = 5. For t = t and t T; ON, and = 5. For t = t t ; OFF, and =. t t T/ T t -5 transition t - - Fig. 3-3 (cont.) Summary: A variety of series and parallel clippers with the resulting output for the sinusoidal input are provided in Fig Fig. 3-4

15 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page 4 of 8 Second Year, Electronics, 9 - Example 3-3: ouble iode Series Clipper, see Fig Fig. 3-4 (cont.) - T/ T t E 3.3 E 7.7 Ge E T i 4 E T 8 i deal deal Fig. 3-5

16 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page 5 of 8 Second Year, Electronics, 9 - For t = t, t t 3, and t 4 T; both and will be OFF, and =. For t = t t ; ON while OFF, and = = 4. i For t = t 3 t 4 ; OFF while ON, and = = 8. i i T/ T 8 t t t 3 t 4-4 t 8-6 i t Fig. 3-5 (cont.) Example 3-4: ouble iode Parallel Clipper, see Fig T T/ t E.3 E tr i d = E.3 tr i = d E 5.3 i d E = ; i d E = ; tr tr d =.7.3 = 3. = = 6. tr Fig. 3-6 tr d

17 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page 6 of 8 Second Year, Electronics, 9 - For t = t, t t 3, and t 4 T; both and will be OFF, and =. For t = t t ; ON while OFF, and = 3. For t = t 3 t 4 ; OFF while ON, and = t T tr t t 3 t 4 T/ t - 6 tr t Fig. 3-6 (cont.) Example 3-5: Special Type Clipper: A Comparator, see Fig T/ T t dial E 5 - Fig. 3-7

18 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page 7 of 8 Second Year, Electronics, 9 - For t = t and t T; OFF, and = E = 5. For t = t t ; ON, and =. 5 E t t T/ T t t Fig. 3-7 (cont.) Exercises:. esign biased parallel clippers (with silicon diodes) to perform the functions indicated in the transfer characteristics of Fig (a) (b) Fig. 3.8

19 iode Clipping Circuits Electrical and Electronic Engineering epartment Lecture Three - Page 8 of 8 Second Year, Electronics, 9 -. Sketch the output voltage ( ) and the transfer characteristics ( against ) for each circuit of Fig. 3-9 for the input ( ) shown. 8 T/ T t deal E 4 (a) - 8 T t vi E 4 E.3 vo (b) 9 vi vo T/ T t Ge E 3.3 E 5.3 (c) T t vi deal vo deal kω kω 5 E E (d) Fig. 3-9

20 iode Clamping Circuits Electrical and Electronic Engineering epartment Lecture Four - Page of 4 Second Year, Electronics, 9 - iode Clamping Circuits Basic efinition: The clamping circuit (clamper) is one will "clamp" a signal to a different dc level. The circuit must have a capacitor, a diode, and a resistive element, but it can also employ an independent dc supply to introduce an additional shift. The magnitude of and C must be chosen such that the time constant τ = C is large enough to ensure that the voltage across the capacitor does not discharge significantly during the interval (T/) the diode is nonconducting. Throughout the analysis we will assume that for all practical purposes the capacitor will fully charge or discharge in five time constants. Therefore, the condition required for the capacitor to hold its voltage during the discharge period between pulses of the input signal is T 5 τ = 5C >> = [4.] f Example 4-: etermine the output ( ) for the circuit of Fig. 4- for the input ( ) shown. f = khz T/ T 3T/ T -5 t vi C.μF E 5 vo 5kΩ - Fig. 4- Solution: The analysis of clamping circuits are started by considering that the part of the input signal that will forward bias the diode. For the circuit of Fig. 4-, the diode is forward bias ("on" state) during the negative half period of the input signal ( ) and the capacitor will charge up instantaneously to a voltage level determined by the circuit of Fig. 4-. C.7 5 5kΩ Fig. 4-

21 iode Clamping Circuits Electrical and Electronic Engineering epartment Lecture Four - Page of 4 Second Year, Electronics, 9 - For the input section KL will result in C.7 5 = => C = 4.3. The output voltage ( ) can be determined by KL in the output section 5.7 = => = 4.3. Now check that the capacitor will hold on or not its establish voltage level during the period (positive half period in case of Example 4-) when the diode is in the "off" state (reverse bias). The total time constant 5τ of the discharging circuit of Fig. 4-3 is determined by the product 5C and has the magnitude 5τ = 5C = 5 (5 3 ) (. -6 ) = 5 ms. The frequency ( f ) is khz, resulting in a period of ms and an interval of.5 ms between levels, that is T/ = /( f ) = /( 3 ) =.5 ms. We find that 5 τ >> T/ ( 5ms /.5ms = 5 times). So that, it is certainly a good approximation that the capacitor will hold its voltage (4.3 ) during the discharge period between pulses of the input signal kΩ Fig. 4-3 The open-circuit equivalent for the diode will remove the 5- battery from having any effect on, and applying KL around the outside loop of circuit will result in 4.3 = => = The resulting output appears in Fig. 4-4, where the input and the output swing are the same T/ T 3T/ T 5T/ t Fig. 4-4

22 iode Clamping Circuits Electrical and Electronic Engineering epartment Lecture Four - Page 3 of 4 Second Year, Electronics, 9 - Example 4-: Using silicon diode, design a clamper circuit that will produce output = nωt 5 when the input is = nωt5. raw the circuit diagram and the input and output signals. Solution: From the input ( ) and output ( ) signals, we have a negative biased clamper. Therefore, the diode is forward bias ("on" state) during the positive half period of the input signal ( ). The output voltage ( ) at this positive period can be determined by KL in the output section of the circuit shown in Fig E.7 = => E = 5.7 = 4.3. For the input section KL will result in 5 C 5 = => C =. Fig. 4-5 = 5 C.7 = 5 E The circuit diagram and the input and output signals are shown in Fig T/ T 3T/ T t vi C E 4.3 vo -5 5 T/ T 3T/ T t -5-5 Fig. 4-6

23 iode Clamping Circuits Electrical and Electronic Engineering epartment Lecture Four - Page 4 of 4 Second Year, Electronics, 9 - Summary: A number of clamping circuits and their effect on the square-wave input signal are shown in Fig Negative Clampers Positive Clampers Clampers with ideal diodes and 5τ = 5C >> T/ Fig. 4-7 Exercise: Sketch the output ( ) for the circuit of Fig. 4-8 for the input ( ) shown. Assume ideal diodes. 5 E C T T/ t v i 3 E 7 E v o -5 Fig. 4-8

24 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page of Second Year, Electronics, 9 - iode ectifier Circuits Basic efinition: A diode circuit that converts an ac voltage to a pulsating dc voltage and permits current to flow in one direction only is called "rectifier" and the ac-to-dc conversion process is termed "rectification". Half-Wave ectifier (HW): N N P T i L P /n = N N n = = ωt P Fig. 5- ωt P T f o = f i =/ T ωt For the half-wave rectifier circuit of Fig. 5-: The average (dc) value of a half-wave rectified sine-wave voltage ( dc ) is T P dc = vo ωt dωt = Pnωt dωt = T ( ) For P close to T, For P >> T, dc =.38( ) [5.a] P T =. 38 [5.b] dc P The root mean square (rms) value of the load voltage ( rms ) is T rms = vo ( t) d t Pn t d t T ω ω = = ω ω For P close to T, For P >> T, P rms =.5( ) [5.a] P T =. 5 [5.b] rms P

25 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page of Second Year, Electronics, 9 - The rms value of the ac component (or the ripple voltage) of the rectified signal [ r (rms)] is rms dc r ( rms) = = (.5 P ) (.38 P ) =. 385 P For P close to T, r ( rms) =.385( P T ) [5.3a] For P >> T, ( rms) =. 385 [5.3b] r P The percent ripple (r) in the rectified waveform (also called the ripple factor) is r ( rms).385 P r = % = % = %.38 dc P Efficiency (η) = [ P dc (load) / P total (circuit) ] % dcl (.38 P ) L 4.5 η = % = % = rms ( rd L ) (.5 P ) ( rd L ) rd / For ideal diode (r d = Ω), η = η max = 4.5 % L % The peak inverse voltage (P) of the diode is P = P [5.4] The frequency of the output rectified signal (f o ) is f = [5.5] o f i Full-Wave ectifiers (FWs):. A Bridge Full-Wave ectifier: P /n = N N n = = N N ωt P (-) P - () Fig. 5-4 ωt (-) i (i ) 3 - () P L i T fo = f i i = i i v o ωt

26 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 3 of Second Year, Electronics, 9 - For the bridge full-wave rectifier circuit of Fig. 5-: P dc = Pnωt dωt = For P close to T, dc =.636( P T ) [5.6a] For P >> T, =. 636 [5.6b] dc P P rms = Pn ωt dωt = For P close to T, For P >> T, rms =.77( ) [5.7a] P T =. 77 [5.7b] rms P rms dc r ( rms) = = (.77P ) (.636 P ) =. 38 P For P close to T, r ( rms) =.38( P T ) [5.8a] For P >> T, ( rms) =. 38 [5.8b] r P r ( rms).38 P r = % = % = 48.4%.636 dc P dcl (.636 P ) L η = % = rms (rd L ) (.77 P ) (rd For ideal diode (r d = Ω), η = η max = 8 % L 8 % = ) r / d L % P = P [5.9] T f = f [5.] o i

27 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 4 of Second Year, Electronics, 9 -. A Center-Tapped (CT) Full-Wave ectifier: N N (-) P - () (-) P - () i (i ) L i v o P /n = N N n = = ωt N P Fig. 5-3 ωt P T fo = f i i = i i ωt For the center-tapped full-wave rectifier circuit of Fig. 5-3: P dc = Pnωt dωt = For P close to T, dc =.636( P T ) [5.a] For P >> T, =. 636 [5.b] dc P P rms = Pn ωt dωt = For P close to T, For P >> T, rms =.77( ) [5.a] P T =. 77 [5.b] rms P rms dc r ( rms) = = (.77P ) (.636 P ) =. 38 P For P close to T, r ( rms) =.38( P T ) [5.3a] For P >> T, ( rms) =. 38 [5.3b] r P

28 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 5 of Second Year, Electronics, 9 - r ( rms).38 P r = % = % = 48.4%.636 dc P dcl (.636 P ) L η = % = rms ( rd L ) (.77 P ) ( rd For ideal diode (r d = Ω), η = η max = 8 % L 8 % = ) r / d L % P = P [5.4] T f = f [5.5] o i Summary: ifferent parameters for the HW and FW circuits are listed in Table 5-. Table 5- Parameter HW Bridge FW CT P P T P T P T dc.38 P.636 P rms.5 P.77 P r.385 P.38 P r % 48.4% η max 4.5% 8% P P P T P T f o f i f i

29 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 6 of Second Year, Electronics, 9 - Example 5-: The input voltage to a full-wave rectifier employing a center-tapped step-down transformer and two silicon diodes is rms, and the transformer has turns ratio n =.5. raw the rectifier circuit diagram when it is connected to a Ω load, and find. the average value of the voltage across the load.. the average power dissipated by the load, and 3. the minimum P rating required for each diode. Solution: The rectifier circuit diagram is shown in Fig P = vin =.5 = =.636 = ( ) ( ). dc P T =. rms =.77( P T ) =.77( ) = 7.. ( 7.) P rms av = = = L 7.3W. 3. P ( ) = ( ) 77.. P T = vi = n =.5 P P L Ω Fig. 5-4

30 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 7 of Second Year, Electronics, 9 - Capacitor Filters: A low-pass filter is connected across the output of a rectifier to suppress the ac components and to pass the dc component. A rudimentary low-pass filter used in power supplies consists simply of a capacitor (C) connected across the rectifier output, that is, in parallel with the load ( L ), as illustrated in Fig P = o C Charging ischarging ( pp) r L C o ON OFF T t Half-wave rectifier with capacitor filter o = C ( pp) r L C o P T t Full-wave rectifier with capacitor filter Fig. 5-5 Operation: uring the positive first quarter-cycle of the input, the diode is forward-biased (when i > C ), allowing the capacitor to charge quickly to within a diode drop of the input peak ( P ). When the input begins to decrease below its peak, the capacitor retain its charge and the diode becomes reverse-biased (when C > i ). uring the remaining part of the cycle, capacitor C can discharge slowly only through load resistance L at a rate determine by L C time constant (τ). The voltage fluctuation in the filtered waveforms is called the peak-to-peak ripple voltage [ r (pp)]. n general, r (pp) in FW is smaller than it is in HW for same L and C values (see Fig. 5-5).

31 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 8 of Second Year, Electronics, 9 - ipple of a Capacitor Filter: We will now derive an expression for the ripple in the output of a rectifier having a capacitor filter (C) and load resistance ( L ). The derivation that follows is applicable to both HW and FW. We can assume that the ripple voltage in a lightly loaded filter ( L C time constant (τ) is large) is a sawtooth wave as illustrated in Fig P dc o = C r ( pp) ( pp) r T Fig. 5-6 t This approximation is equivalent to assuming that the capacitor charges instantaneously and that its voltage decays linearly, instead of exponentially. Assuming that the voltage decays linearly is equivalent to assuming that the discharge current () is constant and equal to = dc / L where dc is the dc value of the filtered waveform. The total charge in capacitor voltage is r (pp) volts, and this charge occurs over the period of time T. Therefore, since Δ Q =. Δt, ΔQ ( dc / L ) T r ( pp) = = C C nce T=/f r, where f r is the frequency of the fundamental component of the ripple, that is, f f = f for HW and f = f = f for FW. So that r = o i r o i or dc r ( pp) = [5.6] f C dc r L = ( pp) f C [5.7] r r L From Fig. 5-6, it is apparent that r ( pp) dc = P Subsuming from Eq. [5.6], we obtain dc dc = P f C r L

32 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page 9 of Second Year, Electronics, 9 - Solving for dc, we obtain an expression for the dc voltage ( dc ) in terms of the peak rectifier voltage ( P ): dc = P [5.8] f C r L The rms value of a sawtooth waveform having peak-to-peak value r (pp) is known to be r ( pp) r ( rms) = [5.9] 3 Therefore, from Eqs. [5.7] and [5.9], the percent ripple is r = r ( rms) % = dc r r ( pp)/( 3) % ( pp) f C r L r = 3 f r L % C [5.] Equation [5.] confirms our analysis of the capacitor filter: a large L C time constant (τ) results in a small ripple voltage, and vice versa. The light-load assumption on which our derivation is based is generally valid for percent ripple (r) less than 6.5%. From a design standpoint, the values of f r and L, are usually fixed, and the designer's task is to select a value of C that keeps the ripple below a prescribed value. Example 5-: A full-wave rectifier is operated from a 5 Hz line and has a filter capacitor connected across its output. What minimum value of capacitance is required if the load is. kω and the ripple must be no greater than.4%? Solution: r = % 3 fr LC.4 = => C C μf.

33 iode ectifier Circuits Electrical and Electronic Engineering epartment Lecture Five - Page of Second Year, Electronics, 9 - Exercises:. A full-wave bridge rectifier isolated from the rms power line by a transformer. Assuming the diode voltage drops are.7. i. What turns ratio should the transformer have in order to produce an average current of A in a Ω load? ii. What is the average current in each diode under the conditions of (i)? iii. What minimum P rating should each diode have? iv. How much power is dissipated by each diode?. A full-wave bridge rectifier is operated from a 5 Hz, rms line. t has a μf filter capacitor and a k Ω load. Neglect diode voltage drops. i. What is the percent ripple? ii. What is the average current in the load?

34 oltage-multiplier Circuits Electrical and Electronic Engineering epartment Lecture x - Page of 3 Second Year, Electronics, 9 - oltage-multiplier Circuits Basic Concepts: iodes and capacitors can be connected in various configurations to produce filtered, rectified voltages that are integer multiples of the peak value of an input sine wave. The principle of operation of these circuits is similar to that of the clamping circuits discussed previously. By using a transformer to change the amplitude of an ac voltage before it is applied to a voltage multiplier, a wide range of dc levels can be produced using this technique. One advantage of a voltage multiplier is that high voltages can be obtained without using a high-voltage transformer. oltage oubler:. Half-Wave oltage oubler: Figure 6- shows a half-wave voltage doubler circuit. vi P P C P C = o P Fig. 6- Operation: uring the positive half-cycle, ON and OFF => Charging C up to P. uring the negative half-cycle, ON and OFF => Charging C to P. The output ( o ) of the half-wave voltage doubler is = [6.] o C = P f a load is connected to the output of the half-wave voltage doubler, the voltage across capacitor C drops during the positive half-cycle (at the input) and the capacitor is recharged up to P during the negative half-cycle. The output waveform across capacitor C is that of a half-wave signal filtered by a capacitor filter. The peak inverse voltage (P) rating of each diode in the half-wave voltage doubler circuit must be at least P.

35 oltage-multiplier Circuits Electrical and Electronic Engineering epartment Lecture x - Page of 3 Second Year, Electronics, 9 -. Full-Wave oltage oubler: Figure 6- shows a full-wave voltage doubler circuit. P P P C o = P C Fig. 6- Operation: uring the positive half-cycle, ON and OFF => Charging C up to P. uring the negative half-cycle, ON and OFF => Charging C up to P. The output ( o ) of the full-wave voltage doubler is = = [6.] o C C P f load current is drawn from the full-wave voltage doubler circuit, the voltage across the capacitors C and C is the across a capacitor fed by a full-wave rectifier. One difference is that of C and C in series, which is less than capacitance of either C and C alone. The lower capacitor value will provide poorer filtering action than the single-capacitor filter circuit. The peak inverse voltage across each diode is P, as it is for filter capacitor circuit. oltage Tripler and Quadrupler: Figure 6-3 shows an extension of the half-wave voltage doubler, which develops three and four times the peak input voltage. t should be obvious from the pattern of the circuit connection how additional diodes and capacitors may be connected so that the output voltage may also be five, six, seven, and so on, times the basic peak voltage ( P ).

36 oltage-multiplier Circuits Electrical and Electronic Engineering epartment Lecture x - Page 3 of 3 Second Year, Electronics, 9 - P P C oubler Tripler( 3 P ) P C C C 4 P P ( P ) Quadrupler( 4 P ) Fig. 6-3 Operation: uring the positive half-cycle, ON and, 3, 4 OFF => Charging C up to P. uring the negative half-cycle, ON and, 3, 4 OFF => Charging C to P. uring the next positive half-cycle,, 3 ON and, 4 OFF => C charges C 3 to P. uring the next negative half-cycle,, 4 ON and, 3 OFF => C 3 charges C 4 to P. The voltage across the combination of C and C 3 is 3 P and that across C and C 4 is 4 P. The P rating of each diode in the circuit must be at least P. Exercises:. A certain voltage doubler has 35 rms on its input. What is the output voltage? Sketch the circuit, indicating the output terminals and P for the diode.. epeat Exercise for a voltage tripler and quadrupler. 3. The output voltage of a quadrupler is 6. What minimum P rating must each diode have?

37 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page of 7 Second Year, Electronics, 9 - Zener iodes: Zener iodes and Applications iodes which are designed with plate power-dissipation capabilities to operate in the breakdown region may be employed as voltage-reference or constant-voltage devices. Such are known as avalanche, breakdown, or zener diodes. The zener diode is made for operation in the breakdown region. By varying the doping level, a manufacturer can produce zener diodes with breakdown voltages from about to 5. When the applied reverse voltage reaches the breakdown value, minority carries in the depletion layer are accelerated and reach high enough velocities to dislodge valence electrons from outer orbits. The newly liberated electrons can then gain high enough velocities to free other valence electrons. n this way, we get an avalanche of free electrons. Avalanche occurs for reverse voltages greater than 6 or so. The zener effect is different. When a diode is heavily doped, the depletion layer is very narrow. Because of this, the electric field across the depletion layer is very intense. When the field strength reaches approximately 3 7 /m, the field is intense enough to pull electrons out of valence orbits. The creation of free electrons in this way is called zener breakdown (also known as high-field emission). The zener effect is predominant for breakdown voltages less than 4, the avalanche effect is predominant for breakdown voltages greater than 6, and both effects are present between 4 and 6. Originally, people thought the zener effect was the only breakdown mechanism in diodes. For this reason, the name "zener diode" came into widespread use before the avalanche effect was discovered. All diodes optimized for operation in the breakdown region are therefore still called zener diodes. Cathode (ma) Z Anode ( ) ZK r Z Z ZM Fig. 7-

38 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page of 7 Second Year, Electronics, 9 - Fig. 7- shows the schematic symbol and the current-voltage curve of a zener diode. Negligible reverse current flows until we reach the breakdown voltage Z. n a zener diode, the breakdown has a very sharp knee, followed by an almost vertical increase in current. Note that the voltage is approximately constant, equal to Z over most of the breakdown region. ata sheets usually specify the value of Z at a particular knee current ZK which is beyond the knee (see Fig. 7-). The power dissipation of a zener diode equals the product of its voltage and current. n symbols, P Z = Z Z As long as P Z is less than the power rating P Z(max), the zener diode will not be destroyed. Commercially available zener diodes have power ratings from.5 W to more than 5 W. ata sheets often specify the maximum current a zener diode can handle without exceeding its power rating. This maximum current is designated ZM (see Fig. 7-). The relation between ZM and power rating is given by ZM PZ (max) = [7.] Z When a zener diode is operating in the breakdown region, a small increase in voltage produces a large increase in current. This implies that a zener diode has a small dynamic resistance (r Z, see Fig. 7-). We can calculate this zener resistance by r Z = Δv Δi The complete equivalent circuit of the zener diode in the zener region includes a small dynamic resistance (r Z ) and dc battery equal to the zener potential ( Z ), as shown in Fig. 7-a. For all applications to follow, however, we shall assume as a first approximation that the external resistors are much larger in magnitude than the zener-equivalent resistor and that the equivalent circuit is simply the dc battery that equal to Z as indicated in Fig. 7-b. Z Z Z r Z Z (a) (b) Fig. 7-

39 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page 3 of 7 Second Year, Electronics, 9 - Zener iode Applications:. AC oltage egulators (Limiters or Clippers): Two back-to-back zeners can be used as an ac regulator or a simple square-wave generator as shown in Examples 7- and 7- respectively. Example 7-: nusoidal ac regulator, see Fig t t t 3 t 4 Z T Z T t vi 5kΩ [ Z = 3.3, ] [ Z = 6.8, ] vo t Fig. 7-3 For t = t and t ; ON and OFF => For t = t t ; ON and BEAKOWN => For t = t 3 and t 4 ; ON and OFF => For t = t 3 t 4 ; ON and BEAKOWN => = vi. v =. o Z T = vi. v =. o Z T Example 7-: mple square-wave generator, see Fig t vi 5kΩ [ Z = ] [ Z = ] vo - Z Z t - 4 Fig. 7-4

40 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page 4 of 7 Second Year, Electronics, 9 -. C oltage eference: Two or more reference levels can be established by placing zener diodes in series as shown in Fig As long as i is grater than the sum of and, both diodes will be in the breakdown state and the three reference voltages will be available. Z Z E 5kΩ 5 [ Z = ] [ Z = ] 3 Fig C oltage egulators: a. Fixed L, ariable i : For the regulator circuit shown in Fig. 7-6; Z L = (Constant) [7.a] L S S S (min) i(min) = = ZK S (min) L S Z [7.b] i Z Z L L S (max) i(max) = = ZM S (max) L S Z [7.c] Fig. 7-6

41 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page 5 of 7 Second Year, Electronics, 9 - b. Fixed i, ariable L : For the regulator circuit shown in Fig. 7-7; S i Z = (Constant) [7.3a] S S S L(min) L(max) = S = Z L(min) ZM [7.3b] i Z Z L L L(max) L(min) = S = Z L(max) ZK [7.3c] Fig. 7-7 c. ariable i and L : For the regulator circuit shown in Fig. 7-8; ZK = S (min) L(max) ZM = S (max) L(min) S (min) = i(min) = Z S L(min) S (max) = i(max) = Z L(max) L(max) S Z L(min) Z [7.4a] S S Z i Z [7.4b] Fig. 7-8 L L

42 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page 6 of 7 Second Year, Electronics, 9 - Example 7-3: The reverse current in a certain,.4 W zener diode must be at least 5 ma to ensure that the diode remains in breakdown. The diode is to be used in the regulator circuit shown in Fig. 7-9, where i can vary from 8 to 4. Find a suitable value for S and the minimum rated power dissipation that S should have. S S i Z L 6Ω Fig. 7-9 Solution: PZ.4 ZK = 5mA and ZM = = = ma. Z L (min) = A (when the switch S is open, L = L (max) = Ω ). Z L(max) = = = ma (when the switch S is closed, L = L(min) = 6Ω ). 6 L(min) ZK ZM = => 5 = => S 5mA. S (min) L(max) S (max) L(min) 3 S (min) 3 S (max) 3 (min) = = => = => S ma. (max) = i(min) Z 8 S (max) = = = 4Ω. 3 5 S (min) i(max) Z 4 S (min) = = = 6Ω. 3 S (max) Thus, we require 6Ω S 4Ω. Choosing or calculating = = 6 4 = Ω. S S (min) S (max) i(max) Z 4 S (max) = = = ma. S 3 ( ). W P S S (max) S = =.

43 Zener iodes and Applications Electrical and Electronic Engineering epartment Lecture Seven - Page 7 of 7 Second Year, Electronics, 9 - Exercises:. Sketch the output ( ) for the circuit of Fig. 7- for the input shown ( ) when m equal to (i) 5, and (ii) 5. m t 5kΩ [ Z =, ] - m Fig. 7-. esign the voltage regulator circuit of Fig. 7- to maintain L at across L with i that will vary between 6 and. That is, determine the proper value of S and the power rating of the zener diode (P Z ). S i L L 4Ω Fig The 6- zener diode in Fig. 7- has a maximum rated power dissipated of 69 mw. ts reverse current must be at least 3 ma to keep it in breakdown. Find a suitable value for S if i can vary from 9 to and L can vary from 5 Ω to. kω. S S i 6 Z L L Fig f S in Exercise 3 is set equal to its maximum permissible value, what is the maximum permissible value of i? 5. f S in Exercise 3 is set equal to its minimum permissible value, what is the minimum permissible value of L? 6. f S in Exercise 3 is set equal to Ω, what is the minimum rated power dissipated that S should have?