Ching-Yuan Yang. (symbol) Called breakdown diode or Zener diode, it can be used as voltage regulator. Breakdown voltage V ZK

Transcription

1 Diodes Read Chapter 3, Section , 3.9 Sedra/Smith s Microelectronic Circuits Ching-Yuan Yang National Chung Hsing University Department of Electrical Engineering Zener diode Operate in the reverse breakdown region (symbol) Called breakdown diode or Zener diode, it can be used as voltage regulator. Breakdown voltage V ZK and reverse I ZK current at V ZK (also called knee current) is the turning point of the diode into the breakdown region. The terminal voltage V Z at a specified test current I ZT, and at this point we can define r z as ΔV = r z ΔI. r z is called incremental resistance at operating point Q. 3C-1 Ching-Yuan Yang / EE, NCHU

2 Model for the zener diode V Z0, the equivalent voltage source, denotes the point at which the straight line of slope 1/r z intersects the voltage axis. Although it is slightly different from the knee voltage V ZK, in practice their values are almost equal. Model: V = V + r I for I > I. Z Z0 z Z Z ZK 3C-2 Ching-Yuan Yang / EE, NCHU Use of the zener as a shunt regulator Example The 6.8-V zener diode in the circuit is specified to have V Z = 6.8 V at I Z = 5 ma, r Z = 20 Ω, and I ZK = 0.2 ma. The supply voltage V + is nominally 10 V but can vary by ±1 V. a) Find V O with no load and with V + at its nominally value. b) Find the change in V O resulting from the ±1V change in V +. (Line regulation = ΔV O / ΔV + ) c) Find the change in V O resulting from connecting a load resistance R L that draws a current I L = 1 ma, and hence find the load regulation (ΔV O / ΔI L ). d) Find the change in V O when R L = 2 kω. e) Find the value of V O when R L = 0.5 kω. f) What is the minimum value of R L for which the diode still operates in the break down region? 3C-3 Ching-Yuan Yang / EE, NCHU

3 Solution V = V + r I V = V r I = = 6.7 V Z Z0 z Z Z0 Z z Z (a) With no load connected, the current through the zener is given by V + VZ IZ = I = = = 6.35 ma R+ r O Z0 z Z z V = V + r I = = 6.83 V (b) For ±1-V change in V +, the change in output voltage + rz 20 Δ VO =Δ V =± 1 =± 38.5 mv R + r z ΔV Line regulation O = 38.5 mv/v ΔV + (c) With I L = 1 ma, the zener current decreases by 1 ma. Δ VO = rzδ IZ = 20 ( 1) = 20 mv ΔVO Load regulation = 20 mv/ma ΔI L 3C-4 Ching-Yuan Yang / EE, NCHU Solution (d) When R L = 2 kω, I L = 6.8V/2kΩ = 3.4 ma. ΔI Z = 3.4 ma, and the corresponding change in zener voltage (output voltage) will thus be (e) When R L = 0.5 kω, I L = 6.8V/0.5kΩ = 13.6 ma. This is not possible because the current I supplied through R is only 6.4 ma (for V + = 10V). Therefore, the zener must be cut off. V O is determined by the voltage divider formed by R L and R, + RL 0.5 Δ VO =Δ V = 10 = 5 V R + R L Since the voltage is lower than the breakdown voltage of the zener, the diode is indeed not operating in the breakdown region. (f) Δ VO = rzδ IZ = 20 ( 3.4) = 68 mv For the zener to be at the edge of the breakdown region, I Z = I ZK = 0.2mA and V Z V ZK 6.7V. At this point the lowest current supplied through R is (9 6.7)/0.5 = 4.6 ma, and thus the load current is = 4.4 ma. The minimum value of R L : 6.7 R L = = 1.5 k Ω 4.4 3C-5 Ching-Yuan Yang / EE, NCHU

4 Temperature effects of the zener diode V Z depends on temperature in term of its temperature coefficient TC (mv/ C). TC depends on V Z, and TC varies with the operating current. Design consideration: For example, TC < 0 if V Z < 5 V; TC > 0 if V Z > 5 V; TC = 0 if V Z = 5 V. TC = 0 can be designed for a 5-V reference voltage. Another commonly used technique for obtaining a reference voltage with low TC is to connect a zener diode with a positive TC of about 2 mv/ C in series with a forward-conducting diode (ΔV = 0.7 V, TC = 2 mv/ C ). V O = V Z with a TC of about zero. 3C-6 Ching-Yuan Yang / EE, NCHU Diode application circuits - Rectifier Block diagram of a dc power supply Half-wave rectifier: Using the battery-plus-resistance diode model, we have R R vo = 0, vs < VD0; vo = vs VD0, vs VD0 R+ rd R+ rd In many applications r R and v v V D O S D0. 3C-7 Ching-Yuan Yang / EE, NCHU

5 Diode application circuits - Rectifier Discussion: The current handling capability required of the diode determined by the largest current flowing through the diode, should be considered. The peak inverse voltage (PIV) that the diode must be able to sustain without breakdown, determined by the largest reverse voltage that is expected to appear across the diode, mush be taken into account in the design. For half wave rectifier, PIV = V S. This kind of rectifier circuit does not function properly when the input signal is small. 3C-8 Ching-Yuan Yang / EE, NCHU Diode application circuits - Rectifier Full-wave rectifier Use center-tapped transformer The full-wave rectifier obviously produces a more efficient waveform than that provided by the half-wave rectifier. V O = V S V D0 PIV = 2V S V D0 In almost all rectifier applications, one opts for the full-wave type. 3C-9 Ching-Yuan Yang / EE, NCHU

6 Diode application circuits - Rectifier Bridge rectifier 3C-10 Ching-Yuan Yang / EE, NCHU Diode application circuits - Rectifier Bridge rectifier V = v 2V O S D0 PIV = V + V = v V ; O D0 S D0 V S > 2V D0 Compared to the previous rectifiers, this one has No need of center-tapped transformer Halved PIV Halved turns for the secondary wing of the transformer V S < 2V D0 3C-11 Ching-Yuan Yang / EE, NCHU

7 Diode application circuits - Rectifier Comparison PIV VO Efficiency Transformer turns Halfwave rectifier vs v V S D0 ~ 40% n : n 1 2 Center - tapped Fullwave rectifier 2v S V D0 v V S D0 ~ 90% n :2n 1 2 Bridge Fullwave rectifier v V S D0 v S 2V D0 ~ 80% n : n 1 2 3C-12 Ching-Yuan Yang / EE, NCHU Diode application circuits Peak Rectifier Peak Rectifier With a filter capacitor connected to the load in parallel. When v I > v O, diode conducts and the C charges up. v O follows v I. After the peak, v I decreases; C holds v O ; v I < v O ; the diode is off. Peak detector is a design example, such as an AM demodulator. 3C-13 Ching-Yuan Yang / EE, NCHU

8 Diode application circuits - Peak Rectifier Half-wave Peak Rectifier During the diode-off interval, the C discharges through R and thus v O decays exponentially with a time constant of RC. The value of C must be selected such that RC >> T. 3C-14 Ching-Yuan Yang / EE, NCHU When the diode is conducting t [t 1, t 2 ]: dvi vo id = ic + il = C + il where il = dt R Observations: Diode conducts for a brief interval, Δt. dvo vo t > t 2, i D = 0: 0 = C + dt R During the diode-off interval, C discharges through R and thus v O decays exponentially with time constant CR. When CR >> T, V r is small and v O is almost constant and equal to the peak of v I. Similarly, i L is almost constant and its dc component I L = V p /R. Output dc voltage: V = V V 1 O p 2 r 3C-15 Ching-Yuan Yang / EE, NCHU

9 During the diode-off interval dv C dt O 0 = + vo R v O = V e p t/ CR At the end of the discharge interval V V V e p r p T/ CR T/ CR CR T Vpe T CR T V p Vr Vp = CR fcr 1 /. V V I V / R constant r p L p 3C-16 Ching-Yuan Yang / EE, NCHU During the diode-on interval Small ωδt and V V cos( ωδ t) = V V p p r and cos( ωδt) 1 ( ωδt) r V : 1 2 p 2 ωδt 2 V / V r p Find the average diode current: Q = i Δ t and Q = CV supplied supplied Cav Since Q = Q, lost lost ( 1 2 / ) i = I +π V V Dav L p r Find the peak diode current: ( π ) i max = I V / V D L p r r For V V, i i r p Dmax 2 Dav 3C-17 Ching-Yuan Yang / EE, NCHU

10 Example Half-wave Peak Rectifier Solution Find C : Conduction angle: ωδ t = 2 Vr / Vp = 2 2/100 = 0.2 rad Diode conducting cycle: (0.2/2π)100% = 3.18 % Average diode current: Peak diode current: Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value V p = 100 V. Let the load resistance R = 10 kω. Find the value of the capacitance C that will result in a peak-to-peak ripple of 2 V. Also calculate the fraction of the cycle during which the diode is considering, and the average and peak values of the diode current. Vp 100 C = = = 83.3 μf VfR r ( π ) i = I 1+ 2 V / V = 324 ma where I = V / R= 10 ma. Dav L p r L p ( π ) idmax = IL Vp / Vr = 638 ma 3C-18 Ching-Yuan Yang / EE, NCHU Diode application circuits - Peak Rectifier Full-wave Peak Rectifier The ripple frequency is twice that of the input. For this case: Vp Vr = idav IL ( 1 π Vp /2Vr ) i max = I 2 fcr 1+ 2 π V /2V = + D L( p r ) Need a capacitor half the size of that required in the half-wave rectifier. Each diode current is approximately half that in the half-wave circuit. 3C-19 Ching-Yuan Yang / EE, NCHU

11 Diode application circuits Half-wave Rectifier Precision half-wave rectifier the super diode Almost identical to the ideal characteristic of a half-wave rectifier: v = v if v 0 v O I I O = 0 if v < 0 The op amp has non-ideal effects. I (No offset) 3C-20 Ching-Yuan Yang / EE, NCHU Diode application circuits Limiting (Clamping) circuits Double Limiter (hard limiter) 3C-21 Ching-Yuan Yang / EE, NCHU

12 Diode application circuits Limiting (Clamping) circuits Double Limiter (soft limiter) Soft limiting is characterized by smoother transitions between the linear region and the saturation region a slop greater than zero in the saturation region. 3C-22 Ching-Yuan Yang / EE, NCHU Diode application circuits A variety of basic limiters 3C-23 Ching-Yuan Yang / EE, NCHU

13 Diode application circuits A variety of basic limiters 3C-24 Ching-Yuan Yang / EE, NCHU Diode application circuits Clamping circuits Clamping shifts an entire signal by a dc level: (clamped capacitor) Assume Ideal diode. vi = 6 V diode conducts, vo = 0 V and vc = 6 V v =+ 4 V diode turns off, v = 6 V and v = v + v = 10 V I C O I C The output voltage has its lowest peak clamped to 0V, which is why the circuit is called clamping circuit. Reversing the diode polarity will provide an output waveform whose highest peak is clamped to 0V. 3C-25 Ching-Yuan Yang / EE, NCHU

14 Diode application circuits Clamping circuits Clamped capacitor with a load resistance R: During [t 0, t 1 ], the output voltage falls exponential with time constant CR. At t 1 the input decreases by V a, and the output attempts to follows. This causes the diode to conduct heavily and to quickly charge the capacitor. At the end of the interval t 1 to t 2, v O would normally be ~( 0.5) V. At t 1 the input rises by V a, and the output attempts to follows. 3C-26 Ching-Yuan Yang / EE, NCHU Diode application circuits Clamping & Rectifying circuits Voltage doubler Two section in cascade: A clamp (C 1 & D 1 ) A peak rectifier (D 2 & C 2 ) 3C-27 Ching-Yuan Yang / EE, NCHU

15 SPICE diode model 3C-28 Ching-Yuan Yang / EE, NCHU SPICE zener diode model 3C-29 Ching-Yuan Yang / EE, NCHU

16 Homework Design a 5-V dc power supply 3C-30 Ching-Yuan Yang / EE, NCHU SPICE format (txt) : 3C-31 Ching-Yuan Yang / EE, NCHU

17 Diode model parameters: 3C-32 Ching-Yuan Yang / EE, NCHU