diode1 - ELEC 101 1 Diode Circuits 1 Diode (or PN junction...

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Unformatted text preview: ELEC 101 1 Diode Circuits 1 Diode (or PN junction diode) First non-linear circuit element b 2009/10 Spring ID - VD Curve (or ID - VD Characteristics) VD Two-terminals passive element with nonlinear I-V characteristics. ID ID VD < 0 a Reverse breakdown region Symbol Cathode K Anode A p n 0 Reverse bias region Not in scale VD > 0 Forward bias region ELEC 101 2 Diode Circuits 2 VD Forward bias region 3 2009/10 Spring Reverse bias region ID ID VD ID I -VZO IS 0 VD IS : reverse saturation current very small and < 1A VD VTH VF VTH : threshold voltage (or called cut-in voltage, turn on voltage ) about 0.6V, voltage when diode just begins to conduct VF : Diode voltage; about 0.7 to 0.8V when a diode is in full conduction Can assume VF VTH ( = diode voltage drop) in some circuits VZO : breakdown voltage, diode voltage when diode just begins to breakdown VTH , VF , IS , VZO are diode dependent Diode is non-linear element and linear diode models are used to simplify diode circuit analysis. ELEC 101 Diode Circuits 3 2009/10 Spring Vin (V) 4 Diode models Example a Ideal diode (or switch) model Find Vout. Assume diode D is ideal. When input voltage is large (>> VF) diode can be modeled by ideal diode model VD 0 Vin Vout 0 -5 Vin 20V Vout 20V Vout 0 0 Vin = -5V, VD < 0 D is OFF, Vout = 0 Vin 0 -5V When diode is reverse bias ( VD < 0 ), diode is like an open switch (or called diode is OFF) Vin Vin = 20V, VD 0 D is ON, Vout = Vin When diode is forward bias ( VD 0 ), diode is like a closed switch (or called diode is ON) VD < 0 20 D Vout 20V (ideal diode) 0 Vin Vout 20V 0 Vout 19.3V (real diode) 0 Vout ELEC 101 5 Diode Circuits 4 Apply KCL Diode Circuits Steps to analyze diode circuits : 1. Assume the mode (region) of diode. 2. Replace diode by the diode model. 3. DC/AC circuit analysis. 4. Check, end, OR repeat steps 1 to 3 Example 12V V1 V V 11V 1 1 5 10 10 35 12V V1 V 9V 4 4 1-3 D 10 10 11V But if V1 = 9V, D is OFF (reverse bias). Hence assumption (diode is ON) is wrong 10 11V 12V Assume diode D is ON, replace D by short 5 V1 D 4 10 10 11V Assume diode D is OFF, replace D by open 5 V1 10 12V 12V 5 9V D Find V1. Assume diode is ideal. 5 V1 1-3 2009/10 Spring D 10 10Ω V1 12V 10 15Ω 11V 8V Assumption is valid and diode is OFF VD = -3V 8V ( Assume diode is NOT breakdown. Usually breakdown voltage > 4V) 11V 11V ELEC 101 6 Diode Circuits 5 Offset and Piecewise Linear diode models a ID VD < 0 Ideal diode (or switch) model 2009/10 Spring VD > 0 0 VTHVF ID VD 0 VD < 0 ID Ideal Diode b Offset diode model VD VF c V D < VF VF V D < VF Diode Voltage VF rD ID Ideal Diode VD VF Piecewise linear (PL) diode model VD VF ID 0 Ideal Diode Diode Voltage ID VD 0 Diode Resistance ID rD VD 0 VF ELEC 101 7 Diode Circuits 6 Diode clipping circuit A diode circuit to clip (cut) or reshape the input waveform. 10k Vi Vi 10k 3V -6V 6V 3V 0V VO D1 0V D2 10k 0-3V VO D1 3V 0V D2 10k Vi 3V VO D1 3V D2 b. Vi = 3V to 6V and 6V to 3V D1 is ON and D2 is OFF, hence VO = 3V Plot VO. Assume ideal diode model. 6V 6V 3V 0V 0V Steps to analyze clipping circuits : 1. Find the mode of diode. 2. Replace diode by model. 3. Find VO. Example 2009/10 Spring a. Vi= 0V to 3V and 3V to 0V D1 is OFF and D2 is OFF Hence VO = Vi Vi 0V Vi 10k VO 10k 3V -6V D1 D2 c. Vi = 0V to -6V and -6V to 0V V 10kΩ VO Vi i D1 is OFF and 10kΩ 10kΩ 2 D2 is ON Vi VO Waveform 3V 0V -3V Vo ELEC 101 Diode Circuits 7 Plot Vo vs t. Given D is a PL diode with VF = 0.6V and rD = 50. Example 2 When VO = 3.6V (i.e. Vi = 10.8V), D is ON 400 3 VO > 3.6V (i.e. Vi > 10.8V), D is ON 15V Vi Vi VO D 3V t 10ms 0 Vi = 1.5kt Ideal diode 200 0.6V 50 When VO < 3.6V (i.e. Vi < 10.8V, t < 7.2ms), D is OFF 1 400 VO VO Vi D 0.6 Vi 50 0 2009/10 Spring 200 When t = 10ms, Vi = 15V, VO ~ 3.98V 400 VO Vi D 0.6 200 50 0 V 200 i 400 200 3 15V 10.8V Vi 3.98V 3.6V VO 0 3 7.2ms 10ms Vi VO VO 3.6V VO 400 50 200 Vi 11VO 28.8V Use KCL 3 1 VO = Vi/3 for t < 7.2ms Vi 28.8V 11 for t 7.2ms 3 VO 10ms t 2 t = 7.2ms, Vi=10.8V, VO=3.6V ELEC 101 Diode Circuits 8 Example Plot i vs Vi and I vs Vi for 15V Vi 0V. Assume ideal diode . 2009/10 Spring 20V I D1 i 1k Vo D2 Vi 10V Method: find i and I for different Vi (Vo) Vi = 0V, D1 is ON, D2 is OFF Vo = Vi = 0V, i = 20mA Vi > 10V, D2 is ON, Vo = 10V, D1 is OFF, i = 10mA Vi = 0V, D1 is ON, Vo = Vi = 0V, I = i = 20mA 20mA i i vs Vi 10mA 10V 20mA I Vi I vs Vi 10mA Vi > 10V, D2 is ON, Vo = 10V, D1 is OFF, I = 0 10V Vi ELEC 101 Diode Circuits 9 Diode in Reverse breakdown region 8 ID VD < -VZO -VZ -VZO VD -IZ slope = 1/rZ VZ , IZ , rZ : operating voltage, current and resistance of breakdown diode iZ a vZ Circuit model of Zener diode Example VZO 14V slope = 1/rZ IS VZO VZ V I IZ 0 i Z = - ID vZ = -VD b Zener diode iZ vZ > VZO Zener diode is a specially designed diode to operate in the reverse breakdown region ( rZ 0 ), with new symbol, iZ and vZ. 0 -IS ID 2009/10 Spring vZ rZ = 1/slope 2V/1A 2 ideal diode (to ensure diode is in reverse breakdown region) VZO = 14V Circuit model of rZ = 2 Zener diode ELEC 101 Diode Circuits 10 Steps to analyze zener diode circuits: 1. Assume the mode (region) of diode. 2. Replace diode by model. 3. DC/AC circuit analysis. 4. Check, end, or repeat step 1 to 3 Example Vi = 32V Find IO and VO. Given VZO= 14V and rZ = 2. 34 VO IO D 2009/10 Spring Example Find IO and VO when Vi = 35.6V. Zener diode is in breakdown. 35.6V 14V 0.6A 36 VO 14V 0.6A 2 15.2V IO VO is roughly a constant voltage. Example Find IO and VO. load 34 32V IO 34 14V D 2 VO VO > VZO , Zener diode is in breakdown. Replace by model. I O 32V 14V 0.5A VO 14V IO 2Ω 34Ω 2Ω 14V 0.5A 2Ω 15V 16V VO D IO 34 VO < VZO , zener diode is NOT in breakdown. D is OFF VO 16V 34Ω 8V and IO IS 34Ω 34Ω ELEC 101 c Diode Circuits 11 Zener diode and voltage regulator When reverse voltage of a diode is increased to high values, internal electric field increases and diode breakdown occurs. Large reverse (breakdown) current flows and reverse voltage of diode VZ is almost constant. If maximum power of diode is not exceeded, the breakdown process is non-destructive and reversible. In breakdown (zener) region of diode, reverse current IZ is large, but VZ is almost a constant. Hence diode can be used as a voltage regulator. IZO and VZO are breakdown (knee) current and voltage I V V V ZO Z VZMAX 1 slope rZ maximum Power point IZO IZ IZMAX IZ and VZ are operating current and voltage of zener diode breakdown (zener) region of diode 2009/10 Spring Find VO when (i) Vi = 4V , (ii) Vi = 6V, (iii) Vi = 8V. Given VZO= 5V, rZ = 0 . Example R Vi D VO Vo (i) Vi = 4V, diode is OFF. VO = Vi = 4V Vi (ii) Vi = 6V, diode is breakdown. VO VZO = 5V. 1V across R . R Vi Vo 5V (iii) Vi = 8V, diode is breakdown. VO VZO = 5V. VO is almost constant. 3V across R. ELEC 101 9 Diode Circuits 12 When diode is in forward bias or reverse ID VD 0 ID ID IS 1) Given in exam k : Boltzmann’s constant = 1.38 x 10-23 J/K T : absolute temperature in K (27oC = 300K) q : charge of electron = 1.6 x 10-19 C n : diode quality factor (or n factor), n = 1 for a good diode a Approximate form I D IS ID IS (eqVD /nkT 1) IS (eqVD /nkT ) IS eVD /nVT VD qVD (e nkT kT/q = VT is about 26 mV. VT is also called the thermal voltage. At 300K, Diode equation bias region 2009/10 Spring VD (en26mV ) When VD >> nkT/e n26mV 26-52mV Example - 5V Find ID . given I D I S (e ID IS (e 5V 26mV VD 26 mV ID 1) 1) IS (0 1) IS V Example Sketch I-V curve of diode when n = 1 and 2 . I 0 I n =1 n =2 V ELEC 101 Example Diode Circuits 13 Find Vi. c For the diode, given that ID = 1mA when VD = 0.7V, n = 2, and I D IS VD VD e n25mV 0.5mA 1k a VD ID = 0.5mA Vi b 1k 0.7V ID = 1mA 1k 0.5mA IS IS VD e 50mV 1mA IS IS VD e 225mV (1) 0.7V e 225mV 700mV e 50mV (2) ( 2) (1) 700 e 50 I 1mA S V e D 0.5mA 50 I Se 700 VD 50 2 700mV VD ln2 50mV VD 700mV 50mV ln2 665mV ID = Vi 2009/10 Spring d Vi VD ID 1k 665mV 0.5mA 1k 1165mV This is also a good method to find the n factor of diode (or to test the quality of an integrated circuit process line). n = 1 is a good diode (a good fabrication process). V1 nVT I1 I Se I V V2 V V2 ln 1 1 n 1 V2 I I2 I2 nVT nVT VT ln 1 I Se I2 If n = 1, VD increased by ~60mV per decade of current. I V V2 V1 V2 VTln10 V2 60mV ln 1 1 I2 nVT ELEC 101 10 Diode Circuits 14 Diode clamping circuit (or DC restorer) A diode circuit to shift the DC level of Vi. C - VC + V O Vi 2009/10 Spring a When Vi = -6V C Vi Vi 0V VO - 6V VC 0V -6V V VO 0V and VO = Vi + VC I flows and charge up C D D is ON and C is charged to VC Vi Diode circuit VO O VC 0 6V 6V Vi C D 0V Assume charging time of C << period of Vi Example C Vi 4V Vi 0V - 6V t 0 Given that C has no initial charge. Find VO. T VO D is OFF and VC is kept at 6V b When Vi = 4V Vi 4V 0V 4V Vi 6V C 10V VO D D 0V VO Vi VC 4V 6V 10V ELEC 101 c Diode Circuits 15 When Vi = -6V Vi 0V - 6V Steps to analyze clamping circuits: D is ON and VC is kept at 6V -6V Vi 6V C 1. 2. 3. 0V D 0V 0V VO C Vi Vi D When Vi = -6V 2 -6V D is ON and C is charged to VC Vi 10V Vi 0V 0V - 6V VO Vi VC VO = Vi + 10V VC = 10V 4V C 0V VO 14V 4V VO I flows and charge up C D 4V - 6V 3 VO 4V 0V - 6V 1 4V Find VO. 4V VO Vi VC 6V 6V 0V Vi Find Vi when D is ON. Find VC ( = VO – Vi ) Find VO Equation: VO = Vi + VC Example VO Hence VO = Vi + 6V 2009/10 Spring ELEC 101 11 Diode Circuits 16 2009/10 Spring Diode voltage regulator ac line 220 Vrms 50Hz Power transformer + + v _ _ S t Power transformer IL Diode rectifier t Voltage +v O regulator _ Filter t t Load t Filter Voltage regulator Diode rectifier ELEC 101 12 a Diode Circuits 17 Diode rectifier 2009/10 Spring Vi = 0V to Vm, D is ON, VO = Vi Vm 0 Half-wave rectifier Vi D VO Vi = 0V to -Vm, D is OFF, VO = 0 Vi Vm 0 Vi D VO Vi R 0 D VO VO 0 R Rectification is the conversion of alternating current (AC) to direct current (DC). Half-wave rectifier is the simplest kind of rectifier circuit. Vm -Vm VO Vm 0 -Vm R -Vm Hence VO waveform Vm VO 0 Vi ELEC 101 b Diode Circuits 18 Full-wave rectifier 2009/10 Spring Hence waveform of VO 2 half-wave rectifiers in parallel. centre-tap transformer VO 0 -Vm 1 D2 D1 Vm VO -Vm 2 -Vm D2 D1 I 0V Vm Vm D2 VO Disadvantages of full-wave rectifier: larger diode and centre-tapped transformer are needed. Current flows in half of transformer only (during each half cycle). Transformer lifetime is shortened. D1 is ON and D2 is OFF. I 0V 0 Vi D1 Vm Vm Vm VO Example Find VO (offset diode model, VF = 0.6V) . When Vi 0.6V, diode is ON 0 D1 is OFF and D2 is ON. VO VF Vm Vi VO VF Vm 0 When Vi < 0.6V, diode is OFF ELEC 101 c Diode Circuits 19 Full-wave bridge rectifier 2009/10 Spring Hence waveform of VO Vm 0 Improved full-wave rectifier. D1 Vm 0 -Vm Vi VO D2 Vi D3 Find VO of bridge rectifier. Vm = 4V. Assume offset diode model and VF is 0.7V. Example Vi R D4 D1 Vm 0 0 -Vm Vm D1 D2 0V Vi D3 -VmD1 D3 VO R D2 0V VO R D4 I Vm 0 VO VF Vi D3 Vm I 0 D4 Vi VO VF Vi = 4V , VO = 4V – 2VF = 2.6V D2 D4 4V 2.6V 1.4V 0 Vi = 1.4V = 2VF , D1, D4 are ON, VO = 0V VO R Vi VO ELEC 101 d Diode Circuits 20 Peak Inverse Voltage (PIV) of diode PIV should be smaller than the diode breakdown voltage. -Vm In half-wave rectifier, PIV of diode = Vm. Example In full-wave rectifier, PIV of diode = 2Vm. -Vm Example In bridge rectifier, PIV of diode = Vm. PIV is the maximum reverse voltage that a diode will block when the diode is reversebiased. Example 2009/10 Spring Vm 2Vm -Vm 0V ELEC 101 13 Diode Circuits 21 Capacitor Filter D is ON D Vm 0 -Vm VO Vi Vi C R D is ON Vm 0 -Vm 2009/10 Spring Vi Vi Vm 0 -Vm R C Vm 0 -Vm VO ( t ) Vm e Vm 0 -Vm VO Vi R C Vm 0 -Vm t CR 0 Vm 0 -Vm Vr : Ripple Voltage Vm VO ( VC ) is discharged to R Vi C VO 0 D is OFF Vi R VO ( VC ) is charged to Vm Vi VO ( VC ) is charged to Vm t T ELEC 101 a Diode Circuits 22 2009/10 Spring For full-wave rectifier, ripple is half. Ripple voltage Vr If T << CR, t T Vr Vm Vm e Vm Vm e Vm [1 e T t T CR Vr Vm CR CR If T/CR << 1, e-T/CR 1 – T/CR T )] Vr Vm [1 (1 CR T Vm CR Vm 0 Example T 2CR VO Vr 0 t T Find Vr of half-wave rectifier. Given Vm = 7V. RC = 20ms, T = 1ms. Assume offset diode and VF is 0.7V. If time constant CR >> period T of Vi Vr Vm T CR If Vr is small, VO is like a DC voltage. Vr Vm (1 e t / RC ) Vm ( T / CR ) (7V 0.7V )(1ms / 20ms) 0.32V ELEC 101 Diode Circuits 23 diode clamper Voltage doublers 14 C1 VO1 VO2 VO1 capacitor filter Example VO1 = Vi shifted up by Vm Vi Vm 0 - Vm VO1 VO1 2Vm Vm 0 VO2 Vi C2 Vi 2009/10 Spring Find VO1 and VO2. Given Vm = 4V. Assume offset diode and VF is 0.7V. VO1 Vi VC Vi Vm VF Vi (4V 0.7 V ) Vi 3.3V VO2 VO1 Vi VO2 is a DC voltage (= 2Vm) VO1 Vi -0.7V VO2 2Vm Vm 0 7.3V VO2 6.6V VO2 VO2 = peak of VO1 - VF 6.6V -1.4V ELEC 101 Diode Circuits 24 Example Voltage multiplier - Vm + V o1 Vi Vm - VF - 2Vm + Vo2 Vi - 2Vm + Vi 0 Vi Vo1 7.3V 3.3V -0.7V Vo1 = Vi + Vm – VF Vm - VF 2Vm – 2VF Vo2 = Vi + 3Vm – 3VF 13.9V 9.9V 5.9V Vo2 Vi 2Vm - VF VF C2 is charged C2 VF If Vm = 4V, VF = 0.7V VF 0 2Vm – 3VF 2Vm – 2VF Vm - VF Vm C3 is charged to 2Vm - 2VF 0 Can give DC voltage of Vm , 2Vm , 3Vm ... -Vm C3 VF Vo3 - 2Vm + C1 is charged to Vm - VF Vm - VF C1 -V F 2Vm – 2VF -VF -Vm Vi Vm 0 -Vm 2009/10 Spring to 2Vm - 2VF 2Vm – 2VF 4V 0 -4V 2Vm – 2VF Vo3 Vo3 6.6V = 2Vm – 2VF ...
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