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BASIC SEMICONDUCTOR ELECTRONIC CIRCUITS Introduction of two basic electronic elements: diode and transistor LEARNING GOALS Diodes structure and four modeling techniques Transistors MOSFETs and BJTs. MOSFETs in switching and amplification PASSIVE DEVICES IN INTEGRATED CIRCUITS Inductor Resistor Capacitor DIODES Structure Symbol I-V convention Ideal diode Forward bias Ideal diode Reverse bias Ideal diode I - V curve Actual diode I - V characteristic I S reverse saturation current q charge of electron k Boltzmann's constant T Temperature in degrees Kelvin qVD I D I S e nkT 1 n 'ideality' factor Typically : I S 1014 A, n 1 I - V curve of an actual diode For high power diodes : I S 1010 A, n 2 q 39 at room temperature kT Next we develop two approximations to the actual I - V curve Constant voltage model Circuit equivalent I - V curve Piecewise linear model Comparison of models Circuit equivalent I - V curve LEARNING EXAMPLE Using the ideal model, find the voltage V 1 V1 12[V ] Diode is forward biased by the source Forward bias V1 4[V ] Diode is reverse biased V1 10[V ] Reverse bias V1 1[V ] LEARNING EXAMPLE Analyze the protection circuit using diode models Simplifying assumption vin vs When diodes are reverse biased Thevenin equivalent of driving network Ideal diode model 0 vin Vdc Under-voltage with piecewise linear model Using source superposition vinu rd Rs vs VF rd Rs Rs rd Using a similar way for over-voltage Von vin Vdc Von Under-voltage protection model using diode constant voltage model vino rd RS vS (Vdc VF ) rd Rs rd Rs The figure in the next slide compares the three models Vdc 3V Von 0.7V (constant voltage model) VF 0.6V (piecewise linear model) rd 10 RS 40 Analysis using non linear diode model Simplifying assumption vin vs When diodes are reverse biased Under-voltage scenario. Top diode open v S vin iD 0 RS i D I S e 39v D 1 I S e 39vin 1 v D vin vin v S RS 1 vin v S vin ln 1 39 Rs I s Implicit function. Must be evaluated numerically Is=1e-14;% MATLAB SCRIPT TO ANALYZE n=1; % UNDER-VOLTAGE Rs=40; vs=[-2:.2:2]'; vin=[]; for k=1:length(vs) x0=[vs(k);vs(k);Rs;Is]; kcl=inline('((x(2)x(1))/x(3)+x(4)*(exp(-39*x(1))-1))^2'); x=fminsearch(kcl,x0); vin=[vin;x(1)]; end plot(vs,vin,'bo',vs,vin,'r'), title('UNDER-VOLTAGE WITH NONLINEAR DIODE') xlabel('Source Voltage(v_S)[V]'), ylabel('Input Voltage(v_{in})[V]'),grid TRANSISTORS STRUCTURE OF MOSFETs The oxide is an insulator allowing zero current into the gate The gate voltage controls resistance from source to drain (in linear range) Cutoff state Bulk internally connected to source threshold conducting N - channel MOSFET in the linear range Complete characteristics of n-channel MOSFET VDS Ron I D Linear characteristic Threshold voltage VT 1V A cutoff region B linear region C saturation region STRUCTURE OF BJTs (Bipolar Junction Tansistors) NPN transistor VBE constant Active mode: IC I E , IC I B 1 common emitter current gain ( 1) commonbase current gain B C IB V BE E Linear model for BJT in active mode MODELING MOSFET SWITCHING APPLICATIONS Ron depends on gate voltage. Cutoff Basic switching circuit Linear Region Typical values range from a few hundred Ohms for low-current applications to milli-ohms for currents in the tens of Amperes Cutoff (switch open) In the next slide we quantify this non-ideal behavior ‘Switch closed’ VS R 2 Ron Ideally,Vo 0 Vo The ratio, Ron , is more important than the actual value of Ron R LEARNING EXAMPLE The MOSFET is used as switch to energize and de-energize an inductor. Determine the output voltage with the “switch closed” R1 RL 1,VS 10V , Ron 0.25 Vo 0.1V Ron 0.01R 10m Vo 0.17 Vo 1.7V VS Circuit after all transients have subsided (R=1 Ohm) Von VS R 2 Ron Ron 0.25 R LEARNING EXAMPLE The MOSFET is used to switch 10A of current. The voltage drop across the FET in the on state must be less than 4% of the supply voltage. Find the maximum value of R_on Von Von 10 A Ron 0.04 40 Ron 0.16 MODELING AMPLIFIER APPLICATIONS MOSFET in saturation Linear model: id gmv gs VGS (by v gs ) I D (by id e) id applied to a resistor,R, will generate vo ( Rgm )v gs Rgm 1 amplification Due to non-zero slope of I - V curves in saturation Linear (small signals) model for MOSFET Common Source Amplifier MOSFET in saturation range Define drain-source current. Together with bias voltage they define operating point Bias voltage. Helps to define location on I - V curve. The operating point determines the values of the parameters gm , rds in the linear model Linear model relating changes in gate voltage to changes in output voltage RD also helps to determine gain LEARNING EXAMPLE Given : gm 50mA / V , rds 2k. Determine RD to produce a gain of - 50 vo (rds || RD ) gmv gs A (rds || RD ) gm 0.05[V / V ] 2k RD 50 RD 2k 2k RD LEARNING EXAMPLE If rds 2k 25%, what are the expected minimum and maximumgains rds min 2k (1 0.25) 1.5k Amin 0.05(1.5k || 2k ) 42.9 rds max 2k (1 0.25) 2.5k Amax 0.05(2.5k || 2k ) 55.5 PROBLEM If one must assure a gain of at least 50 in absolute value what should be the value for Rd? LEARNING EXAMPLE Find the value of R_d such that maximum average power is transferred to the load at 30krad/s 3104 ZTH R jL 1 j3 k LOAD Because there are dependent sources we must determine open-circuit voltage and short-circuit current ZTH VOC I SC Open circuit gm 1.5mS , rds 110k R rds || RD KCL : Small signals model For maximum power transfer the Thevenin impedance of the amplifier must be the complex conjugate of the load V Vgs VOC gmVgs OC 0 1 R jC1 VOC Vgs ( jC1 gm ) R jC1R 1 EXAMPLE (continued) Determination of short-circuit current I SC gmVgs jC1Vgs ( jC1 gm )Vgs VOC Vgs ( jC1 gm ) R jC1R 1 ZTH ZTH VOC R 1 jC1R I SC jC1R 1 1 jC1R R C1R 2 1000 j3000 j 2 1 (C1 ) 1 (C1 )2 C1R 3 R 3 10k 4 9 3 10 10 10k rds || RD 110k || RD RD 11k CASCADING COMMON SOURCE AMPLIFIERS v gs 2 (rds1 || R1 ) gm1v gs1 A1v gs1 vo (rds 2 || R2 ) gm 2v gs 2 A2v gs 2 vo A1 A2v gs1 ANALYSIS OF A DC-DC CONVERTER When S2 is open and S1 closed, the inductor stores energy. A large capacitor helps to keep output voltage constant When S2 is closed and S1 open, the inductor transfers energy to capacitor Idealized circuit 1 2 Vinton D duty cycle W L LI p ; I p 2 L Ts switchingperiod 1 Vin2 D 2Ts2 f ton DTs WL s Ts 2L Practical implementation of booster Turns FET on or off. When FET is off diode becomes direct biased 1 WC C (Vo2 Vo2 ) energy transferred to 2 AssumingW L WC Wload and solving for Vo capacitor Vo voltage at end of cycle Vo voltage at beginningof cycle Energy transferred to load Wload Vo2 Ts Rload 2W L 2Vo2 Vo Vo2 C Rload Cf s USING EXCEL TO COMPUTE OUTPUT OF DC-DC CONVERTER Vin[V] L[H] C[F] D fs Rload 5 1.00E-05 1.00E-06 0.6 1.00E+05 200 n Time(ms) 0.00E+00 1.00E-02 2.00E-02 3.00E-02 4.00E-02 5.00E-02 6.00E-02 7.00E-02 8.00E-02 9.00E-02 1.00E-01 1.10E-01 1.20E-01 1.30E-01 1.40E-01 1.50E-01 1.60E-01 1.70E-01 1.80E-01 1.90E-01 2.00E-01 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Excel formula to compute W_L =$B$1^2*($B$4)^2/(2*($B$2)*($B$5)^2) Excel formula to compute Vo (cellD10) Wl 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 4.50E-05 Vo 0 9.486833 13.0767 15.6173 17.5929 19.19789 20.53541 21.66871 22.64022 23.48024 24.21135 24.85097 25.41286 25.90815 26.34595 26.73384 27.07819 27.3844 27.65709 27.90024 28.11727 =SQRT(2*C10/$B$3-2*D9^2/($B$6*$B$3*$B$5)+D9^2) Plot of DC-DC converter output voltage MATLAB SCRIPT TO COMPUTE AND PLOT THE BOOSTER OUTPUT %booster.m %Solves the booster circuit in BECA 7th Ed, page648 %%%%%%%%% define circuit parameters Vin=5; L=1e-5; C=1e-6; D=0.6; fs=1e5; Rload=200; %%%%%%%%% define iteration array n=[0:30]; tms=1e3*n/fs; %time in msec %%%%%%%%% initialize vectors and do one-time computations Wl=Vin^2*D^2/(2*L*fs^2); t1=2*Wl/C; kt2=1-2/(Rload*C*fs); vo=zeros(size(n)); %initialize output array %%%%%%%%% do the iterations for k=2:length(n) vo(k)=sqrt(t1+kt2*(vo(k-1))^2); end %%%%%%%%% display results plot(tms,vo,'bo',tms,vo,'r'),title('DC-DC CONVERTER OUPUT') xlabel('Time(ms)'),ylabel('V_o(V)'), grid DIODES MOSFETs