I-V Characteristics of BJT Common-Emitter Output Characteristics C i C C i C B v CE B v EC i B E i B E Lecture 26 26-1
To illustrate the I C -V CE characteristics, we use an enlarged β R Collector Current (ma) 2 1 Saturation Region v CE v BE v CE v BE 0 i C = ( β R + 1) i B 0 Forward Active Region v CE i C = v BE β F i B I B = 100 µa I B =80µA I B =60µA I B =40µA I B =20µA I B =0µA Cutoff β F = 25; β R =5-1 -5 Reverse-Active Region Saturation Region v CE v BE 0 V 5 10 CE (V) Lecture 26 26-2
Common Base Output Characteristics E C i C E C i C i E B v CB i E B v BC Lecture 26 26-3
1.0 I E =1.0mA Collector Current (ma) 0.5 0 β F = 25; β R =5 Forward-Active Region I E =0.8mA I E =0.6mA I E =0.4mA I E =0.2mA I E =0mA -2 0 2 4 6 8 10 v CB or v BC (V) Lecture 26 26-4
Common-Emitter Transfer Characteristic i C - v BE. BE voltage changes as -1.8 mv/ o C - this is its temperature coefficient (recall from diodes). Collector Current I C (ma) 10 8 6 4 2 0 v BC =0 60 mv/decade 10-2 10-4 10-6 10-8 10-9 Log(I C ) -2 0.0 0.2 0.4 0.6 0.8 1.0 10-11 Base-Emitter Voltage (V) Lecture 26 26-5
Common-Emitter Transfer Characteristic i C - v BE (p. 180) v BE I C = I S exp -------- V T 1. BE voltage changes as -1.8 mv/ o C - this is its temperature coefficient (recall from diodes). Collector Current I C (ma) 10 8 6 4 2 0 v BC =0 60 mv/decade 10-2 10-4 10-6 10-8 10-9 Log(I C ) -2 0.0 0.2 0.4 0.6 0.8 1.0 10-11 Base-Emitter Voltage (V) Lecture 26 26-6
Junction Breakdown - BJT has two diodes back-to-back. Each diode has a breakdown. The diode (BE) with higher doping concentrations has the lower breakdown voltage (5 to 10 V). In forward active region, BC junction is reverse biased. In cut-off region, BE and BC are both reverse biased. The transistor must withstand these reverse bias voltages. Lecture 26 26-7
Junction Breakdown - BJT has two diodes back-to-back. Each diode has a breakdown. The diode (BE) with higher doping concentrations has the lower breakdown voltage (5 to 10 V). In forward active region, BC junction is reverse biased. In cut-off region, BE and BC are both reverse biased. The transistor must withstand these reverse bias voltages. Lecture 26 26-8
Junction Breakdown - BJT has two diodes back-to-back. Each diode has a breakdown. The diode (BE) with higher doping concentrations has the lower breakdown voltage (5 to 10 V). In forward active region, BC junction is reverse biased. In cut-off region, BE and BC are both reverse biased. The transistor must withstand these reverse bias voltages. Lecture 26 26-9
Minority Carrier Transport in Base Region - + - + v BE i B v BC i E Inj. Holes I F/ β F I R/ β R I i REC T (p no,n po ) Inj. Elec. Coll. Elec. recombined electrons N I REC P N Emitter Base Collector i C n(0) v BE n( 0) = n bo exp -------- V T 0 Space Charge regions n(x) i T = (p no,n po ) dn qad n ----- dx Electron conc. in base (neglects recombination) n(w B ) x W B Lecture 26 26-10
Transport current i T results from diffusion of minority carriers (holes in npn) across base region. Base current i B is composed of holes injected back into E and C and I REC needed to replenish holes lost to recombination with electrons in B. The minority carrier concentrations at two ends of base are and where is the equilibrium electron density in the base region. The junction voltages establish a minority carrier concentration gradient at ends of base region. For a narrow base, we get n bo W B is the B width; A is the cross-sectional area of B region. The saturation current is Lecture 26 26-11
Transport current i T results from diffusion of minority carriers (electrons in npn) across base region. Base current i B is composed of holes injected back into E and C and I REC needed to replenish holes lost to recombination with electrons in B. The minority carrier concentrations at two ends of base are V T v BE v n( 0) n bo --------- BC = exp and nw ( B ) = n bo exp --------- where n bo is the equilib- rium electron density in the base region. The junction voltages establish a minority carrier concentration gradient at ends of base region. For a narrow base, we get V T W B is the B width; A is the cross-sectional area of B region. The saturation current is Lecture 26 26-12
Transport current i T results from diffusion of minority carriers (holes in npn) across base region. Base current i B is composed of holes injected back into E and C and I REC needed to replenish holes lost to recombination with electrons in B. The minority carrier concentrations at two ends of base are V T v BE v n( 0) n bo --------- BC = exp and nw ( B ) = n bo exp --------- where n bo is the equilib- rium electron density in the base region. The junction voltages establish a minority carrier concentration gradient at ends of base region. For a narrow base, we get V T i T v BE dn n qad n ----- bo qad dx n -------- --------- v BC = = W exp exp --------- B V T V T. W B is the B width; A is the cross-sectional area of B region. The saturation current is Lecture 26 26-13
Transport current i T results from diffusion of minority carriers (holes in npn) across base region. Base current i B is composed of holes injected back into E and C and I REC needed to replenish holes lost to recombination with electrons in B. The minority carrier concentrations at two ends of base are V T v BE v n( 0) n bo --------- BC = exp and nw ( B ) = n bo exp --------- where n bo is the equilib- rium electron density in the base region. The junction voltages establish a minority carrier concentration gradient at ends of base region. For a narrow base, we get V T i T v BE dn n qad n ----- bo qad dx n -------- --------- v BC = = W exp exp --------- B V T V T. W B is the B width; A is the cross-sectional area of B region. 2 n bo n The saturation current is I S qad n -------- i = = qad. W n -------------------- B N AB W B Lecture 26 26-14
Base Transit Time Forward transit time is time associated with storing charge Q in Base region and it is τ F Q = ---- with Q qa[ n( 0) n bo ] W B = --------. 2 i T v BE W Using Q qan bo --------- B = exp 1 we get V -------- T 2 Lecture 26 26-15
n(0) n(x) Q = excess minority charge in Base Q n bo 0 W B n(w B )=n bo x v BE W Using Q qan bo --------- B = exp 1 we get V -------- T 2 i T 2 2 qad n v BE W --------------n and. W bo --------- B W = exp 1 B V τ F --------- B = = ---------------- T 2D n 2V T µ n Lecture 26 26-16
1 This defines an upper limit on frequency f ------------. 2πτ F n(0) n(x) Q = excess minority charge in Base Q n bo 0 W B n(w B )=n bo x Lecture 26 26-17
PSPICE EXAMPLE 0Adc I1 Q1 IS=1.0e-15 BF=100 VAF=80 0Vdc V1 *Libraries: * Local Libraries :.LIB ".\example10.lib" * From [PSPICE NETLIST] section of C:\Program Files\OrcadLite\PSpice\PSpice.ini file:.lib "nom.lib" *Analysis directives:.dc LIN V_V1 0 5 0.05 + LIN I_I1 10u 100u 10u.PROBE V(*) I(*) W(*) D(*) NOISE(*).INC ".\example10-schematic1.net" **** INCLUDING example10-schematic1.net **** * source EXAMPLE10 Lecture 26 26-18
PSPICE EXAMPLE (Cont d) Q_Q1 N00060 N00159 0 Qbreakn V_V1 N00060 0 0Vdc I_I1 0 N00159 DC 0Adc **** RESUMING example10-schematic1-example10profile.sim.cir ****.END **** BJT MODEL PARAMETERS ****************************************************************************** Qbreakn NPN IS 1.000000E-15 BF 100 NF 1 VAF 80 BR 3 NR 1 VAR 30 CN 2.42 D.87 JOB CONCLUDED TOTAL JOB TIME.21 Lecture 26 26-19
PSPICE EXAMPLE (Cont d) 12mA 8mA 4mA 0A -4mA 0V 0.5V 1.0V 1.5V 2.0V 2.5V 3.0V 3.5V 4.0V 4.5V 5.0V -I(V1) V_V1 Lecture 26 26-20