ELL 100 - Introduction to Electrical Engineering LECTURE 24: BIPOLAR JUNCTION TRANSISTOR (BJT) Outline Static (dc) characteristics of common-base circuits Static (dc) characteristics of common-emitter circuits Transistor as an amplifier: DC and AC Load lines 2 COMPUTERS . Transistors present everywhere: CPU, memory, I/O units, display, mouse, keyboard, speaker, wi-fi, etc. DISPLAYS (TFT-LCD) . Controlling each pixel & color dot of the flat-panel display . Adjusting display back-light brightness STORAGE DEVICES : SDD . Every bit of (non-magnetic) memory storage: Flash, D-RAM, (E)P-ROM, volatile static RAM. Every bit of cpu register data & addresses & every bit of math or logic or encryption/decryption. STORAGE DEVICES : PEN DRIVE MOBILE PHONES Transistors are part of : . Modulating, transmitting, receiving, de-modulating . WiFi/Bluetooth/cellular WIFI MODULE BLUETOOTH CIRCUIT STATIC (DC) CHARACTERISTICS OF COMMON-BASE CIRCUITS npn . Vary input current IE using the potentiometer R1. For each value of IE, vary output voltage VCB using the potentiometer R2 and note the corresponding output current IC 9 STATIC (DC) CHARACTERISTICS OF COMMON-BASE CIRCUITS IE direction is taken to be flowing into the emitter terminal of BJT 10 STATIC (DC) CHARACTERISTICS OF COMMON-BASE CIRCUITS IC at a constant VCB IE current amplification factor for a common-base circuit 0.95 < α < 0.995 11 STATIC (DC) CHARACTERISTICS OF COMMON-EMITTER CIRCUITS npn . Vary input current IB using the input potentiometer. For each value of IB, vary output voltage VCE using the output potentiometer and note the corresponding output current IC 12 STATIC (DC) CHARACTERISTICS OF COMMON-EMITTER CIRCUITS 13 STATIC (DC) CHARACTERISTICS OF COMMON-EMITTER CIRCUITS IC at a constant VCE IB 1 current amplification factor for a common-emitter circuit 20 < β < 200 14 EFFECT OF REVERSE-BIASED LEAKAGE CURRENT IIIECB If the emitter is not connected, IE = 0. However, we find there is still a small collector current ICBO – this is a leakage current crossing the reverse-biased collector–base junction. IIIC E CBO ICBO IICB IIIB C CBO 1 IIIC1 B CBO IIB ( 1) CBO I II IICBO B CEO CB11 BJT MODES OF OPERATION • The two p-n junctions of the BJT can be independently biased, to result in four possible transistor operating modes as summarized in Table • A junction is forward-biased if the n material is at a lower potential than the p material, and reverse-biased if the n material is at a higher potential than the p material. BJT MODES OF OPERATION • Saturation denotes operation (with VCE ~ 0.2 V and VBC ~ 0.5 V for Si devices) such that maximum collector current flows and the transistor acts much like a closed switch (i.e. short- circuit) from collector to emitter terminals. • Cutoff denotes operation with near zero current, where the transistor acts much like an open switch. Only leakage current (of reverse-biased diode) flows in this mode of operation. Thus, IC = ICEO 0 for Common-Base connection, and IC = ICBO 0 for Common-Emitter connection. • Active or linear mode describes transistor operation in the region to the right of saturation (VCE > 1 V) and above cutoff (IC >> ICEO). Here, near-linear relationships exist between the output and input terminal currents, thus used in amplifier circuits. • Inverse mode is a little-used, inefficient active mode with the emitter and collector interchanged. TRANSISTOR AS AN AMPLIFIER (DC supply) (Small (Amplified a.c. signal) a.c. signal) The transistor is required to operate in a unidirectional (active/linear) mode, otherwise the negative parts of the a.c. would cause, say, the emitter–base junction to be reverse biased and this would prevent normal transistor action occurring. As a result, it is necessary to introduce a d.c. bias. 18 TRANSISTOR AS AMPLIFIER: BIASING & LOAD LINE IE IC AC input source voltage An npn transistor amplifier in common-base configuration with load resistance R 19 LOAD LINE FOR BJT IN IE IC COMMON-BASE CONFIG 1 V 6 V 100 Ω “Load-line” 1 kΩ IC = (B2 ‒VCB)/R = 6 ‒VCB With AC source S off, bias point D is obtained IE = -(B1 ‒VBE)/RS = -(1 – 0.7)/0.1 = -3 mA With AC source S on, input current IE varies between -1 & -5 mA (operating points E and C) 20 LOAD LINE FOR TRANSISTOR AMPLIFIER . The function of capacitor C is to eliminate the d.c. component of the voltage across R from the output voltage. If Ic is the RMS value of the a.c. component of the collector current, IcR gives the RMS value of the a.c. component of the output voltage and 2 Ic R gives the average output a.c. power. Note that a lower-case subscript (Ic) indicates an RMS a.c. value compared with a capital subscript (IC) for the d.c. value. 21 TRANSISTOR AS AMPLIFIER: BIASING & LOAD LINE IC C B IB E Simple common-emitter transistor amplifier 22 LOAD LINE FOR BJT IN COMMON-EMITTER CONFIG IC “Load-line” IB IC = (VS ‒VCE)/RC Ib Bias point Q is obtained IB = (VS ‒VBE)/RB ~ VS/RB With input AC source on, input base current varies as IB I bm sin t (operating points A and B) 23 TRANSISTOR AMPLIFIERS: RELEVANT QUANTITIES III Current gain G o CA CB i II2 ibm V o ΔI Voltage gain Gv o Vi ΔIi V P o o Vi Power gain GP Pi where Po is the output a.c. signal power in the load GGGP v i and Pi is the input a.c. signal power into the transistor. DYNAMIC CHARACTERISTICS & LINEARITY Larger the value of load resistance Dynamic characteristics R, the greater the voltage and power gains, the maximum value of R for a given collector supply voltage being limited by the maximum permissible distortion of the output voltage (due to deviation from linearity of transistor characteristics). 25 DC BIAS (Q-POINT) OF TRANSISTOR AMPLIFIER • Supply voltages and resistors bias a transistor i.e. they establish a specific set of dc terminal voltages and currents, thus determining a point of active-mode operation (called the quiescent point or Q point). • Usually, the quiescent point is unperturbed by the application of a small a.c. signal to the circuit. • e.g. Here D is the Q-point 26 BJT BIASING IN COMMON-EMITTER CONFIG II 1 RR12 EQ BQ RB RR12 R IEQ 1 VRVIR VVBB CC BB B BEQ EQ E RR12 1 27 BJT BIASING IN COMMON-EMITTER CONFIG I VRVIREQ BB1 B BEQ EQ E VVBB BEQ IICQ EQ RRBE/1 (α ~ 1) If component values are such that RB/(β+1) << RE, then ICQ is nearly constant, regardless of changes in β (which can change due to temperature, voltage & current fluctuations) 28 BJT IN COMMON-EMITTER: DC LOAD LINE Bias/quiescent point: Q = (VCEQ, ICQ) DC load line: . IBQ VVCC CE iC Rdc iI BRdc = BQ RC + RE 29 BJT IN COMMON-EMITTER: AC LOAD LINE 1. Coupling capacitors (CC) confine dc quantities to the transistor and its bias circuitry. 2. Bypass capacitor (CE) effectively removes the gain-reducing emitter resistor RE for ac signals, while allowing RE to play its role in establishing -independent dc bias. AC equivalent circuit 30 BJT IN COMMON-EMITTER: AC LOAD LINE The effective resistance seen by the d.c. bias collector current ICQ is Rdc = RC + RE. The a.c. collector signal current ic sees a collector-circuit resistance Rac = RCRL/(RC + RL). Since Rac Rdc in general, the concept of an a.c. load line arises. 31 BJT IN COMMON-EMITTER: AC LOAD LINE vce i c R ac Since ic = iC − ICQ and vce = VCEQ − vCE, the above equation can be written as VvCEQ CE iIC CQ AC load line equation Rac 32 BJT IN COMMON-EMITTER: AC LOAD LINE Bias/quiescent point: Q = (VCEQ, ICQ) VVCC CE DC load line: IC Rdc . IBQ Rdc = RC + RE VvCEQ CE AC load line: iIC CQ Rac Rac = RC || RL 33 NUMERICAL Q-1: In the given circuit, the silicon transistor has = 75 and the collector voltage is VC = 9V. Find the ratio RB / RC NUMERICAL Solution: Given VC = 9 V, = 75 IE Silicon transistor => VBE = 0.7 V 15 9 9 0.7 IE IB RC RB IC I IR6 6 RR B EB 1 BB 105.13 IRBC8.3 8.3 RRCC NUMERICAL Q-2: In the circuit shown, the transistor has β = 30. If the input voltage Vi = +5 V, show that the transistor is operating in the active mode. NUMERICAL Solution: In active mode, VBE = 0.7 V (base-emitter junction forward-biased) => IB = [(Vi – VBE)/15k] − [(VBE + 12)/100k] IB = [(5 – 0.7)/15] − [(0.7 + 12)/100] mA = 0.16 mA IC = βIB = 30×0.16 = 4.8 mA VCE = 12 – (2.2k)IC = 12 – (2.2×4.8) = 1.45 V VCB = VCE – VBE = 1.45 – 0.7 = 0.75 V Thus the collector-base junction is reverse-biased NUMERICAL Q-3: Two perfectly matched silicon transistors are connected as shown below. Assuming the of the transistors to be very high and the forward voltage drop in the diode to be 0.7 V, what is the value of the current I? NUMERICAL 0 0.7 0.7 5 I3.6 mA R 1k Applying KCL at P, I 2 III2 I 2 C I RCB C C Since is very large, IRC I3.6 mA NUMERICAL Q-4: The transistor in the given circuit should be biased in the active region. Take VCE (sat) = 0.2 V; VBE = 0.7 V. What is the maximum value of RC that can be used? NUMERICAL Solution: Applying KVL at the input, 4.3 5 0.7 2kI I 2.15 mA B B 2k Collector current, ICB I 215 mA Applying KVL to the output loop, 5 0.215RVC CE For transistor operation to be in active region, VVCE 0.2 4.8 0.215R 5 0.2 R = 22.32 C C 0.215 NUMERICAL Q-5: In the given circuit, = 100, IBQ = 20 μA, VCC = 15 V, RC = 3 kΩ.