Bipolar Junction Transistors

Transcription

1 Bipolar Junction Transistors Physical Structure & Symbols NPN Emitter (E) n-type Emitter region p-type Base region n-type Collector region Collector (C) B C Emitter-base junction (EBJ) Base (B) (a) Collector-base junction (CBJ) (b) E PNP - similar, but: N- and P-type regions interchanged Arrow on symbol reversed Operating Modes Operating mode EBJ CBJ Cut-off Active Saturation Reverse-active Reverse Forward Forward Reverse Reverse Reverse Forward Forward Active Mode - voltage polarities for NPN VCB > 0 B IC C VBE > 0 IB E IE (Based on Dr Holmes notes for EE1/ISE1 course) 1

2 BJT - Operation in Active Mode n p n IE E { IEn electrons IEp holes recombination C IC IB B I En, I Ep both proportional to exp(v BE /V T ) I C I En I C I S exp(v BE /V T ) (1.1) I B I Ep << I En can write I C = β I B where β large (1.2) I S = SATURATION CURRENT (typ to A) V T = THERMAL VOLTAGE = kt/e 25 mv at 25 C β = COMMON-EMITTER CURRENT GAIN (typ 50 to 250) Active Mode Circuit Model B IB IC C β IB IE = IB + IC E (Based on Dr Holmes notes for EE1/ISE1 course) 2

3 BJT Operating Curves - 1 INPUT-OUTPUT I C vs V BE (for I S = A) IC (ma) 100 CUT-OFF ACTIVE VCB > 0 B IC C VBE E VBE (V) ACTIVE REGION: I C 0 for V BE < 0.5 V I C rises very steeply for V BE > 0.5 V V BE 0.7 V over most of useful I C range I B vs V BE similar, but current reduced by factor β CUT-OFF REGION: I C 0 Also I B, I E 0 (Based on Dr Holmes notes for EE1/ISE1 course) 3

4 BJT Operating Curves - 2 OUTPUT I C vs V CE (for β = 50) IC (ma) SAT ACTIVE I B = 200 µa I B = 160 µa I B = 120 µa I B = 80 µa I B = 40 µa B IB C E IC VCE VCE (V) ACTIVE REGION (V CE > V BE ): I C = β I B, regardless of V CE i.e. CONTROLLED CURRENT SOURCE SATURATION REGION (V CE < V BE ): I C falls off as V CE 0 V CEsat 0.2 V on steep part of each curve In both cases: V BE 0.7 V if I B non-negligible (Based on Dr Holmes notes for EE1/ISE1 course) 4

5 Summary of BJT Characteristics CUT-OFF VCB > 0 ACTIVE IC 0 IB 0 IC = IS exp(vbe/vt) IC = β IB VBE 0.7 V if IC non-negligible VBE < 0 VBE > 0 REVERSE-ACTIVE SATURATION IC < β IB VBE 0.7 V if IB non-negligible VCE < VBE (by definition) VCB < 0 Also I E = I B + I C (always) THIS TABLE IS IMPORTANT - GET TO KNOW IT! For PNP table: Reverse order of suffices on all voltages in table i.e. V CB V BC etc Reverse arrows on currents in circuit i.e. arrows on I B, I C point out of PNP device, while arrow on I E points in. (Based on Dr Holmes notes for EE1/ISE1 course) 5

6 Common-Emitter Amplifier Conceptual Circuit RC IC VCC VIN VOUT Assume active mode: I C = I S exp(v IN /V T ) Apply Ohm s Law and KVL to output side: V OUT = V CC - R C I C (1.3) = V CC - R C I S exp(v IN /V T ) NOTE: Called common-emitter because emitter is connected to reference point for both input and output circuits. Common-Base and Common-Collector also important. (Based on Dr Holmes notes for EE1/ISE1 course) 6

7 C-E Amplifier Input-Output Relationship e.g. V CC = 20 V, R C = 10 kω, I S = A, V T = 25 mv. V OUT (V) 20 ΔV IN 15 ΔV OUT 10 5 Operating Point V IN (V) Plenty of voltage gain i.e. ΔV OUT >> ΔV IN BUT: Highly non-linear Output distorted unless input signal very small Need to BIAS transistor to operate in correct region of graph to get high gain without distortion (Based on Dr Holmes notes for EE1/ISE1 course) 7

8 C-E Amplifier Small-Signal Response - 1 Aim: to get quantitative information about the small-signal voltage gain and the linearity of a C-E amplifier Start with the large signal equations: V OUT = V CC - R C I C = V CC - R C I S exp(v IN /V T ) Suppose we add to V IN a small input signal voltage v in, resulting in a corresponding signal v out at the output. We can relate v out to v in by expanding the above as a Taylor series: V OUT + v out = V CC - R C I C [1 + v in /V T + (v in /V T ) 2 /2 +..] (1.5) Assuming v in << V T, we can neglect quadratic and higher terms, giving: V OUT + v out V CC - R C I C - R C (I C /V T )v in v in << V T This is a LINEAR APPROXIMATION, valid only when v in is small Cont d.. (Based on Dr Holmes notes for EE1/ISE1 course) 8

9 C-E Amplifier Small-Signal Response - 2 Using (1.3), we can separate the output voltage into BIAS and SIGNAL components: V OUT = V CC - R C I C Quiescent O/P Voltage v out - R C (I C /V T )v in Output Signal SMALL-SIGNAL VOLTAGE GAIN: A v = v out /v in = - R C I C /V T = - R C g m (1.10) e.g. If quiescent O/P voltage lies roughly mid-way between the supply rails then R C I C V CC /2. In this case A v = -V CC /(2V T ), so for V CC = 20 V we get A V = The quantity g m = I C /V T is known as the TRANSCONDUCTANCE of the transistor. LINEARITY Include higher order terms from Equation 1.5: v out - R c g m [ v in + v in 2 /2 V T ] Ratio of unwanted quadratic term to linear term is v in /2V T, so expect 10 % distortion when v in /2V T 0.1, or v in 5 mv. Amplifier is linear only for very small signals (Based on Dr Holmes notes for EE1/ISE1 course) 9

10 Bias Stabilisation - 1 Biasing at constant V BE is a bad idea, because I S and V T both vary with temperature, and we require constant I C (or I E ) for stable operation. Also, I S is not a well-defined transistor parameter. We can obtain approximately constant I E as follows: VCC RC VBIAS vin L RE VOUT + vout (a) KVL in loop L (with no signal) gives: I E = (V BIAS - V BE ) /R E (1.11) (V BIAS V) /R E if V BIAS >> V BE I E relatively insensitive to exact value of V BE Get I C from I C = α I E where α = β/(1 + β) 1 α is the COMMON-BASE CURRENT GAIN (Based on Dr Holmes notes for EE1/ISE1 course) 10

11 Bias Stabilisation - 2 R E provides NEGATIVE FEEDBACK i.e. if the emitter current starts to rise as a result of some change in the transistor s characteristics, then the voltage across R E rises accordingly. This in turn lowers the base-emitter voltage of the transistor, tending to bring the emitter current back down towards its original value. STABILISATION BUT R E also: Reduces small-signal voltage gain: A v = - R C g m /(1 + I E R E /V T ) (1.12) - α R C /R E Reduces output swing (Based on Dr Holmes notes for EE1/ISE1 course) 11

12 Bias Stabilisation - 3 Recovery of Small-Signal Voltage Gain We can recover the original value of A v for AC signals by using a BYPASS CAPACITOR: VCC RC VBIAS vin RE VOUT + vout CE (b) Now we have: A v = - R C g m /(1 + I E Z E /V T ) (1.12b) where Z E is the combined impedance of R E and C E : Z E = R E /(1 + jωr E C E ) By making C E large enough, we can make the parallel combination appear like a short circuit (i.e. Z E 0) at all AC frequencies of interest, so that Equation 1.12b reduces to A v - R C g m as for our original common-emitter amplifier. On the other hand, the capacitor has no effect on biasing, because it passes no DC current. NB Technique only really relevant to discrete circuits (no big capacitors inside IC s!) (Based on Dr Holmes notes for EE1/ISE1 course) 12

13 Example 1 Analyze the circuit below to determine the voltages at all nodes and the currents in all branches. Assume β = V BE is around 0.7V (Based on Dr Holmes notes for EE1/ISE1 course) 13

14 Example 2 Analyze the circuit below to determine the voltages at all nodes and the currents in all branches. Assume β = 100. (Based on Dr Holmes notes for EE1/ISE1 course) 14

15 Step 1: Simplify base circuit using Thévenin s theorem. Step 2: Evaluate the base or emitter current by writing a loop equation around the loop marked L. Step 3: Now evaluate all the voltages. VB = VBE + I ERE = = 4. 57V β IC = ( ) I E = = 1. 28mA 1+ β VC = + 15 ICRC = = 8. 6V (Based on Dr Holmes notes for EE1/ISE1 course) 15