BJT Circuit Analysis

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1 BJT Circuit Analysis Assuming that the transistor is in the active region, solve for the voltages and currents --- why this assumption? In general, the problem requires solution of a set of nonlinear equations: RC 1E3Ω VIN 2V + RB 100E3Ω Q1 I S =1e-16 β = V VCC 1

2 BJT Circuit Analysis SPICE solves the system of nonlinear equations to obtain the voltages and currents Is this circuit in the active region? IC ma RC 1E3Ω IB RB 100E3Ω Q1 Default VOUT µa + VIN VBE V - + 2V mv - VCC + 5V 2

3 BE Diode Characteristic We can effectively use a simplified model for the diode if we know the approximate operating range of the BE diode characteristic ma VBE + 0V pa Q1 Default IE 3

4 BE Diode Characteristic Note that V ON changes if we re analyzing an order of magnitude less current So how do we know what the real V ON is? ma VBE + 0V pa Q1 Default IE 4

5 Simplified BJT Circuit Analysis Assuming V BE is 0.78 volts, we can approximate this circuit solution by hand analysis RC 1E3Ω VIN 2V + RB 100E3Ω Q1 I S =1e-16 β = V VCC 5

6 Simplified BJT Circuit Analysis What happens as R C is decreased? Will it remain in the active region? RC 500Ω VIN 2V + RB 100E3Ω Q1 I S =1e-16 β = V VCC 6

7 Simplified BJT Circuit Analysis What happens as R C is increased? Will it remain in the active region? RC 5000Ω VIN 2V + RB 100E3Ω Q1 I S =1e-16 β = V VCC 7

8 Saturation When both the EBJ and CBJ are forward biased, the transistor is no longer in the active region, but it is in the saturation region of operation We can easily solve for the maximum i C that we can have before we reach saturation for this circuit RC VIN + RB Q1 + 5V VCC 8

9 Saturation With both diodes forward biased, the collector-to-emitter voltage, v CE, saturates toward a constant value _ V BC + + V BE _ ma 2 ~0.4 volts ~ volts V BC + + V BE + v CEsat 0.3volts _ V CE IC 9

10 Saturation In saturation, increasing i C shows little increase in i B. Why? V CE ma 1 0 IC -1

11 Regions of Operation The complete i-v characteristic is: I C V CE ma 1 I B =20µA I B =15µA I B =10µA I B =5µA I B =1µA -1

12 Regions of Operation I C ma saturation active 1 cut-off V CE I B =20µA I B =15µA I B =10µA I B =5µA I B =1µA

13 Temperature Variations The collector current vs. the base-emitter voltage follows a diode characteristic, which like a diode, is temperature dependent Q1 Default R4 1E3Ω I C V BE ma T=32ºC T=27ºC VBE + 0V VCC + 15V 2 T=22ºC 0 Does this value of R C significantly impact the values for i C in this example?

14 Temperature Variations In saturation, the collector current no longer increases with increasing V BE. Why not? I C 0.2 V BE Q1 Default R4 100E3Ω ma T=32ºC T=27ºC VBE + 0V VCC + 15V 0.1 T=22ºC 0.0

15 Base Width Variation In the active region, i C does vary somewhat with V CB (hence R C in our previous examples) due to the variation it causes in the base width. Effective base width, W *, decreases with increasing V CB What do you expect would happen to i C as W * decreases? E n-type p-type n-type C V BE B V CB W * x

16 Early Voltage The I C vs. V CE curves in the active region have a finite slope to them due to this i C dependence on V CB Early showed that these slopes all converge to one negative voltage point I C V CE 10 3 I B =20µA ma VAF=20 in SPICE (VA in the book) I B =15µA I B =10µA 1 I B =5µA I B =1µA -1 -VA

17 Early Voltage The finite slope in the active region due to decreasing base width can be approximated by v ce i c I s e v be V T = This means that the output resistance between the collector and emitter is not infinite --- very important for analog design V A v ce 6 i C (ma) i C = g o 0 v ce at some fixed i b point 1-1

18 Early Voltage The output conductance, or resistance, at a fixed i b point represents the slope of the line tangent to that point on the curve: i c I s e v be V T = 1 v ce V A Generally not considered for dc bias point calculations, but r o can have a significant impact on a transistor amplifier gain

19 Early Voltage The equivalent circuit models can be modified accordingly: C i e = B I s ---e v be α i b V T αi e r o = V A i C E or C i b = I s ---e v be β V T B βi b r o = V A i C E

20 dc Bias Point Calculations r o is generally not considered for hand calculations of dc bias point -- why? For hand calculations: use V BE =0.7 and assume that the transistor is in the active region; Later verify that your assumptions were correct. What s the maximum value that R C can be without reaching saturation? Assume β = V R C 4V 3.3kΩ

21 dc Bias Point Calculationsdc Bias Point Calculations 10V R C 4V 3.3kΩ

22 What value of R C saturates the transistor? 10V 2kΩ β = 100 R C -10V

23 dc Bias Point Calculations What value of VCC saturates the transistor for this same circuit? 10V 2kΩ β = 100 1kΩ VCC

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