BJT Circuit Analysis - Carnegie Mellon Universityee321/spring99/LECT/lect12feb24.pdf · BJT Circuit...

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Lecture 12-1 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: Q1 RB 100E3+ 2V VIN RC 1E3+ 5V VCC I S =1e-16 β 100 =

Transcript of BJT Circuit Analysis - Carnegie Mellon Universityee321/spring99/LECT/lect12feb24.pdf · BJT Circuit...

Page 1: BJT Circuit Analysis - Carnegie Mellon Universityee321/spring99/LECT/lect12feb24.pdf · BJT Circuit Analysis ... • Generally not considered for dc bias point calculations, but ro

Lecture 12-1

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:

Q1

RB100E3Ω

+2VVIN

RC1E3Ω

+ 5VVCC

IS=1e-16β 100=

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Lecture 12-2

BJT Circuit Analysis

• SPICE solves the system of nonlinear equations to obtain the voltages and currents

• Is this circuit in the active region?

Q1Default

RB100E3Ω

+ 2VVIN

RC1E3Ω

+ 5VVCC

IB

12.206 µA +

-

VOUT

3.779 V+

-

VBE

779.365 mV

IC

1.221 mA

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Lecture 12-3

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

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0

1

2

3

mA

IE

Q1Default

+ 0VVBE

0.000 pA

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Lecture 12-4

BE Diode Characteristic

• Note that “VON” changes if we’re analyzing an order of magnitude less current

• So how do we know what the real “VON” is?

Q1Default

+ 0VVBE

0.000 pA

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.0

0.1

0.2

0.3

mA

IE

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Lecture 12-5

Simplified BJT Circuit Analysis

• Assuming VBE is 0.78 volts, we can approximate this circuit solution by hand analysis

Q1

RB100E3Ω

+2VVIN

RC1E3Ω

+ 5VVCC

IS=1e-16β 100=

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Lecture 12-6

Simplified BJT Circuit Analysis

• What happens as RC is decreased?

• Will it remain in the active region?

Q1

RB100E3Ω

+2VVIN

RC500Ω

+ 5VVCC

IS=1e-16β 100=

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Lecture 12-7

Simplified BJT Circuit Analysis

• What happens as RC is increased?

• Will it remain in the active region?

Q1

RB100E3Ω

+2VVIN

RC5000Ω

+ 5VVCC

IS=1e-16β 100=

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Lecture 12-8

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 iC that we can have before we reach saturation for this circuit

Q1RB

+VIN

RC

+ 5VVCC

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Lecture 12-9

Saturation

• With both diodes forward biased, the collector-to-emitter voltage, vCE, saturates toward a constant value

_

VBC

+

_VBE

+

~0.4 volts

~0.7-0.9 volts

VBC

VBE

_

+

_

+ _

+

vCEsat 0.3volts≅

0 1 2 3 4 5

-1

0

1

2

mA

IC

VCE

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Lecture 12-

Saturation

• In saturation, increasing iC shows little increase in iB. Why?

0.0 0.2 0.4 0.6 0.8 1.0

-1

0

1

2

mA

IC

VCE

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Lecture 12-

Regions of Operation

• The complete i-v characteristic is:

0 1 2 3 4 5

-1

1

3

mA

VCE

IB=1µAIB=5µA

IB=10µA

IB=15µA

IB=20µA

IC

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Lecture 12-

Regions of Operation

VCE

IB=1µAIB=5µA

IB=10µA

IB=15µA

IB=20µA

0.0 0.2 0.4 0.6 0.8 1.0

1

3

mA saturation active

cut-off

IC

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Lecture 12-

Temperature Variations

• The collector current vs. the base-emitter voltage follows a diode characteristic, which like a diode, is temperature dependent

0.6 0.7 0.8

0

2

4

mA

VBE

T=32ºCT=27ºC

T=22ºC

Q1Default

+ 0VVBE

R41E3Ω

+ 15VVCC

• Does this value of RC significantly impact the values for iC in this example?

IC

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Lecture 12-

Temperature Variations

• In saturation, the collector current no longer increases with increasing VBE. Why not?

VBE

T=32ºCT=27ºC

T=22ºC

Q1Default

+ 0VVBE

R4100E3Ω

+ 15VVCC

0.6 0.7 0.8

0.0

0.1

0.2

mA

IC

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Lecture 12-

Base Width Variation

n-typep-typen-typeCE

BVBE VCB

x

• In the active region, iC does vary somewhat with VCB (hence RC in our previous examples) due to the variation it causes in the base width.

• Effective base width, W*, decreases with increasing VCB

• What do you expect would happen to iC as W* decreases?

W*

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Lecture 12-

Early Voltage

• The IC vs. VCE curves in the active region have a finite slope to them due to this iC dependence on VCB

• Early showed that these slopes all converge to one negative voltage point

VCE

IB=1µA

IB=5µA

IB=10µA

IB=15µA

IB=20µA-20 -10 0 10

-1

1

3

mA VAF=20 in SPICE(VA in the book)

-VA

IC

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Lecture 12-

Early Voltage

• The finite slope in the active region due to decreasing base width can be approximated by

ic Isevbe VT⁄

1vceVA-------+

=

• This means that the output resistance between the collector and emitter is not infinite --- very important for analog design

vce∂∂iC go 0≠=

0 1 2 3 4 5 6

-1

1

3

iC (mA)

at some fixed ib point

vce

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Lecture 12-

Early Voltage

• The output conductance, or resistance, at a fixed ib point represents the slope of the line tangent to that point on the curve:

ic Isevbe VT⁄

1vceVA-------+

=

• Generally not considered for dc bias point calculations, but ro can have a significant impact on a transistor amplifier gain

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Lecture 12-

Early Voltage

αie

ieIsα----e

vbe VT⁄=

E

B

C

ib

• The equivalent circuit models can be modified accordingly:

ro

VAiC-------=

βib

E

B

C

ibIsβ----e

vbe VT⁄=

ro

VAiC-------=

or

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Lecture 12-

dc Bias Point Calculations

• ro is generally not considered for hand calculations of dc bias point -- why?

• For hand calculations: use VBE=0.7 and assume that the transistor is in the active region; Later verify that your assumptions were correct.

4V

10V

3.3kΩ

RC

What’s the maximum value that RC can be without reaching saturation? Assume β =100.

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Lecture 12-

dc Bias Point Calculationsdc Bias Point Calculations

4V

10V

3.3kΩ

RC

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Lecture 12-

• What value of RC saturates the transistor?

10V

2kΩ

RC

β 100=

-10V

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Lecture 12-

dc Bias Point Calculations

• What value of VCC saturates the transistor for this same circuit?

10V

2kΩ

1kΩ

β 100=

VCC