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Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009

Recitation 18: BJT-Regions of Operation & SmallSignal Model

BJT: Regions of Operation

System of equations that describes BJT operation:

Ic = Is exp(qV

kT

BE) exp(qV

kT

BC) Is

(exp(qV

kT

BC) 1)

R

IB =

I

F

s(exp(qV

kT

BE) 1) +

I

R

s(exp(qV

kT

BC) 1)

IE = Is(exp(qV

kT

BE

) exp(qV

kT

BC

))

I

F

s

(exp(qV

kT

BE

) 1)

Is =qAEni

2DnB

NaBwBNdFDnB wE

F =NaB DpE wB

NdC DnB wCR =

NaB DpC wB

This set of equations can describe all four regimes of operation for BJT

Forward Active: VBE > 0, VBC < 0

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Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009

Reverse Active (RAR)

VBE

0

Cut-off

VBE < 0, VBC < 0

Saturation

VBE > 0, VBC > 0

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Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009

Understanding the IC vs. VCE curve: IC drops rapidly below VCE,SAT 0.1 t o 0.2 V.

Why?

Each curve IB is fixed VCE = VBE VBC, = VBC = VBE VCE When VCE is large, VBC < 0, FAR. As we reduce VCE, VBC reduces, at some point, VBC

starts to become forward biased. Now, hole flux from B C increases exponentially

from Law of Junction; to keep IB constant, hole flux into emitter must be reduced,

= VBE drops, = IC drops quickly.

Small Signal Model of BJT

(Next week we will be using BJT & MOSFET for amplifier circuits) Want to know the

small signal circuit model of BJT1. Transconductance gm =

ic

VBE Q

q qVBE/kT IcqVBE/kT IseIc = Ise = gm = =kT Vth

wNote, different from MOSFET: gm 2 ID (depends upon device size), but not for

Lbipolar case.

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Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009

2. Input resistance:

Is qVBE/

kT

IB = eF1 iB IB gm

g = = = = VBE Vth FF

or =gm

The input resistance of MOSFET is . In order to have a high input resistance for

BJT, need high current gain F

Example: npn with F = 150, Ic = mA

gm

=Ic

=1 103 A

= 40mSVth 0.025Vgm 40mS

g = = = 0.267mS ( = 3.7k)F 150

3. Output resistance: Ebers-Moll model have perfect current source in FAR. Real char-

acteristics show some increase in ic with VCE

go =ic

where does go come from?VCE

qAEn2DnB

In FAR: Ic = IseqVBE/kT = i eqVBE/kT

NaBwB

wB shrinks as |VBC| , thus Ic .

Ic IcModel: go = slope =

VCE + VAVA

(VA VCE)

1 Icgo = =o VA

Example: Ic = 100A, VA = 35 V, = o = 350 k

VA increases with increasing base width and increasing base doping. This is also why NaBusually NdC

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Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009

Now what do we have so far? Need to add capacitances...

Junction Capacitance (depletion capacitance)

(B-E): CjE =qsNaBNdE

( NdE NaB)2(NaB +NdE)(BE VBE)

qsNaBNdC qs NdC(B-E): CjC =2(NaB +NdC)(BC VBC)

2 (BC VBE)

Both are functions of bias

Since NaB NdC, CjE CjC. CjC is often called C.

Diffusion Capacitance

=CbVBE

|QnB|

|QnB| =21qAEwBnpBOeqVBE/kT

w2=

1wBwB qAEDnB

npBOeqVBE/kT = B Ic

2 DnB wB 2DnB 2 2 wB wBCb = ic = gm

VBE 2DnB 2Dn2w

2DB

nF= base diffusion transit time

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Recitation 18 BJT: Regions of Operation & Small Signal Model 6.012 Spring 2009

Cb is in between base and emitter:

Cb +CjE = C

Add the following

depletion capacitance: collector to bulk CCS

parasitic resistances: b of base, ex of emitter, c of collector

Complete small signal model

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6.012 Microelectronic Devices and Circuits

Spring 2009

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