ECE:3410 Electronic Circuits - University of Iowas-iihr64.iihr.uiowa.edu/.../Lectures/BJTReview.pdf · A. Kruger BJT Review, Page- 4 Calculating Bias Currents & Voltages Key to solving

A. Kruger BJT Review, Page- 1

ECE:3410 Electronic Circuits

BJT

Sections in Chapter 5 & 6 in Textbook


npn Output Family of Curves


npn BJT dc Equivalent

VBE ~ 0.6 – 0.7 V for Si transistors at room temperature

Forward biased

Reverse biased

1

TBE Vv

SB eIi

Diode relationship


Calculating Bias Currents & Voltages

Key to solving circuits such as this, is that if the transistor is in the forward active

mode, then VBE ~ 0.7 (for Si)

DC model for npn BJT

What is IC, IE, VCE, etc., for the circuit below? Assume 𝛽 = 200

Example 5.3


Circuit with BJT dc model Circuit

220K

4 V

𝐼𝐵

2K

10 V

𝐼𝐶

−4 + 220K 𝐼𝐵 + 0.7 = 0

⇒ 𝐼𝐵 = 15 𝜇A

KVL around the red loop gives KVL around the green loop, using a loop current 𝐼𝑥, gives

⇒ 𝐼𝐶 = 𝛽 𝐼𝐵 = 3 mA

−𝑉𝐶𝐸 + 2K 𝐼𝑥 + 10 = 0

𝑰𝒙

However, we know that 𝐼𝑥 = −𝐼𝐶 = −3 mA

⇒ −𝑉𝐶𝐸 + 2K −3 m𝐴 + 10 = 0

⇒ 𝑉𝐶𝐸 = 4 V

Further 𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 (KCL) ⇒ 𝐼𝐸 = 3.02 mA

0.7 V


Calculating Bias Currents Figure 5.20

See example 5.3 in textbook for another style of solving this circuit


Voltage Transfer Characteristic for npn Circuit


Voltage Transfer Characteristic for pnp Circuit


Digital Logic

Inverter NOR gate

Figure 5.46

𝑽𝟏 𝑽𝟐 𝑽𝑶

0 0 5

0 1 0

1 0 0

1 1 0

𝑉𝑂 as function of 𝑉1, 𝑉2 NOR Gate Truth Table


Bipolar Inverter as Amplifier

Determine the Q-point


Improper Biasing

Input voltage

Output voltage is clipped


Single, Base-Resistor Biasing

VBE decreases by about 2 mV/oC. This increases IC, which heats up the

transistor, which further decreases VBE, , which increases IC even more , …

Generally a bad idea

Reason 1: β varies significantly from device to device

Reason 2: Thermal runaway

Figure 5.51


Stabilizing Q-Points

Stabilizes Q-point against

variations in β, temperature,…

Provides a

bias voltage

Bad

Figure 5.51

Good


npn BJT dc Equivalent

CCTH VRRRV )]/([ 212

VBE ~0.7 V

0BI

Find IB, IC, etc.

IB not zero, so

CCB VRRRV )]/([ 212

Replace base network with Thevenin equivalent

21 || RRRTH

0)1(7.0 EBTHBTH RIRIV

THBTHB RIVV


Voltage Divider Biasing and Emitter Resistor

0bi

CCTH VRRRV )]/([ 212

VBE ~0.7 V


Example

Find Q-point voltages and currents. For the Si transistor, = 50

VBE ~0.7 V


Example

Find Q-point voltages and currents. For the Si transistor, = 75

VBE ~0.7 V


Integrated Circuit Biasing

21

1

I

II QC

Constant Current Source

Current mirror

Question: what is I1? 1

1

3.4

RI


Cascade Circuit

npn transistor

pnp transistor


Moving On to Chapter 6 Material


CE with Time-Varying Input

Slope = -1/RC

vs changes vBE, which moves the Q-point along the load line

TBE Vv

SC eIi

Sect. 6.2.1


IB Versus VBE Characteristic

TBE VvSB e

Ii

vbe or vπ

ptQBE

B

v

i

r

1

ptQ

VvS

BE

TBEeI

v

ptQ

VvS

T

TBEeI

V

1

BQ

T

IV

1

TBE Vv

SC eIi

CQ

T

BQ

T

I

V

I

Vr

TBE Vv

SC eIi

What is the incremental resistance?

Or, if vBE changes by a small amount,

by how much will iB change?

The symbol vπ is going to be shorthand for ΔvBE


TBE Vv

SC eIi

CQI

IC Versus VBE Characteristic

vbe or vπ

ptQBE

Cm

v

ig

ptQ

Vv

S

BE

TBEeIv

ptQ

Vv

S

T

TBEeIV

1

CQ

T

IV

1

TBE Vv

SC eIi

T

CQ

mV

Ig

TBE Vv

SC eIi

What is the incremental increase

in IC as a function of an

incremental change in VBE?


AC Equivalent Circuit for CE


Small Signal Equivalent Circuits

AC Equivalent

Small Signal Model I

Small Signal Model II

Sect. 6.2.2 Original


Small-Signal Hybrid- Model for npn BJT

rg

I

Vr

IV

Ig

m

CQ

T

CQ

T

CQ

m 40

VT = ? VT ~ 26 mV at room temp., => gm ~ 40 ICQ at room temp.

vπ


Why “ “?


Small Signal Equivalent Circuits

AC Equivalent

Small Signal Model I

Small Signal Model II

Sect. 6.2.2


Hybrid- Model for npn with Early Effect

CQ

Ao

I

Vr

Sect. 6.2.3

How do we account for the slope?


Hybrid- Model for pnp with Early Effect


Small-Signal Equivalent Circuit for npn CE

B

CmvRr

rRgA

)(

)( VgRV mCo

?vA

S

Ov

V

VA

B

SRr

rVV

B

SmCoRr

rVgRV

𝑉𝑂

𝑅𝐶+ 𝑔𝑚𝑉𝜋 = 0 KCL at Output


Sect. 6.4.2

E B

C

E

B C

CE with RE

B C

E


CE with RE

Si

i

E

Cv

RR

R

Rr

RA

)1(

Si

E

C

Si

i

E

Cv RR

R

R

RR

R

R

RA

if

b

inib

I

VR

b

Ebbb

b

inib

I

RIIrI

I

VR

Eib RrR )1(

ibi RRRR21

S

Ov

V

VA

S

Cb

V

RI

0 Ebbbin RIIrIV KVL:

Ebbbin RIIrIV

Sib

inC

VR

VR

1


CE with RE

Si

E

Cv RR

R

RA if

50.4

2

E

Cv

R

RA


CE with RE

?vA

100.1

1

E

Cv

R

RA


CC or Emitter-Follower Amplifier

Av < 1

RO is low Ri is high (for BJT)

Sect. 6.6


Common-Base Amplifier

E

B

C

E

B C

Note sign of Vπ and direction of Io

Sect. 6.7


CB Small Signal Model

LCmo RRVgV ||

0

S

sm

E R

VVVg

r

V

R

V

KCL for node E: sum of currents flowing away from node is 0

E

mgr

1

SE

S

S RRr

R

VV ||

1

SE

S

LCmv RR

r

R

RRgA ||

1

||

0 as || SLCmv RRRgA

Sect. 6.7.1


Input Resistance: CB

Sect. 6.7.2

rV

Vgr

V

VgII

m

mbi

1

1

r

I

VR

i

ie


CB Output Resistance

Sect. 6.7.2

Previous Exam Question

?oR

x

xo

I

VR

0 VgR

VI m

C

xx

?Vgm

VgR

VI m

C

xx

KCL at collector (C):

Why? constantVgm

Vx cannot change the current flowing through the current source, so gmVπ is fixed and

we can remove it from the circuit just like any normal current source.

Turn off independent sources

and add test source Vx

Thus C

xx

R

VI C

x

xO R

I

VR

In other words, looking into the current source

from C, one sees and infinite resistance.


CB Output Resistance

If the previous result seems counterintuitive, below is an alternative approach. Start with a

BJT model that includes the output resistance ro. Then turn off independent sources, add a

test source, etc., to determine the output resistance.

KCL equations at the collector and emitter

nodes, using the convention that currents

flow away from the node (carefully note the

sign of Vπ) give

−𝐼𝑥 + 𝑔𝑚𝑉𝜋 +𝑉𝑥 − −𝑉𝜋

𝑟𝑜= 0

−𝑔𝑚𝑉𝜋 −𝑉𝜋

𝑅𝑆 𝑅𝐸 𝑟𝜋

+−𝑉𝜋 − 𝑉𝑥

𝑟𝑜= 0

𝑉𝑥

𝐼𝑋= 𝑅𝑂𝐶 = 𝑟𝑜 1 + 𝑔𝑚 𝑅𝑆 𝑅𝐸 𝑟𝜋 + 𝑅𝑆 𝑅𝐸 𝑟𝜋 Combining gives

Letting 𝑟𝑜 → ∞ results in 𝑅𝑂𝐶 → ∞ Same result as before: looking into the current

source, once sees an infinite resistance

See section 6.7.2 in 4th Edition of Neaman’s text


21iA ?vA?iR

Darlington Pair

Sect. 6.9.2

We will see how to answer the input impedance

question using BJT scaling a little bit later.


DC and AC Load Lines

Sect. 6.5

DC load line

AC load line

At operating frequencies

capacitors are shorts, and load

line is determined by RC || RL

At dc, capacitors are open, and

load line is determined by RC


Cascode Circuit

Voltages are in blue

Currents are in green

Common Emitter

Common Base


BTJ Impedance Scaling or Resistance Reflection Rules

Sections in Chapter 6 in

Textbook + Additional Material not In Textbook


Emitter Follower Analysis Sect. 6.6

Ri = ?

Ro = ?

Rib = ?

Notice that ro is in parallel with RE


Emitter Follower Analysis

Eooo RrIV

))(1( Eo

b

inib Rrr

I

VR

bo II 1

obin VrIV )||()1( Eobbin RrIrIV

Eobo RrIV 1

oES

o rRRRRr

R

1

|||| 21

Rib = ?

Ro = ?

Somewhat

involved to

derive each

time

} Notice that Rib is rπ plus resistance

at emitter, scaled up with (1+β)

Notice that Ro is base resistance

scaled down with (1+β), in parallel

with other resistance at emitter

Rib = ?

Ro = ?


Emitter Follower Small Signal Model

rπ

ro

ro

))(1( Eoib RrrR Example of BJT resistance scaling: looking

from the base, the emitter impedance is

scaled up by (1+

Rib = ?


BJT Resistance Scaling

rπ

ro

ro

?oR

oES

o rRRRRr

R

1

|||| 21

Resistance in base circuit divided by β+1


Resistance Scaling Example

Rib

Eib RrR )1(

ES

o RRRRr

R

1

|||| 21

The transistor in the circuit below has β =150. Neglect ro , and estimate the input

and output resistances shown. The collector current is IC = 0.8 mA, VA = ∞

mA/V 3240 Cm Ig

K7.4mg

r

K302K7.4

K306

rπ

Ro

8.33K2151

0.49KK7.4


BJT Resistance Scaling

)1( RgrRR moo

EmoEo RrgrRrR ||1)||(

Emo Rrgr ||1

The following sub-circuits often appears in small signal circuits

Question: can you derive this?


21iA ?vA?iR

Darlington Pair

Sect. 6.9.2


Composite Transistors

Sect. 8.5.4

BnBPpCp iii

BpnpBppn iii )1(2

Bnn ii )1(2

Bpnp ii 2


Example

Worked example on board

Darlington


Documents

ECE:3410 Electronic Circuits - University of Iowas-iihr64.iihr.uiowa.edu/.../Lectures/BJTReview.pdf · A. Kruger BJT Review, Page- 4 Calculating Bias Currents & Voltages Key to solving