ELECTRONICS II

Engr. Ryann AlimuinINSTRUCTOR

BJT Modelings• 2 models

1. RE Model

2. Hybrid Model

Model – is the combination of the circuit elements properly chosen that best appropriate the actual behavior of the semiconductor devices under specific operating condition.

Steps in getting the AC Equivalent Circuit

• Set all the DC source to zero and replace them by a short circuit equivalent.

• Replace all the capacitors by a circuit equivalent.• Remove all the elements bypassed by the short circuit

equivalent introduced by steps 1 and 2.• Redraw the networks In a more convenient and logical

form.

1. Input Impedance, Zi

• for small signal analysis, once the input impedance has been determined the same numerical value can be used for changing levels of applied signals.

• The input impedance of a BJT amplifier is purely resistive in nature and depending in the manner in which the transistor is employed, can vary form a few ohms to mse.

Note:

An ohmmeter can be used to measure the small signal ac input impedance since the ohmmeter operates in dc mode.

2. Output impedance, Zo

• The output impedance is determined at the output terminals looking back into the system with the applied signal set to zero.

3. Voltage Gain – Av

• One of the most important characteristics of an amplifier is the signal AC voltage gain as determined by

• Condition is if Rl is approaching infinity.

• AVNL – No Load Voltage Gain

• The load has not been converted to the output terminals• • For Transistor amplifiers , The no load voltage gain is

greater than the loaded voltage gain

Depending on the configuration, magnitude of the

voltage gain for a loaded single stage transitive amplifier

typically ranges from just less than 1 to a few hundred.

A multi-stage system, however can have a voltage gain in

thousands.

4. Current Gain- Ai

• For BJT Amplifier, the current typically range from a level , just less than 1 to a level that may exceed 100

• Voltage gain

Example:•Vi•Zi•Avnl•Avs (A.)Vi

Vs-IiRs-Vi=018mV-(10µA)(.65KΩ)-Vi

Vi= 11.5mV

(B.)ZiZi=Vi/Ii

=11mV/10mA=1.15KΩ

(C.) Avnl

Avnl=Vo/Vi

=3.6V/11.5mV

=313V

(D.) Avs

= Vo/Vs

=3.6/18mV

=200V

Systems Approach

• Effects of Rs and Rl

• Using Voltage Divider

Example

In the figure , a load impedance has been applied to the fixed bias transistor amplifier.Determine Av and Ai using two port systems. Determine Av and Ai using the RE Model and compare results.

• Rc= Ro• Ro=10.71• Avnl= -280.11• Zi= 1.071KΩ• Zo= 3KΩ

• Av using current divider theorem

2. For prefixed bias,

Determine:

A. Avnl,Zi and Zo

B. Sketch the two port model and parameters

C. Calculate Av.

D. Determine Ai

E. Calculate Av and Ai using AC analysis

• DC Analysis

• (B.)Sketch two port model and parameter• (C.)Calculate Av

• (D.)Determine Ai

• (E.)Calculate Av and Ai using AC Analysis

Effects of Source Impedance (Rs)• - the effect of an internal resistance on the gain of the

amplifier • - The parameter Zi and Avnl of a two port system are

unaffected by an internal resistance of the applied source.

• In Fig. A source w/ an internal resistance has been applied to the fixed bias transistor

• Determine the voltage gain Avs= Vo/Vs.What % of the applied signal

• Determine the voltage gain Avs= Vo/Vs using the RE model

DC Analysis

Combined Effects Of Rs and Rl

• A source of Rs and a load Rl have been applied in a two port systems for which the parameters Zi, Avnl, and Zo have been specified.

• Assume Zi and Zo are unaffected by Rl and Rs

• Using Voltage Divider Bias

• Example.

For a single stage amplifier with Rl= 4.7KΩ and Rs= 0.3KΩ.

BJT And JFET Frequency Response

• Logarithms

• Example. B=10 , X= 2

Decibels

• the relationship of logarithms to power and audio levels.

• the term (BEL) was derived from the surname of Alexander Grahambell.

Power = 10• Voltage = 20• Current = 20

Examples.

1.) Find the magnitude gain corresponding to a decibel gain of 100.

2.) The input power to a device is 10000v at avoltage of 1000v. The output power is 500w, while the output impedance is 20Ω.

• (a.)Find the power gain dB

• b.)Find the voltage gain in dB

CASCADED AMPLIFIER

AVT= (Av1)(Av2)

V4.77V

15KΩ4.7KΩ

4.7KΩ20V

RR

R VV

21

2

ccTH

3.58KΩ15KΩKRTH //7.4

A(201)1KΩ3.58KΩ

0.7V4.77V

1)R(βR

VVI

ETH

BETH

B 89.19

3.94mAμA(201)19.89IE

6.50Ω3.94mA

26mVre

953.69ΩΩ)(200)(6.50KBRZ reTHi //58.3//

102.3346.50Ω

953.69ΩK

re

ZiRcAv1

//2.2//

mVVAvVo i11 56.225334.102

338.466.50Ω

2.2KΩ

re

RcAv2

34,635338.46)102.334)((ΑvΤ

865.9mVV)5(34,635)(2ViAVo VT2 If a 10KΩ resistor is connected across the output. What is Vi?

709.75mV856.9mV2.2KΩ10KΩ

10KΩVo

RCRL

RLVi

JFET(JUNCTION FIELD EFFECT TRANSISTOR)

A type of FET that operates with a reversed-biased junctio.n to control current in a channel.

GM= Forward transconductance change in drain current

for a given change in rate to source voltage

with the drain to source voltage constant. DI GSV

2

GS(OFF)

GS

DSS

GS

B

V

V1IIB

SIEMENSΔV

ΔIGM

gm or Forward Transfer Admittance

DEPLETION MOSFET ( D-MOSFET)- The drain & source are diffused into the substrate material & then connected by a narrow channel adjacent to the insulated gate.

||V

2Igm

V

V1gmgm

GS(off)

DSS

o

GS(off)

GS

o

n-channel-operates in the depletion mode when a negative gate-to-source voltage is applied & in enhancement mode when a positive gate to source voltage is applied

FET AMPLIFIERS

AC Equivalent

gmRs1

gmRdAv

2

GS(Off)

D

SDSD

V

RsI1II

RsIV DGS

FET EQUIVALENT CIRCUIT

If no RL: Id

gmIdRd

gmId

IdRd

V

Vout

Vo

ViAv

Gs

gmRDAv

gmRdAv

Internal Drain Resistance

Effects of the source resistance on gain

By KVL:

)gm(rds//RdAv

)//Rgm(rds//(RAv CD

IdRdVo

IdRsVgsVi

IdRsVgsVi

0

( If there is Rs )

gmRs1

gmRdAv

IdRsgmId

IdRd

IdRsVgs

IdRd

Vi

VoAv

gmRs1

gmRdAv

Example: Bypassed source Resistance

D

LO

gmRAv

)//Rgm(RAv

RdZo

RZi G

gmRs1

gmRdAv

Example: The JFET has a gm=4mS w/ external ac drain resistance of 1.5K Ω , is the ideal voltage gain?

-6Av

4mS(1.5KAv

gmRDAv

)

Example: An FET equivalent circuit is shown. Determine the volatage gain when the output is taken across Rd.

1.852Av

4mS(560Ωm1

4mS(1.5KΩm

gmRs1

gmRAv

D

JFET SELF-BIAS CONFIGURATION

AC Analysis

yosrd

1

yos

1rd

Example: The fixed bias configuration has an operating point defined by VGSQ=-2V and

IDQ=5.625mA with IDSS=10mA & Vp=-8V, The network should be redrawn w/ an applied signal Vi. The Value of yos is provided as 40µS.

Calculate:

a. gm e. Av

b. rd f. Av if rd is IGNORED

c. Zi

d. Zo

Circuit Diagram

d.)2KΩ

c.)1MΩ

25KΩ40μ0

1

yos

1b.)rd

1.875mS8

212.5mgm

2.5mS8

0.02

//V

2Ia.)gm

p

DSS

o

75.3)2)(875.18(.)

475.3

)25//2)(875.1()//(.)

KmSAvf

KKmSRdRDgmAve

CASCODE CONNECTION (BJT)

- Cascode Connection has a transistor on top or in series with another.-A Common Emitter ( CE ) stage feeding a Common Base

(CB ) stage

Ex.Calculate the Av for the cascode amplifier

6.74Ω3.86mA

26mV

Ie

26mvre

3.86mA1.1KΩ

0.74.95

R

VVI

0RIVVusing:

10.89V6.8KΩ4.7KΩ5.6KΩ

4.7KΩ.18v(5.6KΩ

RRR

)RVcc(RVB2

4.95V18V6.8KΩ5.6KΩ4.7KΩ

4.7KΩVcc

RRR

RV

E

BEB1

E

EEBEB1

B3B2B1

B3B2

B3B2B1

B3

B1

sdsdsdsds

267.29267.29(1)))(A(AA

267.296.74KΩ

1.8KΩ

re

RcA

1re

re

re

RcA

V2V1VT

V2

V1

Darlington Connection- A super beta transistor-The composite transistor acts as a single unit with a current gain that the product of the current gain of the individual transistor.

if

21 D

2

21

D

What current gain is provided by a darlington connection of two identical transistor each having a current gain of

200

000,40

20022

D

D

EEE

BDE

BEEB

EBB

BECCB

EBBBEcc

EBBeBBcc

EEBeBBcc

RIV

II

VVV

RR

VVI

RRIVV

RIVRIV

RIVRIV

1

0)1((

0)1(

0

DC Bias of a darlington circuit

Calculate the dc bias voltage and current ECCB VVII ,,,

VV

IV

RIVV

VV

mARIV

mAI

AIII

AI

MRBR

VVI

CC

CCC

CCCCC

E

EEE

C

BDEC

B

EDB

BECCB

18

0

957.7

390403.20

403.20

55.2180001

55.2

39080003.3

6.118

VVBE

D

6.1

8000

AC equivalent circuit

- For a darlington emitter-follower the ac input signal is applied to the base of the darlington transistor through capacitor C1. with the ac output Vo obtained from the emitter through capacitor Cr

- The darlington transistor is replaced by an ac equivalent circuit composed of an input resistance, ri and an output current source BDIB

BDI

EDBO

DRBO

EDEBO

EEDEBO

OiBC

RIV

RIV

RRIV

RIRIV

VrIV

1

0

0

BEDi

BDi

DEiB

B

i

B

i

oD

BDBo

i

B

B

o

i

oi

DEiBi

DEiB

i

DEiBi

DEBiBi

oiBi

RRr

RA

RrR

R

I

I

I

I

III

I

I

I

I

I

IA

RrRZ

RrI

V

RrIV

RIrIV

VrIV

)//(

0

If ri is so small if not given = 0

BED

BDi RR

RA

VVBE

D

6.1

8000

4112

3.3)390(8

)3.7)(8(

61.1Z

)390)(8(5//3.3Z

5r if impedanceinput theCalculate

i

i

i

i

i

A

Mk

MkA

M

kkM

k

1

0given not is r i

v

EDi

EDv

EDiB

EDB

i

ov

A

if

Rr

RA

RrI

RI

V

VA

998.0A

)390)(8000(5

)390(8000A

Av Calculate

v

v

k

Feedback Pair

2121

2222

1111

CCCEC

EBC

EBC

IIIII

III

III

180

140

2

1

CB

EBB

CBBEB

BBCBEB

BBEBCB

BBEBCC

RR

VI

RRIV

RIRIV

RIVRI

RIVRI

21

CC1

21CC

21CC

21CC

CC

V

V

V

0V

0V

:KVLby

Calculate the dc bias current and voltages

AI

MRR

VVI

B

CDB

EBCCB

465.4

)75)(180)(140(2

7.18

1

1

mAI

mAAIII

mAI

AI

AI

III

C

CCC

C

C

C

BBC

143.113

518.1121.625

518.112

1.625180

1.625

)465.4(140

21

2

2

1

1121

VV

VVV

VVV

VV

mA

RIVV

dci

EBodc

i

oEBdc

i

dco

CCCCdc

o

81.8

7.051.9

0

51.9

)75)(143.113(18

)(

)(

)(

)(

\)(

AC Operation

11 BI 22 BI

BCi

CB

o

CBo

CB

CB

CBBo

CBBCCo

RRZ

RI

V

RIV

RI

RI

RIIV

RIIRIV

//

1

21

211

22

211

211

11121

1122

211

211

211

211

1121

111112

11122

1

11

1

B

o

Bo

B

B

B

BBB

BBBo

I

I

II

I

I

I

III

IIII

iC

CV

iBCB

CBoV

iBiiBoiB

Bi

i

B

B

oi

iB

B

i

B

rR

RA

rIRI

RIVA

rIVrIVZR

RA

I

I

I

IA

ZR

R

I

I

21

21

21

21

i

oi21

V

V , V

Calculate the ac circuit values of Zi , Zo and Av and Ai assume that ri = 3kΩ

KZ

MZ

i

i

72.971

2//)75)(180)(140(180

140

2

1

048.119

)180)(140(

3

)180)(140(

o

io

Z

KrZ

88.1695929.9712

)2)(140(180

KM

MAi

9998.03)750)(180)(140(

)750)(180)(140(

KAV

Differential Amplifier Circuit

-If an output signal is applied to either input with the other input connected to ground, the operator is referred to as “single ended”.

-If two opposite polarity input signals are applied, the operation is referred to as “double-ended”.

-If the same input is applied to both inputs, the operator is called “common mode”.

-In double ended operation two input signals are applied the difference of the inputs resulting in outputs from both collectors due to the difference of the signals applied is both input.

- In common-mode operation, the common-input signal results in opposite signals to each collector, these signals canceling so that the resulting output signal is zero.

- The main future of the differential amplifier is the very large gain when opposite signals are applied to the inputs as compared to the very small gain resulting common inputs.

Common-mode rejection ratio- ratio of the difference gain to common gain DC bias.

2E

C

II

0BVE

EEEE

EEEEE

R

VVI

VRIV

0

7.0

E

BEE

BEBE

V

VV

VVV

CE

CCC

CCCCC

RI

VV

RIVV

2

Solve for Ie and Vc

mVIK

I

E

E

515.23.3

)7.0(9

VV

Km

V

C

C

096.4

)3.3(2

515.29

Single Ended

11 BI 22 BI

eii

iii

BBB

rZr

rrr

III

21

21

i

iB

iBi

iBiBi

r

VI

rIV

rIrIV

2

02

0

e

CV

i

C

i

OV

Ci

iO

CBO

CCO

BC

r

RA

r

R

V

VA

Rr

VV

RIV

RIV

II

2

2

2

Calculate the single ended output voltage = Vo

AK

I

R

VVI

E

E

EEEE

02.19343

7.09

mAI

II

C

EC

51.962

02.193

2

39.26951.96

26

mA

mVre

mVV

mVV

VAV

KA

o

o

iVo

V

56.174

)2(23.87

23.87)39.269(2

47

Low Frequency Reponse- BJT Amplifier

• Effect of Cs on low frequency response

• Getting AC Equivalent

• Effect of Cc on the Low Frequency Response

Low Frequency Response

• Can be establish for each capacitive element and the frequency at which the output voltage drops to 0.707 of its maximum value

• -3db drop in gain from the midband level w hen f=f an RC network will determine the low-frequency cut-off frequency for a BJT transistor, f will be

• Example.– Determine the lower out off frequency using:

– Solution:

OP-AMP BASICS

• A very high gain differential amplifier with very high input ompedance and low output impedance.

1. Provide voltage amplitude amplitude changes

2. Oscillators

3. Filter circuits

4. Instrumentation circuit

• Single-ended input (mode)– Results when the input signal is connected to one

input with the order input connected to the ground.

• Differential mode, two opposite-polarity(out of phase signals are appliedto the inputs. Refered as double-ended).

• Common mode input voltage range– Range of input voltages which, when applied to both

inputs will not cause clipping or other output distortion.

• Input offset voltage– Differential dc voltage required between the inputs to

force the output to zero volts.

• Input offset voltage drift– Specifies how much change occurs in the input offset

voltage for each degree change in temperature.

• Input bias current– DC current required by the inputs of the amplifier to

properly operate the first stage.

• Input impedance ( Differential input impedance )– Total resistance between the inverting and non-

inverting inputs.

• Input offset current– Difference of the input bias currents.

• Output impedance– Resistance viewed from the output terminal.

• Double Ended Differential Input

• Double Ended output

• Common Mode Operation– Two inputs one equally amplified and since they result

in opposite polarity signals at the output, these signals cancel and results in 0V output.

• CMR ( Common Mode Rejection )

– Amplifier the difference signal while rejecting the common signal at the two inputs.

Common and Differential Mode Operation

• Differential Inputs– difference of the two signals.

• Common Input– Average of the sum of the two signals.

• Output Voltage

• CMMR – Common Mode Rejection Ratio

• The output for Vo

• Example.– Determine the output voltage of an op-amp for input

voltage of Vi=150μV, Vi2=140μV. The amplifier has a differential gain of Ad= 4000 and CMRR is

(a)100

(b)10^5

Solution:

• Slew Rate– Maximum rate of change of the output voltage in

response to a step input voltage.

• Negative Feedback– The inverting (-) input effectively makes the

feedback signal 180 degrees out of phase with the input signal.

• Closed-Loop Voltage Gain, Acl– Voltage gain of an op-amp with external feedback.

• Non-Inverting Amplifier– Op-amp connected in a closed-loop configuration.

• Example.– Determine the gain of the amplifier. The open-loop

voltage gain of the op-amp is 100,000.

– Solution:

• Voltage Follower– A special case of a non-inverting amplifier where all of

the output voltage is fed back to the inverting (-) input by a straight connection.

• Inverting Amplifier– Configuration where there is a controlled amount of

voltage gain.

• Example.– Given the op-amp configuration determine the value

of Rf required to produce a closed-loop voltage gain of -100.

– Solution:

• Impedance of Non-Inverting Amplifier– Input impedance

– Output impedance

• Example.– Determine the input and output impedance of the

amplifier . The op-amp datasheet gives Zin=2MΩ, Zout=75Ω and Acl= 200,000.

– Solution:

BASIC OP-AMP CIRCUITS

• Comparator– To determine when an input voltage exceeds a

certain level the (-) inverting input is grounded to produce a zero level and that the input signal voltage is applied to the non-inverting (+) input.

• Non-Zero Level Detection

– The zero level detector can be modified to detect voltages other than zero by connecting a fixed reference voltage source to the (-) inverting input.

• Example. The input signal is applied to the comparator circuit make a sketch of the output showing its proper relationship to the input signal. Assume the maximum output levels of the op-amp are 12V.

• Summing Amplifier– Has two or more inputs, its output voltage is

proportional to the negative of the algebraic sum of its input voltages.

• Summing Amplifier with Unity Gain

• Example.– Determine the output voltage.

• Summing Amplifier with Gain Greater than Unity.

Since

• Example.– Determine the output voltage for the summing

amplifier.

– Solution :

• Averaging Amplifier– A summing amplifier can be make to produce the

mathematical average of the input voltages. This is done by setting the ratio equal to the reciprocal of the number of inputs(n).

• Example.– Show that the amplifier produces an output whose

magnitude is the mathematical average of the input voltages.

– Solution :

• Scaling Adder

– Example. Determine the weight of each input and the output voltage.

• Solution:

• Multiple-Stage Gain

• Example– Determine the output voltage using the circuit for

resistor components of value Rf = 470kΩ, R1= 4.3kΩ, R2 = 33kΩ, and R3 = 38kΩ for an input of 80μΩ

• Voltage Subtraction

• Example.– Determine the output for the circuit with components

Rf= 1MΩ, R1= 100kΩ, and R3= 500kΩ.

solution:

Submitted by:

Octavo, Antonio

Pasquile, Ronald

Pineda, John

Ricarde, Julius Clarrence Ricarde B.

Rogador, Mark

Trinanes, Nomeer

EC32FB1