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Advance Electronics

Prof. Rajput SandeepAssist. Prof., EC Dept.HCET ,Siddhpur

Chapter 1: Transistor at High Frequency

Lecture : 1

The hybrid- pi CE Transistor Model

The hybrid π model at low frequencies

Chapter 1: Transistor at High Frequency

Lecture : 2

The Input Conductance - gb’e To obtain the expression for the input conductance gb’e , refer to the

low frequency hybrid – π model shown in figure 1.

Figure 1:The hybrid π model at low frequencies

Note that all the capacitor from the high frequency hybrid- π model have been removed because the capacitances at low frequencies are of negligible values.

The Input Conductance - gb’e Figure 2 shows the h parameter model at low frequency, for the same

transistor.

Now, rb’c > rb’e therefore almost all current Ib will flow through rb’e .

so, V b’e = Ib x rb’e (1)

Figure 2:The hybrid model at low frequencies

The Input Conductance - gb’e So that, the short circuit collector current in figure 1 is given by ,

Ic = gm Vb’e = gm ( Ib x rb’e ) (2)

The short circuit current gain is defined as,

hfe = Ic / Ib = gm rb’e

or,

rb’e = hfe / gm = (hfe VT) / | Ic | (3)

Now, input conductance , gb’e = Ib / Vb’e

= 1 / rb’e

By substituting the value of rb’e we can get,

gb’e = gm / hfe (4)

This is the required expression for the input conductance.

The Input Conductance - gb’e Conclusion :

By referring the below equation ,

rb’e = hfe / gm = (hfe VT ) / | Ic | (5)

It can be concluded that if hfe is constant over a certain range of current, then rb’e is directly proportional to temperature (due to V T ) and it is inversely proportional to current Ic .

Therefore input conductance gb’e will be inversely proportional to temperature and directly proportional to current if hfe is constant.

The feedback Conductance – gb’c

Let us define the reverse voltage ratio hre using the low frequency hybrid – pi circuit of figure 1 as follows:

hre = (Vb’e / V ce ) | for Ib = 0 (6)

Let’s see the modified low frequency hybrid – pi model in figure 3.

Figure 3: Modified hybrid – pi model at low frequencies

The feedback Conductance – gb’c

From the figure 3 it is evident that rb’e and rb’ c form a potential divider across Vce .

So that , Vb’e = { rb’e / (rb’e + r b’c ) } x Vce (7)

so, hre = Vb’e / V ce = { rb’e / (rb’e + r b’c ) } (8)

Rearrange this expression ,

h re rb’e + hre rb’c = rb’e so, h re rb’c = rb’e - hre rb’e

= rb’e ( 1 – hre ) (9)

The feedback Conductance – gb’c

But, hre < < 1. Hence (1- hre ) = 1.

so that , hre rb’c = rb’e

The feedback conductance gb’c is defined as,

gb’c = Ib / Vce

But , Ib = Vb’e / r b’e so that gb’c = Vb’e / ( Vce * rb’e )

But, Vb’e / Vce = hre and 1 / rb’e = gb’e

So that, The feedback conductance = hre * gb’e . (11)

The Base Spreading Resistance – rb’b

Let us define hie from the low frequency hybrid- pi model.

From the figure 3 it is evident that rb’e and rb’ c from a potential divider across Vce and rb’c > > r b’e so that,

(rb’e || rb’c ) = rb’e (12)

Figure 4: Modified hybrid – pi model at low frequencies

The Base Spreading Resistance – rb’b

Hence hie = (Vbe / Ib ) | Vce =0

= {Ib rb’b + Ib * (rb’e || rb’c )}/ Ib

So that , hie = rb’b + (rb’e || rb’c ) = rb’b + rb’e (13)

This is the expression for hie in terms of hybrid – pi parameter.

The base spreading resistance is given by,

rb’b = hie – rb’e

In the expression for hie, Substitute the expression of

rb’e = (hfe VT ) /|Ic | to get,

hie = rb’b + rb’e = rb’b +{ hfe VT / |Ic |} (14)

Chapter 1: Transistor at High Frequency

Lecture : 3

The output conductance – gce The output conductance hoe can be obtained from the low frequency

hybrid – pi equivalent circuit by open circuiting the input terminal, i.e. Ib = 0.

Figure 5 : Modified hybrid – pi model at low frequencies

From figure 5 , hoe = ( Ic /Vce ) | Ib = 0 , (16)

But Ic = I1 + I2 + I3

Now, I1 = Vce / r ce , I2 = gm V b’e = gm hre Vce and I3 = Vce / (rb’e + rb’c)

Now Substituting (1 / r ce ) = gce and (1/rb’c ) = gb’c to get,

hoe = gce +gm hre + gb’c (17)

Hence output conductance, gce = hoe – gm hre – gb’c

Substitute gm = hfe gb’e and hre = gb’c / gb’e to get,

gce = hoe – [ hfe gb’e * (gb’c / gb’e ) ] – gb’c (18)

so that , gce = hoe – hfe gb’c – gb’c

gce = hoe – ( 1 + hfe ) gb’c (19)

This is the required expression for the output conductance.

The output conductance – gce

The Hybrid – pi Capacitance The Basic high frequency hybrid – pi model includes two

capacitance.

Cb’c or Cc and Ce.

Capacitance Cb’c or Cc :

Cb’c or Cc is the collector junction capacitance. It is measured CB capacitance between collector ( C ) and base ( B ) with the input (E) open.

Cb’c or Cc is usually specified by the manufacturer as Cob.

In the active region of the transistor operation , the CB junction is reverse biased. Hence Cc represents the transition capacitance CT.

Cc is inversely propositional to VCE, because as VCE is increased, the width of the depletion region also increases and Cc decreases.

The Hybrid – pi Capacitance

Capacitance Ce : Ce capacitance appears between base and emitter, in the hybrid – pi model.

Ce represents the sum of emitter diffusion capacitance (CDe ) and the emitter junction capacitance CTe.

But the emitter junction is forward biased and the diffusion capacitance of a forward biased junction is much higher than its transition capacitance.

i.e. CDe > > CTe

Therefore, Ce = (CDe + CTe ) = CDe (20)

Ce or CDe is propositional to the emitter bias current Ic and it is almost independent of temperature.

The Hybrid – pi Capacitance The diffusion capacitance CDe is mathematically expressed as,

CDe = gm (W2 / 2DB ) (21)

Where DB = Diffusion constant for minority carriers in the base.

Experimentally Ce or CDe is determined from the measurement of fT , a frequency at which the CE short circuit current gain reduces to 1.

The value of Ce in terms of fT is given by, Ce = gm / 2 π fT (22)

But, Ce = CDe + Cte (23)

Therefore , (CDe + CTe ) = gm / 2 πfT

So that fT = gm / 2 π (CDe + CTe ) . (24)

Simplified Hybrid – pi Model

Figure 6 : Simplified hybrid – pi model

If we neglect the resistance rb’b because it is too small and rb’c because it is too high, Then we get simplified hybrid model as shown in figure 6.

Validity of Hybrid – pi Model Consider the CDe and Ce equations, CDe = gm (W2 / 2DB )

(25)

= (W2 / 2DB r‘e)

= (W2 IE/ 2DB VT )

And g m = α ∂IE / ∂ VE , (26)

In both the equations we have assumed that VBE varies slowly. This is done to maintain the minority carrier distribution in the base region triangular as shown in figure 7.

Validity of Hybrid – pi Model If the distribution remains triangular under the condition of varying

VBE then the slope at x =0 and x = W are equal. Hence the emitter and collectors currents will also be equal.

Figure 7 : Modified- carrier charge distribution in the base region

If the rate of change of VBE is small such that the base incremental current IB is small in comparison to the collector incremental current IC then under such dynamic condition the hybrid – pi model is valid.

Validity of Hybrid – pi Model Scientist Giacollecto has shown that the network elements of hybrid –

pi model are frequency independent provided that,

2πf W2 / 6DB < < 1 (27) Now we know that,

CDe = gm (W2 / 2DB )(28)

= (W2 IE / 2DB VT)

And Ce = gm / 2 π fT

Therefore, W2 /6DB = Ce / 3gm = 1/ 6πfT

So that f < < 6πfT / 2π = 3fT (29)

Thus the hybrid π model is valid for frequencies approximately up to fT / 3.

Chapter 1: Transistor at High Frequency

Lecture : 4

The CE Short Circuit Current Gain Consider the single stage CE transistor amplifier or the last stage of

cascade configuration with load RL on this stage.

The hybrid – pi equivalent circuit for a single transistor with a resistive load RL is show in figure 8.

Figure 8 : The hybrid – pi equivalent circuit for a single transistor with resistive load RL

The CE Short Circuit Current Gain In this as we want to obtain the short circuit current gain. So we will

short the output terminals, C and E. Therefore RL = 0.

The approximate equivalent circuit with RL = 0 is as shown in figure 9.

Figure 9 : Equivalent circuit with RL = 0

Note that the resistance rb’c has been neglected as it is of very high value, and rce is short circuited.

The capacitance Ce and Cc will appear in parallel with each other.

The CE Short Circuit Current Gain The short circuit load current IL is given by,

IL = - gmVb’e (30)

Now let us calculate Vb’e from figure 9. Let the parallel combination of rb’e and Cc+Ce be represented by Z.

Z = { rb’e * 1/(jw ( Ce + Cc ) )} / { rb’e + 1/(jw ( Ce + Cc )) } (31)

= rb’e / {rb’e * (jw ( Ce + Cc ) + 1)}

= rb’e / {1 + rb’e * (jw ( Ce + Cc ))} (32)

Including this term into the equivalent circuit we et the simplified hybrid – pi equivalent circuit as shown in figure 10

The CE Short Circuit Current Gain

Figure 10 : Equivalent circuit with Z included

From figure 10 we can write that,

Vb’e = Ii x Z = (Ii * rb’e ) / (1+ jw rb’e (Cc + Ce)) (33)

Substituting this value into equation (31) we get,

IL = - gm rb’e Ii /{1 + rb’e * (jw ( Ce + Cc ))}

The CE Short Circuit Current Gain

Figure 11: The short circuit CE current gain Vs frequency (plotted on the Log- Log scale

Therefore, the current gain is given by,

AI = IL / Ii = - gm rb’e /{1 + rb’e * (jw ( Ce + Cc ))}

But hfe = gm rb’e So that, AI = - hfe /{1 + rb’e * (jw ( Ce + Cc ))} where ,w = 2πf

The CE Short Circuit Current Gain That means the current gain is inversely propositional to frequency.

At very low frequencies, the term jw rb’e (Cc + Ce ) is very small as compared to 1.Thereforecurrent gain AI ~ hfe.

But as the frequency increases the, current gain goes on decreasing.

Figure 11 shows the variation in the current gain AI expressed in decibels (i.e. 20log|AI |) against frequency on a logarithmic frequency scale.

Note that AI (dB) = 0 dB at frequency f = fT and for f > > fB the graph is a straight line having a slope of 6 dB / octave or 20 dB/decade.

Let us now define various frequencies as fα , fβ and fT .

Chapter 5: Operational Amplifiers

Lecture : 5

“operational” was used as a descriptor early-on because this form of amplifier can perform operations of

• adding signals

• subtracting signals

• integrating signals, dttx )(

The applications of operational amplifiers ( shortened to op amp ) have grown beyond those listed above.

Operational Amplifiers

At this level of study we will be concerned with how to use the op amp as a device.

The internal configuration (design) is beyond basic circuit theory and will be studied in later electronic courses. The complexity is illustrated in the following circuit.

Operational Amplifiers The op amp is built using VLSI techniques. The circuit diagram of an

LM 741 from National Semiconductor is shown below.

Internal circuitry of LM741

Fortunately, we do not have to sweat a circuit with 22 transistors and twelve resistors in order to use the op amp.

The circuit in the previous slide is usually encapsulated into a dual in-line pack (DIP). For a single LM741, the pin connections for the chip are shown below.

Pin connection, LM741.

Operational Amplifiers

i n ve r t i n g i n p u t

n o n i n ve r t i n g i n p u to u t p u t

V -

V +

The basic op amp with supply voltage included is shownin the diagram below.

Basic op am diagram with supply voltage.

Operational Amplifiers

In most cases only the two inputs and the output areshown for the op amp. However, one should keep inmind that supply voltage is required, and a ground.The basic op am without a ground is shown below.

Outer op am diagram.

Operational Amplifiers

A model of the op amp, with respect to the symbol, isshown below.

V 1

V 2

_

+

V d R i

R o

A V d

V o

Op Amp Model

Operational Amplifiers

The previous model is usually shown as follows:

R i

R i

A V d

_

+

V d

V 1

V 2

V o

+

_

Working circuit diagram of op amp.

Operational Amplifiers

Application: As an application of the previous model,consider the following configuration. Find Vo as a function of Vin and the resistors R1 and R2.

+

_

R 2

R 1

+

_

+

_

V i nV o

Op amp functional circuit.

Operational Amplifiers

In terms of the circuit model we have the following:

R i

R i

A V i

_

+

V iV i n V o

+

_

+

_

R 1

R 2

ab

Total op amp schematic for voltage gain configuration.

Operational Amplifiers

R i

R i

A V i

_

+

V iV i n V o

+

_

+

_

R 1

R 2

ab

Circuit values are:

R1 = 10 k R2 = 40 kA = 100,000 Ri = 1 meg

Operational Amplifiers

We can write the following equations for nodes a and b.

ioin

o

oiiiin

AVk

)VV(V

k

VV

meg

V

k

)VV(

4050

40110

Operational Amplifiers

Equation simplifies to;

inio VVV 10012625

simplifies to;

010410005.4 95 io VxVx

Operational Amplifiers

From Equations we find;

ino VV 99.3

This is an expected answer.

Fortunately, we are not required to do elaborate circuitanalysis, as above, to find the relationship between theoutput and input of an op amp. Simplifying the analysisis our next consideration.

Eq 8.5

Operational Amplifiers

For most all operational amplifiers, Ri is 1 meg orlarger and Ro is around 50 or less. The open-loop gain, A, is greater than 100,000.

Ideal Op Amp:The following assumptions are made for the ideal op amp.

i

o

RohmsinputInfinite

RohmsoutputZero

AgainloopopenInfinite

;.3

0;.2

;.1

Operational Amplifiers

Ideal Op Amp:

_

+ ++

+

+

_

__ _

V i

V 1

V 2 = V 1V o

i 1

i 2

= 0

= 0

(a) i1 = i2 = 0: Due to infinite input resistance.

(b) Vi is negligibly small; V1 = V2.

Figure 8.9: Ideal op amp.

Operational Amplifiers

Ideal Op Amp:

Find Vo in terms of Vin for the following configuration.

+

_

R 2

R 1

+

_

+

_

V i nV o

Figure 8.10: Gain amplifier op amp set-up.

Operational Amplifiers

Ideal Op Amp:

+

_

R 2

R 1

+

_

+

_

V i nV o

a

V i

Writing a nodal equation at (a) gives;

21

)(

R

VV

R

VV oiiin

Eq 8.6

Operational Amplifiers

Ideal Op Amp:

21

)(

R

VV

R

VV oiiin

With Vi = 0 we have;

With R2 = 4 k and R1 = 1 k, we have

ino VV 4 Earlierwe got ino VV 99.3

Eq 8.7

inVR

RV

1

20

Operational Amplifiers

Ideal Op Amp:

When Vi = 0 in Eq 8.7 and we apply the Laplace Transform;

1

20

R

R

)s(V

)s(V

in

Eq 8.8

In fact, we can replace R2 with Zfb(s) and R1 with Z1(s) andwe have the important expression;

)s(Z

)s(Z

)s(V

)s(V

in

fb

in

0 Eq 8.9

Operational Amplifiers

Ideal Op Amp:

At this point in circuits we are not able to appreciate theutility of Eq 8.9. We will revisit this at a later point incircuits but for now we point out that judicious selectionsof Zfb(s) and Zin(s) leads to important applications in

• Analog Filters

• Analog Compensators in Control Systems

• Application in Communications

Operational Amplifiers

Ideal Op Amp:

Example 8.1: Consider the op amp configuration below.

+

+

+

_

__

3 VV in

6 k

1 k

V 0

a

Figure 8.11: Circuit for Example 8.1.

Assume Vin = 5 V

Operational Amplifiers

+

+

+

_

__

3 VV in

6 k

1 k

V 0

a

At node “a” we can write;

k

V

k

)V( in

6

3

1

3 0

From which; V0 = -51 V

Eq 8.10

Example 8.1 cont.

Operational Amplifiers

Example 8.2: Summing Amplifier. Given the following:R fb

R 1

R 2

V 2

V 1V 0

a

Figure 8.12: Circuit for Example 8.2.

fbR

V

R

V

R

V 0

2

2

1

1 Eq 8.11

Operational Amplifiers

Example 8.2: Summing Amplifier. continued

Equation 8.11 can be expressed as;

2

21

10 V

R

RV

R

RV fbfb Eq 8.12

If R1 = R2 = Rfb then,

210 VVV Eq. 8.13

Therefore, we can add signals with an op amp.

Operational Amplifiers

Example 8.3: Isolation or Voltage Follower.

Applications arise in which we wish to connect one circuitto another without the first circuit loading the second. This requires that we connect to a “block” that has infinite inputimpedance and zero output impedance. An operational amplifier does a good job of approximating this. Considerthe following:

T he" B lo c k "

C irc u it 1 C irc u it 2+

_

+

_V in V out

Figure 8.13: Illustrating Isolation.

Operational Amplifiers

Example 8.3: Isolation or Voltage Follower. continued

C i r c u i t 1 C i r c u i t 2

Th e B l o c k

+

_

+V i n V 0_

Figure 8.14: Circuit isolation with an op amp.

It is easy to see that: V0 = Vin

Operational Amplifiers

Example 8.4: Isolation with gain.

+

_

__

+

+

2 0 k

V in

V in

V 0

1 0 k

1 0 k

a

+

_

Figure 8.15: Circuit for Example 8.4:

Writing a nodal equation at point “a” and simplifying gives;

inVV 20

Operational Amplifiers

Example 8.5: The noninverting op amp.

Consider the following:

R 0

R fb

V 0V 2_

+

+

_

a+

_

Figure 8.16: Noninverting op am configuration.

Operational Amplifiers

Example 8.5: The noninverting op amp. Continued

Writing a node equation at “a” gives;

20

0

02

0

02

0

2

1

,

11

0)(

VR

RV

giveswhich

RRV

R

V

so

R

VV

R

V

fb

fbfb

fb

Remember this

Operational Amplifiers

Example 8.6: Noninverting Input.

Find V0 for the following op amp configuration.

+

_

+

+

_

_ 4 V

2 k

6 k

5 k

1 0 k

V 0

a

V x

Figure 8.17: Op amp circuit for example 8.6.

Operational Amplifiers

Example 8.6: Noninverting Input.

The voltage at Vx is found to be 3 V.

Writing a node equation at “a” gives;

0105

0

k

)VV(

k

V xx

or

VVV x 930

Operational Amplifiers

THANKS