Small Signals Modeling of BJT and Their Analysis (Part - 3) - Electrical Engineering (EE) PDF Download

h-Parameter Model vs. re Model 

•  CB re vs. h-Parameter Model 

Common-Base h-Parameters

 h_ib = r_e
h_fb = -α ≌ -1

• Small signal ac analysis includes determining the expressions for the following parameters in terms of Z_i, Z_o and AV in terms of – b

– r_e
– r_o and – R_B, R_C

•  Also, finding the phase relation between input and output
• The values of β, ro are found in datasheet
• The value of re must be determined in dc condition as r_e = 26mV / I_E

Common Emitter Fixed bias configuration

Removing DC effects of VCC and Capacitors

r_e model

Small signal analysis – fixed bias Input impedance Z_i: From the above re model, is,
Z_i = [R_B || βr_e] ohms
If R_B > 10 βr_e, then,
[R_B || βr_e] ≌ βr_e
Then,
Z_i ≌ βr_e

Ouput impedance Z_oi: Z_o is the output impedance when V_i = 0. When V_i = 0, i_b = 0, resulting in open circuit equivalence for the current source.

Z_o = [R_C|| r_o ] ohms

Voltage Gain A_v:
V_o = - βI_b( RC || ro)
From the r_e model, I_b = V_i / βr_e
thus, V_o = - β (V_i / β r_e) ( R_C || r_o)
A_V = V_o / V_i = - ( R_C || r_o) / r_e

If r_o >10R_C, A_V = - ( R_C / r_e)

Phase Shift: The negative sign in the gain expression indicates that there exists 180^o phase shift between the input and output.

Problem: 

Common Emitter - Voltage-Divider Configuration

Equivalent Circuit:

The re model is very similar to the fixed bias circuit except for R_B is R₁|| R ₂ in the case of voltage divider bias.

Expression for AV remains the same.

Z_i = R₁ || R₂ || β r_e

Z_o = R_C
:
Voltage Gain, AV:

From the r_e model,
Ib = V_i / β r_e
V_o = - I_o ( R_C || r_o),
I_o = β I_b
thus, V_o = -β (V_i / β r_e) ( R_C || r_o)
A_V = V_o / V_i = - ( R_C || r_o) / r_e
If r_o >10R_C,
A_V = - ( R_C / r_e)

Problem: Given: β = 210, r_o = 50kΩ.
Determine: r_e, Z_i, Z_o, A_V. For the network given:

To perform DC analysis, we need to find out whether to choose exact analysis or approximate analysis.

This is done by checking whether βR_E > 10R₂, if so, approximate analysis can be chosen.

Here, βR_E = (210)(0.68k) = 142.8kΩ.

10R₂ = (10)(10k) = 100k.

Thus, βR_E > 10R₂.

Therefore using approximate analysis, V_B = V_ccR₂ / (R₁+R₂)

= (16)(10k) / (90k+10k) = 1.6V

V_E = V_B – 0.7 = 1.6 – 0.7 = 0.9V 

I_E = V_E / R_E = 1.324mA

r_e = 26mV / 1.324mA = 19.64Ω

Effect of ro can be neglected if r_o ≥ 10( R_C). In the given circuit, 10R_C is 22k, ro is 50K.

Thus effect of r_o can be neglected.

Z_i = ( R₁||R₂||βR_E)

= [90k||10k||(210)(0.68k)]

= 8.47kΩ Z_o = R_C = 2.2 kΩ

 A_V = - R_C / R_E = - 3.24

If the same circuit is with emitter resistor bypassed, Then value of re remains same.

Z_i = ( R₁||R₂||βr_e) = 2.83 kΩ

Z_o = R_C = 2.2 kΩ

A_V = - R_C / r_e = - 112.02

Common Emitter Un bypassed Emitter - Fixed Bias Configuration

 Equivalent Circuit:

Applying KVL to the input side:

V_i = I_bβr_e + I_eR_E
V_i = I_bβr_e +(β +1) I_bR_E

 Input impedance looking into the network to the right of R_B is
Z_b = V_i / I_b = βr_e+ (β +1)R_E

Sinceβ>>1,   (β +1) = β
Thus, Z_b = V_i / I_b = β (r_e+R_E)

Since R_E is often much greater than r_e, Z_b

= βR_E, Z_i
= R_B||Z_b

Z_o is determined by setting V_i to zero, I_b = 0 and bI_b can be replaced by open circuit equivalent. The
result is, We             Z_o = R_C 
know that                   V_o = - I_oR_C

= - b_IβR_C
= - β(V_i/Z_b)R_C
A_V = V_o / V_i = -β(R_C/Z_b)
Z_b = β(r_e + R_E)
A_V = V_o / V_i = - β[R_C /(r_e + R_E)]
A_V = V_o / V_i = - β[R_C /R_E]

Phase relation: The negative sign in the gain equation reveals a 180 ^o phase shift 
between input and output.

 Problem:

Given: β = 120, r_o = 40kΩ . Determine: r_e, Z_i, Z_o, A_V.

To find r_e, it is required to perform DC analysis and find IE as r_e = 26mV / I_E To find I_E, it is required to find I_B.
We know that,

I_B = (V_CC – V_BE) / [R_B + (β+1)R_E]

I_B = (20 – 0.7) / [470k + (120+1)0.56k] = 35.89μA

I_E = (β+1)I_B = 4.34mA
r_e = 26mV / I_E = 5.99Ω

Effect of ro can be neglected,  if r_o ≥ 10( R_C + R_E)

10( R_C + R_E) = 10( 2.2 kΩ + 0.56k) = 27.6 kΩ

and given that ro is 40 kΩ , thus effect of ro can be ignored.

Z _i = R_B|| [β ( r_e + R_E)] = 470k || [120 ( 5.99 + 560 )] = 59.34Ω

Z_o = R_C = 2.2 kΩ

A_V = - βR_C / [β ( r_e + R_E)] = - 3.89

Analyzing the above circuit with Emitter resistor bypassed i.e.,

Common Emitter Iβ = (V_CC – V_BE) / [R_B + (β+1)R_E]

I_B = (20 – 0.7) / [470k + (120+1)0.56k] = 35.89mA

I_E = (β+1)I_B = 4.34mA

r_e = 26mV / I_E = 5.99Ω

Z_i = R_B|| [βr_e] = 717.70Ω

Z_o = R_C = 2.2 kΩ

A_V = - R_C / r_e = - 367.28  ( a significant increase)

Emitter – follower

r_e model

Z_i = R_B || Z_b
Z_b = βr_e+ (β +1)R_E
Z_b = β(r_e+ R_E)

Since R_E is often much greater than re, Z_b = βR_E

To find Z_o, it is required to find output equivalent circuit of the emitter follower at its input terminal.

This can be done by writing the equation for the current I_b. I_b = V_i / Z_b
I_e = (β +1)I_b
= (β +1) (V_i / Z_b)

We know that,

Z_b = βr_e+ (β +1)R_E 
substituting this in the equation for Ie we get, I_e = (β +1) (V_i / Z_b) = (β +1) (V_i / βr_e+ (β +1)R_E )

dividing by (β +1), we get,

I_e = V_i / [βr_e/ (β +1)] + R_E

Since (β +1) = β,

I_e = V_i / [r_e+ R_E]

Using the equation I_e = V_i / [r_e+ R_E], we can write the output equivalent circuit as,

As per the equivalent circuit,
Z_o = R_E||r_e
Since R_E is typically much greater than r_e,
Z_o ≌ r_e