The document summarizes the collector-to-base bias circuit with voltage feedback (voltage divider bias). It provides the following key points: 1) Derives expressions for the stability factors S, S', and S" by taking derivatives of the key equations for the circuit. 2) S is approximately 1, providing good stability, due to the resistor RE which provides negative feedback. 3) The voltage divider formed by R1 and R2 provides bias voltage VTH, while RE improves stability by reducing the effect of variations in the collector current IC.

1. COLLECTOR TO BASE BAIS
SELF BIAS
Dr.N.G.Praveena
Associate Professor/ECE
R.M.K. COLLEGE OF ENGINEERING
AND TECHNOLOGY

2. COLLECTOR-TO-BASE BIAS (DC BIAS
WITH VOLTAGE FEEDBACK)
IB + IC

3.  For dc analysis
ac input signal = 0
C1 and C2 – open circuit (∵f = 0)
IB + IC IB + IC

4. STEP 1: OBTAIN AN EXPRESSION FOR IB
Apply KVL to the base
𝑽𝑪𝑪 = 𝑰𝑩 + 𝑰𝑪 𝑹𝑪 + 𝑰𝑩𝑹𝑩 + 𝑽𝑩𝑬
𝑽𝑪𝑪= 𝑰𝑩𝑹𝑪 + 𝑰𝑪𝑹𝑪 + 𝑰𝑩𝑹𝑩 + 𝑽𝑩𝑬
𝑽𝑪𝑪 = (𝑹𝑩 + 𝑹𝑪 )𝑰𝑩 + 𝑰𝑪𝑹𝑪 + 𝑽𝑩𝑬
(𝑹𝑩+𝑹𝑪 )𝑰𝑩 = 𝑽𝑪𝑪 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑪
𝑰𝑩 =
𝑽𝑪𝑪 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑪
𝑹𝑩+𝑹𝑪
--- (1)
IB+IC

5. STEP 2: TO OBTIAN S , DIFFERENTIATE IB WITH RESPECT TO IC AND
SUBSITUTE IN THE STANDARD EQUATION OF S
Differentiate eqn (1) with respect to 𝑰𝑪
𝝏𝑰𝑩
𝝏𝑰𝑪
= -
𝑹𝑪
𝑹𝑪+ 𝑹𝑩
----------- (2)
We know that the standard equation of S is
S =
(1 + β)
1 – β ∂ 𝑰𝑩 / ∂ 𝑰𝑪
---(3)
Sub (2) in (3)
S =
(1 + β)
1 – β − 𝑹𝑪
𝑹𝑪+ 𝑹𝑩
S =
(1 + β)
1 + β 𝑹𝑪
𝑹𝑪+ 𝑹𝑩
𝑰𝑩 =
𝑽𝑪𝑪 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑪
𝑹𝑩+𝑹𝑪
--- (1)

6. STEP 3: TO OBTAIN S′, REPLACE IB INTERMS OF VBE IN THE
STANDARD EQUATION OF IC AND DIFFERENTIATE IT WITH RESPECT
TO VBE
The Standard Equation of Ic is
𝑰𝑪 = 𝜷𝑰𝑩 + (1+𝛃) 𝑰𝑪𝑶
Sub (1)
𝑰𝑪 = β (
𝑽𝑪𝑪 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑪
𝑹𝑩+𝑹𝑪
) (∵ 𝑰𝑪𝑶<<)
𝑰𝑪 (𝑹𝑩 + 𝑹𝑪 ) = β ( 𝑽𝑪𝑪− 𝑽𝑩𝑬 − 𝑰𝑪 𝑹𝑪)
𝑰𝑪 (𝑹𝑩 + 𝑹𝑪 ) + β 𝑰𝑪𝑹𝑪 = β ( 𝑽𝑪𝑪 − 𝑽𝑩𝑬)
𝑰𝑪 (𝑹𝑩 + 𝑹𝑪 + β 𝑹𝑪 ) = β ( 𝑽𝑪𝑪 − 𝑽𝑩𝑬)
𝑰𝑪 =
β( 𝑽𝑪𝑪 − 𝑽𝑩𝑬)
𝑹𝑩+𝑹𝑪 +β𝑹𝑪
𝐈𝐂 =
β(𝐕𝐂𝐂 − 𝐕𝐁𝐄)
𝐑𝐁+𝐑𝐂 (1 +β)
----------- (4)
𝑰𝑩 =
𝑽𝑪𝑪 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑪
𝑹𝑩+𝑹𝑪
--- (1)

7. 𝑰𝑪 =
𝜷(𝑽𝑪𝑪 − 𝑽𝑩𝑬)
𝑹𝑩+𝑹𝑪 (1 + 𝜷)
----------- (4)
Differentiate eqn (4) with respect to VBE
𝝏𝑰𝑪
𝝏𝑽𝑩𝑬
= -
𝜷
𝑹𝑩+𝑹𝑪 (𝟏+𝜷 )
∴ S′ =
𝝏𝑰𝑪
𝝏𝑽𝑩𝑬
= -
𝜷
𝑹𝑩+𝑹𝑪 (𝟏+𝜷 )

8. STEP 4 : TO OBTAIN S′′ DIFFERENTIATE THE EQUATION OBTAINED
IN STEP 3 WITH REPSPECT TO β
𝑰𝑪 =
𝜷(𝑽𝑪𝑪 − 𝑽𝑩𝑬)
𝑹𝑩+𝑹𝑪 (1 + 𝜷)
----------- (4)
Differentiate eqn (4) with respect to β
𝛛𝐈𝐂
𝛛𝛃
=
𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
𝐕𝐂𝐂
−𝐕𝑩𝑬
−𝛃(𝐕𝐂𝐂
−𝐕𝐁𝐄
)𝐑𝐂
𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
𝟐
=
𝐕𝐂𝐂
−𝐕𝑩𝑬
[ 𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
−𝛃𝐑𝐂)
𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
𝟐
=
𝐕𝐂𝐂
−𝐕𝑩𝑬
[𝐑𝐂
+𝐑𝐁
]
𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
𝟐
=
𝐕𝐂𝐂
−𝐕𝑩𝑬
[𝐑𝐂
+𝐑𝐁
]
[ 𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
][ 𝟏+ 𝛃 𝐑𝐂
+𝐑𝐁
]
S’’ =
𝛛𝐈𝐂
𝛛𝛃
=
𝑰𝑪(𝑹𝑩+𝑹𝑪)
𝜷[ 𝟏+𝜷 𝑹𝑪
+𝑹𝑩]
[using eqn 4]

9. Advantages
 Simple method as it requires only one resistance
RB.
 Provides some stabilization of the operating point
Disadvantages
 Does not provide good stabilization because
stability factor is fairly high
 Negative feedback which reduces the gain of the
amplifier

10. VOLTAGE DIVIDER BIAS
• R1 and R2 are connected across the supply
voltage VCC and provide biasing.
• RE provides stabilization

11.  For dc analysis
ac input signal = 0
C1 and C2 – open circuit (∵f = 0)

12. Equivalent circuit
𝑽𝑻𝑯 = 𝑽𝑪𝑪
𝑹𝟐
𝑹𝟏+ 𝑹𝟐
RB =

13. STEP 1:OBTAIN AN EXPRESSION
FOR 𝑰𝑩
 Apply KVL to the base
𝑽𝑻𝑯 = 𝑰𝑩𝑹𝑩 + 𝑽𝑩𝑬 + 𝑰𝑬𝑹𝑬
= 𝑰𝑩𝑹𝑩 + 𝑽𝑩𝑬 + (𝑰𝑩 +𝑰𝑪)𝑹𝑬
𝑰𝑩𝑹𝑩 + 𝑰𝑩𝑹𝑬= 𝑽𝑻𝑯 - 𝑽𝑩𝑬 - 𝑰𝑪𝑹𝑬
𝑰𝑩(𝑹𝑩 + 𝑹𝑬)= 𝑽𝑻𝑯 - 𝑽𝑩𝑬 - 𝑰𝑪𝑹𝑬
𝑰𝑩 =
𝑽𝑻𝑯 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑬
𝑹𝑩 + 𝑹𝑬
------(1)

14. STEP 2: TO OBTIAN S , DIFFERENTIATE IB
WITH RESPECT TO IC AND SUBSITUTE IN
THE STANDARD EQUATION OF S
𝑰𝑩=
𝑽𝑻𝑯 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑬
𝑹𝑩 + 𝑹𝑬
------(1)
Differentiate eqn (1) with respect to Ic
𝝏𝑰𝑩
𝝏𝑰𝑪
= -
𝑹𝑬
𝑹𝑩 + 𝑹𝑬
------(2)
We know that the standard equation of S is
S =
(1 + β)
1 – β ∂ 𝐈𝐁 / ∂ 𝐈𝐂
---(3)
Sub (2) in (3)
S =
(1 + β)
1 – β − 𝐑𝐄
𝐑𝐄+ 𝐑𝐁
=
(1 + β)
1 + β 𝐑𝐄
𝐑𝐄+ 𝐑𝐁
By rearranging the above expression we get
S =
(1 + β)
1 + β 𝐑𝐄
𝐑𝐄+ 𝐑𝐁
≈ 𝟏 (∴
𝑹𝑩
𝑹𝑬
<< 1 )

15. 𝑰𝑩 =
𝑽𝑻𝑯 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑬
𝑹𝑩 + 𝑹𝑬
------(1)
The Standard Equation of Ic is
𝑰𝑪 = β𝑰𝑩 + (1+β) 𝑰𝑪𝑶
Sub (1)
𝑰𝑪 = β
𝑽𝑻𝑯 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑬
𝑹𝑩 + 𝑹𝑬
(∵
𝑰𝑪𝑶 << )
𝑰𝑪(𝑹𝑩 + 𝑹𝑬 ) = β (𝑽𝑻𝑯 − 𝑽𝑩𝑬 − 𝑰𝑪𝑹𝑬)
𝑰𝑪(𝑹𝑩 + 𝑹𝑬 + β 𝑹𝑬 ) = β (𝑽𝑻𝑯 − 𝑽𝑩𝑬 )
𝑰𝑪 =
β (𝑽𝑻𝑯 − 𝑽𝑩𝑬 )
𝑹𝑩 + 𝑹𝑬 (1+ β )
---------(4)
STEP 3: TO OBTAIN S′, REPLACE IB INTERMS OF
VBE IN THE STANDARD EQUATION OF IC AND
DIFFERENTIATE IT WITH RESPECT TO VBE

16. 𝑰𝑪 =
β (𝑽𝑻𝑯 − 𝑽𝑩𝑬 )
𝑹𝑩 + 𝑹𝑬 (1+ β )
---------(4)
Differentiate eqn (4) with respect to 𝑽𝑩𝑬
𝝏𝑰𝑪
𝝏𝑽𝑩𝑬
= -
𝜷
𝑹𝑩+𝑹𝑬 (𝟏+𝜷 )
∴ S′ =
𝝏𝑰𝑪
𝝏𝑽𝑩𝑬
= -
𝜷
𝑹𝑩+𝑹𝑬 (𝟏+𝜷 )

17. STEP 4 : TO OBTAIN S′′ DIFFERENTIATE
THE EQUATION OBTAINED IN STEP 3
WITH REPSPECT TO β
 Differentiate eqn (4) with respect to β
𝛛𝐈𝐂
𝛛𝛃
=
𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
𝐕𝑻𝑯
−𝐕𝑩𝑬
−𝛃(𝐕𝑻𝑯
−𝐕𝐁𝐄
)𝐑𝑬
𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
=
𝐕𝑻𝑯
−𝐕𝑩𝑬
[ 𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
−𝛃𝐑𝑬
)
𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
𝟐
=
𝐕𝑻𝑯
−𝐕𝑩𝑬
[𝐑𝑬
+𝐑𝐁
]
𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
𝟐
=
𝐕𝑻𝑯
−𝐕𝑩𝑬
[𝐑𝑬
+𝐑𝐁
]
[ 𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
][ 𝟏+ 𝛃 𝐑𝑬
+𝐑𝐁
]
S’’ =
𝛛𝐈𝐂
𝛛𝛃
=
𝑰𝑪
(𝐑𝑬
+𝐑𝐁
)
𝜷[ 𝟏+𝜷 𝐑𝑬
+𝐑𝐁
]
[using eqn 4]
𝑰𝑪 =
β (𝑽𝑻𝑯 − 𝑽𝑩𝑬 )
𝑹𝑩 + 𝑹𝑬 (1+ β )
---------(4)

18. Advantages
 Provides good stability.