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4 DC BIASING—BJTS (CONT’D II )

KOM2751 Analog Electronics :: Dr. Muharrem Mercimek :: YTU - Control and Automation Dept. 1

Most of the content is from the textbook:

Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11th ed, 2013

4.11 Current Sources

• The concept of a power supply provides the starting point in our consideration of current source

circuits.

• A practical voltage source is a voltage supply in series with a resistance.

• An ideal voltage source has 𝑅 = 0, whereas a practical source includes some small resistance.

• A practical current source is a current supply in parallel with a resistance.

• An ideal current source has 𝑅 = ∞, whereas a practical current source includes some very large

resistance.


• An ideal current source provides a constant current regardless of the load connected to it. There are

many uses in electronics for a circuit providing a constant current at a very high impedance.


𝐼𝐸

𝑅𝐸

𝐼𝐶

-

- +

+

𝑅1

𝑉𝐵𝐸

𝑉𝐶𝐸 𝐼𝐵

𝑅2

+

𝐼1

𝐼2 𝐼1 = 𝐼2 + 𝐼𝐵 ≅ 𝐼2

𝐼1 ≅ 𝐼2 =𝑉𝐸𝐸

𝑅1 + 𝑅2

𝐼2𝑅2 = 𝑉𝐵𝐸 + 𝐼𝐸𝑅𝐸

𝐼𝐸 =𝐼2𝑅2 − 𝑉𝐵𝐸

𝑅𝐸

𝐼𝐶 = 𝐼𝐸

𝛽

𝛽 + 1≅ 𝐼𝐸

𝑉𝐸𝐸

𝑉𝐸𝐸 value is > 0

4.12 Switching of the Transistors • The application of transistors is not limited solely to the amplification of signals. Through proper

design, transistors can be used as switches for computer and control applications. The network of

next example can be employed as an inverter in computer logic circuitry

Example 4.18 𝑉𝐶 ?

𝑉𝐶𝐶 = 5V 𝑅𝐵 = 68 k 𝑅𝐶 = 0.82 k 𝛽=125 𝑉𝐵𝐸 = 0.7 V

𝑉𝑖 𝑅𝐵

𝐼𝐸

𝐼𝐶

-

- +

+ +

𝑅𝐶

𝑉𝐶𝐶

𝑉𝐵𝐸

𝑉𝐶𝐵

- 𝑉𝐶𝐸

𝐼𝐵

𝑉𝐶

• Proper design for the inversion process requires that the operating

point switch from cutoff to saturation along the load line.

• For our purposes we will assume that 𝐼𝐶 = 𝐼𝐶𝐸𝑂 ≅ 0 mA when

𝐼𝐵 = 0 𝜇𝐴. • In addition, we will assume that 𝑉𝐶𝐸 = 𝑉𝐶𝐸𝑠𝑎𝑡 ≅ 0 𝑉 rather than the

typical 0.1 V 𝑡𝑜 0.3 V level.

Transistor switches between

𝐼𝐶 = 0 mA, 𝑉𝐶𝐸 = 5 V 𝑎𝑛𝑑 𝐼𝐶 = 𝐼𝐶𝑠𝑎𝑡 , 𝑉𝐶𝐸𝑠𝑎𝑡 = 0 V


When 𝑉𝑖 = 5 V :

𝐼𝐶𝑠𝑎𝑡 =5

0.82 x 103= 6.098 mA

If 𝐼𝐶𝑠𝑎𝑡 < 𝛽𝐼𝐵 or 𝐼𝐵 >𝐼𝐶𝑠𝑎𝑡

𝛽

then the transistor is in the saturation region

𝐼𝐵 =𝑉𝑖 − 0.7

68 x 103= 63 𝜇A >

6.098 𝑥 10−3

125

𝑉𝐶𝐸 ≅ 0 𝑉

𝑉𝐶 ≅ 0 𝑉

When 𝑉𝑖 = 0 V :

𝐼𝐵 = 0 𝜇A

𝐼𝐶 = 0 mA

𝑉𝐶𝐸 ≅ 5 𝑉

𝑉𝐶 ≅ 5 𝑉

48.78 𝜇A


When 𝑉𝑖 = 5 V, the transistor will be “on” and the

design must ensure that the network is heavily

saturated by a level of 𝐼𝐵 greater than that associated

with the 𝐼𝐵 curve appearing near the saturation level.

5 BJT AC ANALYSIS


Most of the content is from the textbook:

Electronic devices and circuit theory, Robert L. Boylestad, Louis Nashelsky, 11th ed, 2013

5.1 Introduction

• We will examine the ac response of the BJT amplifier by reviewing the models most frequently used to represent the transistor in the sinusoidal AC domain.

• One of our first concerns in the sinusoidal ac analysis of transistor networks is the magnitude of the input signal.

• It will determine whether small-signal or large-signal techniques should be applied. There is no set dividing line between the two, but the application—and

• the magnitude of the variables of interest relative to the scales of the device characteristics will usually make it quite clear which method is appropriate.

• The small-signal technique is considered right now.

• There are three models commonly used in the small-signal ac analysis of transistor networks:

1. the 𝑟𝑒 model,

2. the hybrid 𝜋 model,

3. and the hybrid equivalent model.

• We will emphasize the 𝑟𝑒 model.


5.2 Amplification in the AC Domain

𝑖𝑎𝑐 𝑝−𝑝 ≫ 𝑖𝑐 𝑝−𝑝 due to current amplification

• The peak value of the oscillation in the output circuit is controlled by the established DC level.

• In general, therefore, proper amplification design requires that the DC and AC components be

sensitive to each other’s requirements and limitations.

• However, it is extremely helpful to realize that: The superposition theorem is applicable for the

analysis and design of the DC and AC components of a BJT network, permitting the separation of

the analysis of the DC and AC responses of the system.

• We can make a complete dc analysis of a system before considering the AC response. Once the

DC analysis is complete, the AC response can be determined using a completely AC analysis. It

happens, however, that one of the components appearing in the AC analysis of BJT networks will

be determined by the DC conditions, so there is still an important link between the two types of

analysis.


5.3 BJT Transistor Modeling

• A model is a combination of circuit elements, properly chosen, that best approximates the actual

behavior of a semiconductor device under specific operating conditions.

• Once the ac equivalent circuit is determined, the schematic symbol for the device can be replaced

by this equivalent circuit and the basic methods of circuit analysis applied to determine the desired

quantities of the network.

• Let us assume for the moment that the small-signal ac equivalent circuit for the transistor has

already been determined. Because we are interested only in the AC response of the circuit, all the

DC supplies can be replaced by a zero-potential equivalent (short circuit) because they determine

only the DC (quiescent level) of the output voltage and not the magnitude of the swing of the AC

output.

−

+

𝑅𝐸

𝐼𝑖

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅1

𝑅2

𝑉𝑖

𝑉𝑠

𝑅𝑠

+

𝐶3 −

+

𝑉𝑜

𝐼𝑜

𝑍𝑖 𝑍𝑜


• All DC supplies can be replaced by a zero potential equivalent

• Use short circuits for DC voltage sources

• Use open circuits for DC current sources

• Coupling capacitors 𝐶1 and 𝐶2, by-pass capacitor 𝐶3 chosen to have very small reactance at the

frequency of application.

• Therefore they can be replaced by short circuits in many applications.

𝐼𝑜

−

+

𝑅𝐸

𝐼𝑖

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅1

𝑅2

𝑉𝑖

𝑉𝑠

𝑅𝑠

+

𝐶3 −

+

𝑉𝑜

𝑍𝑖 𝑍𝑜

𝐼𝑜

−

+

𝑅𝐸

𝐼𝑖

𝑅𝐶

𝑅1

𝑅2

𝑉𝑖

𝑉𝑠

𝑅𝑠

+

−

+

𝑉𝑜

zero-potential equivalent of the network

𝑍𝑖 𝑍𝑜


𝐼𝑜: output current

𝐼𝑖 : input current

𝑍𝑖: input impedance (impedance when looking into the system.)

𝑍𝑜: output impedance (impedance when looking back to the system.)

Small signal modeling: very common analysis technique which is used to approximate behavior of

non-linear devices with linear equations.

𝐴𝑉 =𝑉𝑜

𝑉𝑖 : Voltage gain of the system

𝐴𝑖 =𝐼𝑜

𝐼𝑖 : Current gain of the system

−

+

−

+

𝑉𝑖 𝑉𝑜

𝑍𝑖 𝑍𝑜

𝐼𝑖 𝐼𝑜

−

+

−

+

𝑉𝑖 𝑉𝑜 𝑅𝑖 𝑅𝑜


𝐼𝑜

−

+

𝑅𝐸

𝐼𝑖

𝑅𝐶

𝑅1

𝑅2

𝑉𝑖

𝑉𝑠

𝑅𝑠

+

𝑍𝑖

−

+

𝑉𝑜

𝑍𝑜

zero-potential equivalent of the network

𝐼𝑜 −

+

𝐼𝑖

𝑅𝐶

𝑅1||𝑅2

𝑉𝑖

𝑉𝑠

𝑅𝑠

+

−

+

𝑉𝑜

𝑍𝑖 𝑍𝑜

*

* Now, we can insert small signal AC model of BJT

zero-pot. eq. of the network (re-drawn)

AC equivalent of the network can be found after:

1. All DC supplies can be replaced by a zero potential equivalent.

• short circuits for DC voltage sources

• open circuits for DC current sources

2. Replace all capacitors with short circuits.

3. Remove bypassed components.

4. Redraw the network in a more convenient form.


𝐼𝑜

5.4 𝑟𝑒 model

𝛽𝐼𝑏

𝑉𝑏𝑒

𝐼𝑒

𝐼𝑏

𝐼𝑐

+

¯

𝑉𝑐𝑒

+

¯

BJT equivalent circuit.

Diode can be replaced

with 𝑟𝑒 resistor

𝑟𝑒

𝑍𝑖 =𝑉𝑖

𝐼𝑏=

𝑉𝑏𝑒

𝐼𝑏

𝑉𝑏𝑒 = 𝐼𝑒𝑟𝑒 = 𝐼𝑐 + 𝐼𝑏 𝑟𝑒 = 𝛽 + 1 𝐼𝑏𝑟𝑒

𝑍𝑖 = 𝛽 + 1 𝑟𝑒

𝑟𝑒 = 26𝑚V/𝐼𝐸

𝐼𝐸 : emitter current we obtained from DC analysis

𝑟𝑜 =∆𝑉𝐶𝐸

∆𝐼𝐶

(given in the data sheet)

𝑉𝑖 ,

𝑍𝑖 𝑍𝑜

−

+

𝑉𝑜 𝛽𝑟𝑒 𝛽𝐼𝑏 𝑟𝑜

𝐼𝑐 𝐼𝑏 c b

−

+

e

𝑉𝑖 𝑟𝑒 model of CE

𝐼𝑖

𝑍𝑖 𝑍𝑜


𝑟𝑒 = 26𝑚V/𝐼𝐸

𝑍𝑖 = 𝑟𝑒

𝐼𝐸 : emitter current we obtained from DC analysis

𝑧𝑜 ≅ ∞

• The equivalent circuit of will be used throughout the analysis to follow for the common-emitter

configuration.

• Typical values of 𝛽 run from 50 to 200, with values of 𝛽𝑟𝑒 typically running from a few hundred ohms

to a maximum of 6 k to 7 k . The output resistance is typically in the range of 40 k to 50 k.

This is a pnp transistor !

𝐼𝑐 𝐼𝑒 𝐼𝑖 𝐼𝑜

• The direction of the collector current in the output circuit is now opposite to that of the defined output current.

• Because the output current is opposite to the defined 𝐼𝑜 direction, you will find in the analysis to follow that

there is no phase shift between the input and output voltages.

• For the common-emitter configuration there is a 180 ° phase shift.

𝐼𝑜 𝐼𝑖

−

+

𝑉𝑜 𝑟𝑒

𝛼𝐼𝑒

𝑟𝑜

𝐼𝑐 𝐼𝑒 c e

−

+

b

𝑉𝑖 𝑟𝑒 model of CB

𝑍𝑖 𝑍𝑜


5.5 Fixed Bias Configuration (CE) • Note that the input signal 𝑉𝑖 is applied to the base of the transistor, whereas the output 𝑉𝑜 is off the

collector. In addition, recognize that the input current 𝐼𝑖 is not the base current, but the source

current, and the output current 𝐼𝑜 is the collector current.

• The small-signal AC analysis begins by removing the DC effects of 𝑉𝐶𝐶 and replacing the DC

blocking capacitors 𝐶1 and 𝐶2 by short-circuit equivalents

𝐼𝑜 −

+

𝐼𝑖

𝑅𝐶

𝑅𝐵 𝑉𝑖

𝑍𝑖

−

+

𝑉𝑜

𝑍𝑜

𝐼𝑖 𝐼𝑐

𝑟𝑜

𝐼𝑜

−

+

𝑅𝐶 𝑅𝐵 𝑉𝑖

𝑍𝑖

-

+

𝑉𝑜

𝑍𝑜

𝛽𝑟𝑒 𝛽𝐼𝑏

𝐼𝑏

small-signal AC equivalent of the network using 𝑟𝑒 model

zero-potential equivalent

of the network


𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅𝐵

𝐼𝑜

𝐼𝑖

−

+

𝑉𝑖

−

+

𝑉𝑜

𝑍𝑖 𝑍𝑜

𝑍𝑖 = 𝑅𝐵 ∥ (𝛽𝑟𝑒) 𝑍𝑜 = 𝑟𝑜 ∥ 𝑅𝐶

𝑍𝑜 is determined when 𝑉𝑖 = 0

Mostly 𝑟𝑜 is greater than 𝑅𝐶 more than a factor of 10.

𝑉𝑜 = −𝛽𝐼𝑏(𝑟𝑜 ∥ 𝑅𝐶)

𝐼𝑏 =𝑉𝑖

𝛽𝑟𝑒

𝑉𝑜 = −𝛽𝑉𝑖

𝛽𝑟𝑒 (𝑟𝑜 ∥ 𝑅𝐶 = −

𝑉𝑖

𝑟𝑒(𝑟𝑜 ∥ 𝑅𝐶

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −

(𝑟𝑜 ∥ 𝑅𝐶)

𝑟𝑒≅ −

𝑅𝐶

𝑟𝑒

𝐴𝑣 is negative there is a 180° phase shift between 𝑉𝑜 and 𝑉𝑖

Example 5.1

a) Determine 𝑟𝑒

b) Calculate 𝑍𝑖

c) Calculate 𝑍𝑜 and 𝐴𝑣 for 𝑟𝑜 = ∞

d) Calculate 𝑍𝑜 and 𝐴𝑣 for 𝑟𝑜 = 50 k

𝑟𝑜 𝑅𝐶

-

+

𝑉𝑜

𝑍𝑜

If 𝑟𝑜 ≥ 10 𝑟𝑒

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅𝐵

𝐼𝑜

𝑉𝑖

𝑉𝑜

𝐼𝑖 𝑉𝐶𝐶 = 12 V 𝑅𝐵 = 470 k 𝑅𝐶 = 3 k 𝐶1 = 10 μF 𝐶2 = 10 μF 𝛽=100 𝑉𝐵𝐸 = 0.7 V


a) Regarding DC conditions ;

𝐼𝐵 = 24.04 μA

𝐼𝐸 = 2.428 mA

𝑟𝑒 =26 mV

𝐼𝐸= 10.71

b) 𝛽𝑟𝑒 = 100 ∙ 10.71 = 1.071 k

c) 𝑍𝑜 = (𝑟𝑜 ∥ 𝑅𝐶) → (∞ ∥ 3 k) = 3k

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −


𝑟𝑒= −

3x103

10.71= −280.11

d) 𝑍𝑜 = 𝑟𝑜 ∥ 𝑅𝐶 → (50k ∥ 3 k) = 2.83k

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −


𝑟𝑒= −

2.83x103

10.71= −264.24


5.6 Voltage Divider Configuration (CE)

𝐼𝑜

−

+

𝑅𝐸

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅1

𝑅2 𝑉𝑖 𝐶3 −

+

𝑉𝑜

𝑍𝑖 𝑍𝑜

𝐼𝑖

𝐼𝑖 𝐼𝑐

𝑟𝑜

𝐼𝑜

−

+

𝑅𝐶 𝑅1||𝑅2 𝑉𝑖

𝑍𝑖

-

+

𝑉𝑜

𝑍𝑜


𝐼𝑏


𝑍𝑖 = 𝑅1 ∥ 𝑅2 ∥ (𝛽𝑟𝑒) 𝑍𝑜 = 𝑟𝑜 ∥ 𝑅𝐶

Mostly 𝑟𝑜 is greater than 𝑅𝐶 more than a factor of 10.

𝑉𝑜 = −𝛽𝐼𝑏(𝑟𝑜 ∥ 𝑅𝐶)

𝐼𝑏 =𝑉𝑖

𝛽𝑟𝑒

𝑉𝑜 = −𝛽𝑉𝑖

𝛽𝑟𝑒 (𝑟𝑜 ∥ 𝑅𝐶) = −

𝑉𝑖

𝑟𝑒(𝑟𝑜 ∥ 𝑅𝐶)

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −


𝑟𝑒≅ −

𝑅𝐶

𝑟𝑒



Example 5.2

𝐼𝑜

−

+

𝑅𝐸

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅1

𝑅2 𝑉𝑖 𝐶3 −

+

𝑉𝑜

𝑍𝑖 𝑍𝑜

𝐼𝑖



c) Calculate 𝑍𝑜 and 𝐴𝑣 for 𝑟𝑜 = ∞

d) Calculate 𝑍𝑜 and 𝐴𝑣 for 𝑟𝑜 = 50 k

𝑉𝐶𝐶 = 22 V 𝑅1 = 56 k 𝑅2 = 8.2 k 𝑅𝐶 = 6.8 k 𝑅𝐸 = 1.5 k 𝐶1 = 10 μF 𝐶2 = 10 μF 𝐶3 = 20 μF 𝛽=90 𝑉𝐵𝐸 = 0.7 V


𝐼𝐸 = 1.41 mA

𝑟𝑒 =26 mV

𝐼𝐸= 18.44

b) 𝛽𝑟𝑒 = 90 ∙ 18.44 = 1.66 k

𝑍𝑖 = 𝛽𝑟𝑒 ∥ (𝑅1 ∥ 𝑅2) = 1.35 k

c) 𝑍𝑜 = 𝑟𝑜||𝑅𝐶 → (∞ ∥ 6.8 k ) = 6.8 k

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −


𝑟𝑒= −

6.8x103

18.44= −368.76

d) 𝑍𝑜 = 𝑟𝑜 ∥ 𝑅𝐶 → (50k ∥ 6.8 k) = 5.98 k

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −


𝑟𝑒= −

5.98x103

18.44= −324.3


5.7 Emitter Bias Configuration (CE)

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅𝐵

𝐼𝑜

𝐼𝑖

𝑅𝐸

Emitter resistor is not by-passed

With no 𝑟𝑜 effect. In most application for the sake of simplification

its effect can be ignored.

𝑟𝑜 ≥ 10 𝑅𝐶 + 𝑅𝐸 𝑟𝑜

𝑟𝑒≫ 1 if 𝑟𝑜 can be ignored


𝐼𝑖 𝐼𝑐

𝐼𝑜

−

+

𝑅𝐶 𝑅𝐵

𝑉𝑖

𝑍𝑖

-

+

𝑉𝑜

𝑍𝑜


𝐼𝑏

𝑅𝐸

𝑍𝑏


−

+

𝑉𝑖

−

+ 𝑉𝑜

𝑍𝑖 𝑍𝑜

𝑉𝑖 = 𝛽𝑟𝑒𝐼𝑏 + 𝐼𝑏 + 𝛽𝐼𝑏 𝑅𝐸

𝑍𝑏 =𝑉𝑖

𝐼𝑏= 𝛽𝑟𝑒 + 1 + 𝛽 𝑅𝐸

𝑍𝑖 = 𝑅𝐵 ∥ 𝑍𝑏

𝛽 is much greater than 1, 𝑅𝐸 is greater than 𝑟𝑒

𝑍𝑏 ≅ 𝛽𝑅𝐸

𝑍𝑖 = 𝑅𝐵 ∥ 𝛽𝑅𝐸

When 𝑉𝑖 is zero, 𝐼𝑏 = 0 → 𝛽𝐼𝑏 = 0, 𝑍𝑜 = 𝑅𝑐

𝑉𝑜 = −𝛽𝐼𝑏𝑅𝐶

𝑉𝑖 = 𝛽𝑟𝑒𝐼𝑏 + 𝐼𝑏 + 𝛽𝐼𝑏 𝑅𝐸

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −

𝛽𝐼𝑏𝑅𝐶

𝛽𝑟𝑒𝐼𝑏 + 𝐼𝑏 + 𝛽𝐼𝑏 𝑅𝐸≅ −

𝑅𝐶

𝑟𝑒 + 𝑅𝐸



Example 5.3

𝐼𝑜

−

+

𝑅𝐸

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2

𝑅𝐵

𝑉𝑖

−

+

𝑉𝑜

𝑍𝑖 𝑍𝑜

𝐼𝑖



c) Calculate 𝑍𝑜 and 𝐴𝑣

𝑉𝐶𝐶 = 20V 𝑅𝐵 = 470 k 𝑅𝐶 = 2.2 k 𝑅𝐸 = 0.56 k 𝐶1 = 10 μF 𝐶2 = 10 μF 𝛽=120 𝑉𝐵𝐸 = 0.7 V

𝑟𝑜 = 40 k


𝐼𝐵 = 35.89 𝜇A , 𝐼𝐸 = 4.35 mA

𝑟𝑒 =26 mV

𝐼𝐸= 5.99

𝑟𝑜 ≥ 10 𝑅𝐶 + 𝑅𝐸 … . ? 𝑟𝑜

𝑟𝑒≫ 1 … . ?

40x 103 ≥ 10 ∙ 2.2 + 0.56 x 103 → True

40x 103

5.99≫ 1 → True

We can ignore 𝑟𝑜

b)

𝑍𝑏 =𝑉𝑖

𝐼𝑏= 𝛽𝑟𝑒 + 1 + 𝛽 𝑅𝐸 = 67.92 k

𝑍𝑖 = 𝑅𝐵 ∥ 𝑍𝑏 = 470 k ∥ 67.92 k = 59.34 k c) 𝑍𝑜 = 𝑅𝐶 = 2.2 k

𝐴𝑣 =𝑉𝑜

𝑉𝑖≅ −

𝑅𝐶

𝑟𝑒 + 𝑅𝐸= −3.69 a) 𝑟𝑒 =

26 mV

𝐼𝐸= 5.99

b) 𝑍𝑖 = 𝑅𝐵 ∥ (𝛽𝑟𝑒) = 717.70 c) 𝑍𝑜 = 𝑅𝐶 = 2.2 k

𝐴𝑣 =𝑉𝑜

𝑉𝑖≅ −

𝑅𝐶

𝑟𝑒= −367.28

If 𝑅3 is by-passed


5.8 Collector Feedback Configuration (CE)


With no 𝑟𝑜 effect. In most application for the sake of simplification

its effect can be ignored.

𝑟𝑜 ≥ 10 𝑅𝐶

𝑅𝐵 ≫ 𝑅𝐶 if 𝑟𝑜 can be ignored

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2 𝑅𝐵

𝐼𝑜

𝐼𝑖

𝐼𝑏 𝐼𝑜

−

+

𝑅𝐶 𝑉𝑖

𝑍𝑖

-

+

𝑉𝑜

𝑍𝑜


𝐼𝑐

𝐼𝑖 𝐼1 𝑅𝐵

𝑍𝑖 𝑍𝑜


−

+

𝑉𝑖

−

+

𝑉𝑜

𝐼𝑜 = 𝐼1 + 𝛽𝐼𝑏

𝛽𝐼𝑏 ≫ 𝐼1 𝐼𝑜 ≅ 𝛽𝐼𝑏

𝐼𝑏 =𝑉𝑖

𝛽𝑟𝑒

We have to relate 𝑉𝑖 with 𝐼𝑖 to find 𝑍𝑖

𝐼1 =𝑉𝑜 − 𝑉𝑖

𝑅𝐵 → 𝑉𝑜 = −𝛽𝐼𝑏𝑅𝐶 = −𝛽

𝑉𝑖

𝛽𝑟𝑒𝑅𝐶 = −

𝑅𝐶

𝑟𝑒𝑉𝑖

𝐼1 = −𝑅𝐶

𝑟𝑒𝑉𝑖 − 𝑉𝑖

1

𝑅𝐵= −

1

𝑅𝐵1 +

𝑅𝐶

𝑟𝑒𝑉𝑖

𝑉𝑖 = 𝐼𝑏𝛽𝑟𝑒 = 𝐼𝑖 + 𝐼1 𝛽𝑟𝑒 = 𝐼𝑖 −1

𝑅𝐵1 +

𝑅𝐶

𝑟𝑒𝑉𝑖 𝛽𝑟𝑒

𝑉𝑖 1 +𝛽𝑟𝑒

𝑅𝐵1 +

𝑅𝐶

𝑟𝑒= 𝐼𝑖𝛽𝑟𝑒 → 𝑍𝑖 =

𝑉𝑖

𝐼𝑖=

𝛽𝑟𝑒

1 +𝛽𝑟𝑒𝑅𝐵

1 +𝑅𝐶𝑟𝑒

𝐼𝑏 𝐼𝑜

−

+

𝑅𝐶 𝑉𝑖

𝑍𝑖

-

+

𝑉𝑜

𝑍𝑜


𝐼𝑐

𝐼𝑖 𝐼1 𝑅𝐵 If 𝑅𝐶 ≫ 𝑟𝑒

𝑍𝑖 =𝑉𝑖

𝐼𝑖=

𝛽𝑟𝑒

1 +𝛽𝑟𝑒𝑅𝐵

𝑅𝐶𝑟𝑒

=𝛽𝑟𝑒

1 +𝛽𝑅𝐶𝑅𝐵

When 𝑉𝑖 is zero, 𝐼𝑏 = 0,𝛽𝐼𝑏 = 0, 𝐼𝑒 = 0 𝑍𝑜 = 𝑅𝐶 ∥ 𝑅𝐵

𝐴𝑉 =𝑉𝑜

𝑉𝑖= −

𝑅𝐶

𝑟𝑒

𝑅𝐶 𝑉𝑖 = 0

𝑍𝑜

𝛽𝑟𝑒

𝑅𝐵


Example 5.4



c) Calculate 𝑍𝑜 and 𝐴𝑣

𝑉𝐶𝐶 = 9V 𝑅𝐵 = 180 k 𝑅𝐶 = 2.7 k 𝐶1 = 10 μF 𝐶2 = 10 μF 𝛽=200 𝑉𝐵𝐸 = 0.7 V

𝑟𝑜 ≅ ∞


𝐼𝐵 = 11.53 𝜇A , 𝐼𝐸 = 2.32 mA

𝑟𝑒 =26 mV

𝐼𝐸= 11.21

𝑟𝑜 ≥ 10 𝑅𝐶 𝑇𝑟𝑢𝑒 𝑅𝐵 ≫ 𝑅𝐶 𝑇𝑟𝑢𝑒

We can ignore 𝑟𝑜

b)

𝑍𝑖 =𝛽𝑟𝑒

1 +𝛽𝑅𝐶𝑅𝐵

= 560.5

c) 𝑍𝑜 = 𝑅𝐶 ∥ 𝑅𝐵 = 2.6 k

𝐴𝑣 =𝑉𝑜

𝑉𝑖= −

𝑅𝐶

𝑟𝑒= −240.86

𝑉𝑖

𝑉𝑜

𝑅𝐶

𝑉𝐶𝐶

𝐶1

𝐶2 𝑅𝐵

𝐼𝑜

𝐼𝑖

𝑍𝑖 𝑍𝑜


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