Solving Dc Analysis And Plotting Load Line With Q Point

Electronics Principles & Applications Sixth Edition

Solving DC Analysis and Plotting Load Line with Q Point

INTRODUCTION •Load Line •Q Point •Common-Emitter Amplifier •Voltage Divider Bias

1.5 V

In

5V

Amplifier

Out

The units cancel Gain =

5V Out InV 1.5

= 3.33

A small-signal amplifier can also be called a voltage amplifier. Common-emitter amplifiers are one type. The emitter terminal is grounded Next, load resistor a base bias resistor A coupling capacitor issupply often required Add aapower Start with an NPN junction transistor Connect abipolar signal source andThen common to the input and output signal circuits.

RB

RL C

CC

B

E

VCC

The output is phase inverted.

RB

RL C

CC

B

E

VCC

The maximum value of VCE for this circuit is 14 V. The maximum value of IC is 14 mA. These are the limits for this circuit.

14 V IC(MAX) = 1 kΩ

350 kΩ

1 kΩ C

CC

B

E

14 V

The load line connects the limits. This end is called saturation.

The linear region is between the limits. 100 μA

14 12 10 IC in mA 8 6 4 2

SAT.

LINEAR

80 μA 60 μA 40 μA 20 μA

0 2 4 6

8 10 12 14 16 18

VCE in Volts This end is called cutoff.

CUTOFF

0 μA

Use Ohm’s Law to determine the base current: 14 V IB = 350 kΩ

350 kΩ

1 kΩ C

CC

= 40 μA

B

E

14 V

An amplifier can be operated at any point along the load line. The base current in this case is 40 μA.

100 μA

14 12 10 IC in mA 8 6 4 2

Q

80 μA 60 μA 40 μA 20 μA

0 2 4 6

8 10 12 14 16 18

VCE in Volts Q = the quiescent point

0 μA

The input signal varies the base current above and below the Q point.

100 μA

14 12 10 IC in mA 8 6 4 2

80 μA 60 μA 40 μA 20 μA

0 2 4 6

8 10 12 14 16 18

VCE in Volts

0 μA

Overdriving the amplifier causes clipping. 100 μA

14 12 10 IC in mA 8 6 4 2

80 μA 60 μA 40 μA 20 μA

0 2 4 6

8 10 12 14 16 18

VCE in Volts The output is non-linear.

0 μA

What’s wrong with this Q point? 100 μA

14 12 10 IC in mA 8 6 4 2

80 μA 60 μA 40 μA 20 μA

0 2 4 6

8 10 12 14 16 18

VCE in Volts How about this one?

0 μA

14 V IB = = 40 μA 350 kΩ IC = β x IB = 150 x 40 μA = 6 mA VRL = IC x RL = 6 mA x 1 kΩ = 6 V VCE = VCC - VRL = 14 V - 6 V = 8 V

This is a good Q point for linear amplification. 350 kΩ

1 kΩ C

CC

B

E

14 V β = 150

14 V IB = = 40 μA (IB is not affected) 350 kΩ IC = β x IB = 350 x 40 μA = 14 mA (IC is higher) VRL = IC x RL = 14 mA x 1 kΩ = 14 V (VR is higher) L

VCE = VCC - VRL = 14 V - 14 V = 0 V (VCE is lower)

This is not a good Q point for linear amplification. 350 kΩ

1 kΩ C

CC

B

E

14 V β = 350 β is higher

The higher β causes saturation. 100 μA

14 12 10 IC in mA 8 6 4 2

80 μA 60 μA 40 μA 20 μA

0 2 4 6

8 10 12 14 16 18 VCE in Volts

The output is non-linear.

0 μA

This common-emitter amplifier is not practical.

It’s β dependent! It’s also temperature dependent. RB

RL C

CC

B

E

VCC

Basic C-E Amplifier Quiz The input and output signals in C-E are phase ______________.

inverted

The limits of an amplifier’s load line are saturation and _________.

cutoff

Linear amplifiers are normally operated near the _________ of the load line. center The operating point of an amplifier is also called the ________ point.

quiescent

Single resistor base bias is not practical since it’s _________ dependent.

β

This common-emitter amplifier is practical. RB1

RL C

CC B RB2

VCC

E RE

It uses voltage divider bias and emitter feedback to reduce β sensitivity.

+VCC

Voltage divider bias RB1

RL

RB1 and RB2 form a voltage divider RB2

RE

+VCC

Voltage divider bias analysis: RB1 RB2 VB = VCC RB1 + RB2 The base current is normally much smaller than the divider current so it can be ignored.

+VB

RB2

Solving the practical circuit for its dc conditions:

VCC = 12 V RB1 22 kΩ

B RB2 2.7 kΩ

VB =

RB2 RB1 + RB2

x VCC

RL= 2.2 kΩ C

2.7 kΩ x 12 V VB = 2.7 kΩ + 22 kΩ

E

VB = 1.31 V

RE = 220 Ω

Solving the practical circuit for its dc conditions:

VCC = 12 V RB1 22 kΩ

RL= 2.2 kΩ C

B RB2 2.7 kΩ

VE = VB - VBE

VE = 1.31 V - 0.7 V = 0.61 V

E RE = 220 Ω

Solving the practical circuit for its dc conditions:

VCC = 12 V RB1 22 kΩ

RL= 2.2 kΩ C

B RB2 2.7 kΩ

IE =

IE =

VE RE

0.61 V 220 Ω

E RE = 220 Ω

IC ≅ IE

= 2.77 mA

Solving the practical circuit for its dc conditions: VRL = IC x RL

VCC = 12 V

VRL = 2.77 mA x 2.2 kΩ

RB1 22 kΩ

RL= 2.2 kΩ C

B RB2 2.7 kΩ

E

VRL = 6.09 V VCE = VCC - VRL - VE

VCE = 12 V - 6.09 V - 0.61 V

RE = 220 Ω

VCE = 5.3 V

A linear Q point!

Review of the analysis thus far: 1. Calculate the base voltage using the voltage divider equation. 2. Subtract 0.7 V to get the emitter voltage. 3. Divide by emitter resistance to get the emitter current.

4. Determine the drop across the collector resistor. 5. Calculate the collector to emitter voltage using KVL. 6. Decide if the Q-point is linear. 7. Go to ac analysis.

Solving the practical circuit for its ac conditions:

VCC = 12 V The ac emitter resistance is rE:

RB1 22 kΩ

RL= 2.2 kΩ C

B RB2 2.7 kΩ

E RE = 220 Ω

25 mV rE = IE 25 mV rE = = 9.03 Ω 2.77 mA

Solving the practical circuit for its ac conditions:

VCC = 12 V The voltage gain from base to collector:

RB1 22 kΩ

RL= 2.2 kΩ C

B RB2 2.7 kΩ

E

AV =

RE = 220 Ω

AV =

RL RE + rE

2.2 kΩ 220 Ω + 9.03 Ω

= 9.61

Solving the practical circuit for its ac conditions:

VCC = 12 V RB1 22 kΩ

An emitter bypass capacitor can be used to increase AV:

RL= 2.2 kΩ

AV =

C B RB2 2.7 kΩ

E RE

AV =

CE

2.2 kΩ 9.03 Ω

RL rE = 244

Practical C-E Amplifier Quiz β-dependency is reduced with emitter feedback and voltage _________ bias.

divider

To find the emitter voltage, VBE is subtracted from ____________.

VB

To find VCE, VRL and VE are subtracted from _________.

VCC

Voltage gain is equal to the collector resistance _______ by the emitter resistance. divided Voltage gain can be increased by ________ the emitter resistor.

bypassing