BJT Biasing

BJT Biasing Circuits LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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Base Bias •

VCC

RC

In a simple biasing circuit, VBB is eliminated by connecting the resistor RB to the supply VCC. VCC

RB VBB

RB

•

RC

As shown in the above circuit, two dc voltage supplies are needed to bias a BJT which is not practical. •

This biasing circuit is called base bias, or fixed bias.

LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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VCC

RB

C B VBE

•

•

Since IC = βDCIB ,

IC =

IC

RC

IB

•

β DC (VCC -VBE ) RB

•

Applying KVL, VCC = ICRC + VCE

•

Therefore, VCE = VCC – ICRC

•

βDC varies from device to device resulting in the variation of IC. A good biasing circuit is to maintain a constant IC while βDC varies.

VCE E

Applying KVL, VCC = IBRB + VBE Therefore,

VCC -VBE IB = RB

•

LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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DC Load Line •

IC

The dc load line is a graph that represents all the possible combinations of IC and VCE for a given BJT circuit.

IC(sat)

VCC

RB

IB

VCE(off)

IC

RC

A generic dc load line

C B VBE

VCE E

VCE

•

Ideally, Saturation: VCE = 0 VCC = IC R C IC(sat) =

LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

Cut-off: IC = 0 VCE(off) = VCC

VCC RC 4

Example 1: Plot the dc load line for the circuit shown below. +12 V

IC 2 kΩ

6 mA

RB

IC(sat) =

VCC 12 = = 6 mA R C 2k

12 V

VCE

The dc load line

VCE(off) = VCC = 12 V LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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Q-Point +8 V

•

IC and VCE at Q-point are called ICQ and VCEQ respectively.

•

Determine IC and VCE if VBE is assumed to be 0.7 volt.

2 kΩ 360 kΩ

βDC = 100

•

• •

When a BJT does not have an ac input, it will have specific dc values of IC and VCE. These values will correspond to a specific point on the dc load line. This point is called the Q-point. The letter Q comes from the word quiescent, meaning at rest.

IC = =

β DC (VCC -VBE ) RB (100)(8-0.7) =2.028 mA 360k

VCE = VCC - IC R C = 8 - (2.028mA)(2k) = 3.94 V

LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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•

Determine IC(sat) and VCE(off).

•

Construct the dc load line and plot the Q-point. IC

+8 V

2 kΩ

4 mA

360 kΩ

βDC = 100

2.028 mA

IC(sat)

3.94 V

V 8 = CC = = 4 mA R C 2k

VCE(off) = VCC = 8 V

Q

8V

VCE

The dc load line

•

The circuit is said to be midpoint biased since the values of IC and VCE at Q-point are one-half of their maximum values.

LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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Feedback Bias: Collector-Feedback Bias VCC

•

Applying KVL,

VCC = VRC +VRB +VBE RC

= (IC +I B )R C +I B R B +VBE = β DC I B R C +I B R C +I B R B +VBE

RB

= (β DC +1)I B R C +I B R B +VBE

IC

IB

C B

VBE

VCE E

IB =

VCC -VBE R B +(β DC +1)R C

ICQ =

β DC (VCC -VBE ) R B +(β DC +1)R C

VCEQ = VCC - (ICQ +I B )R C ≈ VCC -ICQ R C ; β DC >> 1 LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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Feedback Bias: Emitter-Feedback Bias VCC

•

Applying KVL,

RB

VCC = VRB +VBE + VRE IB

= I B R B +VBE + ( β DC + 1) I B RE

IC

RC

C B

VCE

IB =

VCC -VBE R B +(β DC +1)RE

ICQ =

β DC (VCC -VBE ) R B +(β DC +1)RE

E

VBE

VCEQ = VCC - ICQ R C + I E R E RE

IE

≈ VCC -ICQ (R C +R E ) ; β DC >> 1

LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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Voltage-Divider Bias VCC

VCC RC

R1

IC

RC

I1

IB RBB

C IB

R2

I2

IC

VBB

B

VCE

VBE

RE

IE

E

RE

IE

• •

Replace circuit A with the thevenin equivalent circuit. Find VBB and RBB.

Circuit A LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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VCC

VCC RC

IC

I1

R1

VBB = + I2

R2

IB

R1 VCC R1 +R 2

RBB VBB RE

IE

VBB=VOC -

•

Applying KVL, VBB = I B R BB + VBE + I E R E

R1

R BB = R1 // R 2 R2

= I B R BB + VBE + (β DC + 1)I B R E IB =

VBB - VBE R BB +(β DC + 1)R E

ICQ =

β DC (VBB - VBE ) R BB +(β DC +1)R E

RBB

VCEQ = VCC - (R C + R E )ICQ ; β DC >> 1 LE210 Lecture Notes 2002 Edition: BJT Biasing Circuits

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