lesson #8 chapter 4joelsd/electronics/classwork/... · transistor characteristics, the following...

BME 372 Electronics I –J.Schesser

225

Transistors

Lesson #8Chapter 4


226

Configurations of the BJT

• Common Emitter and Emitter Follower

o

o

o

Base

Emitter

npn

Collector

iE

iCiBB

E

C

vBE

vCE++

--


227

Configurations of the BJT

• Common Collector

• Common Base

o

o

o

Base

Collector

npn

Emitter

ic

iEiB

B

C

E

vBE

vCE

+--

+


228

Characteristics of the BJT npn

Common Emitter Configuration

.6 .8 1

25

20

15

iB

vBE

Base-emitter junction looks like a forward biased diode

Collector-emitter is a family of curves which are a function of

base current

10mA

20V vCE

iC

iB

100μA

5mA 50μA


229

BJT Equations

1

1

)1(

BBC

EB

E

C

BCE

iii

iiii

iii

o

o

o

Base

Emitter

Collector

iE

Ic=β iB =α iEiB

B

E

C

vBE

vCE++

--

Since α is less than unity then βwill be greater than unity and there is current gain from base to collector.


230

Example

• Calculate the values of β and α from the transistor shown in the previous graphs.

β = ic / ib

= 5m/50µ = 100α = β /(β+1) = 100/101

=.99

10mA

20V vCE

iC

iB

100μA

5mA 50μA


231

BJT Analysis

VBB

+ -

VCC10V

RC 2k

Vin+

-

RB

50k

1.6V

+

-

Here is a common emitter BJT amplifier:

What are the steps?


232

BJT Analysis – Inputs and Outputs

VBB

+ -

VCC10V

RC 2k

Vin+

-

RB

50k

1.6V

+

-

We would want to know the collector current (ic), collector-emitter voltage (VCE), and the voltage across RC.

To get this we need to fine the base current (ib) and the base-emitter voltage (VBE).

iC

VCE

+

-ib

VBE

+

-


233

BJT Analysis – Input Equation

VBB

+ -

VCC10V

RC 2k

Vin+

-

RB

50k

1.6V

+

-

To start, let’s write KVL around the base circuit.

Vin(t) + VBB = iB(t)RB + VBE(t)

iC

VCE

+

-iB

VBE

+

-


234

BJT Analysis – Output Equations

VBB

+ -

VCC10V

RC 2k

Vin+

-

RB

50k

1.6V

+

-

Likewise, we can write KVL around the collector circuit.

VCC= iC(t)RC + VCE(t)

iC

VCE

+

-iB

VBE

+

-


235

BJT Analysis Use Superposition: DC & AC sources

• Note that both equations are written so as to calculate the transistor parameters (i.e., base current, base-emitter voltage, collector current, and the collector-emitter voltage) for both the DC signal and the AC signal sources.

• Let’s use superposition, calculate the parameters for each separately, and add up the results– First, the DC analysis to calculate the DC Q-point

• Short Circuit any AC voltage sources• Open Circuit any AC current sources

– Next, the AC analysis to calculate gains of the amplifier.• Depends on how we perform AC analysis

– Graphical Method– Equivalent circuit method for small AC signals


236

BJT DC Analysis

• Using KVL for the input and output circuits and the transistor characteristics, the following steps apply:1. Draw the load lines on the transistor characteristics2. For the input characteristics determine the Q point for the

input circuit from the intersection of the load line and the characteristic curve (Note that some transistor do not need an input characteristic curve.)

3. From the output characteristics, find the intersection of the load line and characteristic curve determined from the Q point found in step 2, determine the Q point for the output circuit.


237

-1

4

9

14

19

24

29

34

39

-0.3

-0.2

-0.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

v BE volts

i B

amps

VBEQ = 0.6 V

IBQ = 20A

BJT DC Analysis Base-Emitter Circuit Q point

First let’s set Vin(t) =0 to get the Q-point for the BJT.

We start with the base circuit.

VBB = iBRB + VBE

And the intercepts occur at iB = 0; VBE = VBB = 1.6 V

and at VBE = 0; iB = VBB / RB = 1.6 /50k = 32A

The Load Line intersects the Base-emitter characteristics at VBEQ = 0.6 V and IBQ = 20A

iB = 0; VBE = VBB = 1.6 V

VBE = 0; iB = VBB / RB = 1.6 /50k = 32A

Q POINT


238

0.E+00

1.E-03

2.E-03

3.E-03

4.E-03

5.E-03

6.E-03

7.E-03

0 2 4 6 8 10

v CE volts

i C amps

BJT DC Analysis Collector-Emitter Circuit Q point

0 μa

iB=50 μa

40 μa

30 μa

10 μa

maRVCC

C 5

vVCC 10

Q POINT maICQ 5.2 20 μa

Now that we have the Q-point for the base circuit, let’s proceed to the collector circuit.

VCC= iCRC + VCE

The intercepts occur at iC = 0; VCE = VCC = 10 V; and at VCE = 0; iC = VCC / RC = 10 /2k = 5mA

The Load Line intersects the Collector-emitter characteristic, iB=20mA at VCEQ = 5.9 V and ICQ = 2.5mA

β = 2.5m/20m = 125

VCEQ = 5.9 V


239

BJT DC AnalysisSummary

• Calculating the Q-point for BJT is the first step in analyzing the circuit

• To summarize:– We ignored the AC (variable) source

• Short circuit the voltage sources• Open Circuit the current sources

– We applied KVL to the base-emitter circuit and using load line analysis on the base-emitter characteristics, we obtained the base current Q-point

– We then applied KVL to the collector-emitter circuit and using load line analysis on the collector-emitter characteristics, we obtained the collector current and voltage Q-point

• This process is also called DC Analysis• We now proceed to perform AC Analysis


240

BJT AC Analysis

• How do we handle the variable source Vin(t) ?• When the variations of Vin(t) are large we will use

the base-emitter and collector-emitter characteristics using a similar graphical technique as we did for obtaining the Q-point

• When the variations of Vin(t) are small we will shortly use a linear approach using the BJT small signal equivalent circuit.


241

BJT AC Analysis

• Let’s assume that Vin(t) = 0.2 sin(ωt).• Then the voltage sources at the base vary from a

maximum of 1.6 + .2 = 1.8 V to a minimum of 1.6 -.2 = 1.4 V

• We can then draw two “load lines” corresponding the maximum and minimum values of the input sources

• The current intercepts then become for the:– Maximum value: 1.8 / 50k = 36 µa– Minimum value: 1.4 / 50k = 28 µa


242

-1

4

9

14

19

24

29

34

39

-0.3

-0.2

-0.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

v BE volts

i B

amps

VBEQ = 0.6 V

IBQ = 20μA

BJT AC Analysis Base-Emitter Circuit

From this graph, we find:

At Maximum Input Voltage: VBE = 0.63 V, iB = 24μa

At Minimum Input Voltage: VBE = 0.59 V, iB = 15μa

Recall: At Q-point: VBE = 0.6 V, iB = 20μa

Note the asymmetry around the Q-point of the Max and Min Values for the base current and voltage which is due to the non-linearity of the base-emitter characteristics

ΔiBmax=24-20=4μa; ΔiBmin=20-15=5μa

“load line” for maximum input voltage

“load line” for minimum input voltage


243

00.20.40.60.8

11.21.41.61.8

2

0 10 20 3000.0000050.00001

0.0000150.000020.0000250.00003

0.0000350.00004

vinib

-1

4

9

14

19

24

29

34

39

-0.3

-0.2

-0.1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

v BE volts

i B

amps

VBEQ = 0.6 V

IBQ = 20A

BJT AC Analysis Base-Emitter Circuit

“load line” for maximum input voltage

“load line” for minimum input voltage


244

0.E+00

1.E-03

2.E-03

3.E-03

4.E-03

5.E-03

6.E-03

7.E-03

0 2 4 6 8 10

v CE volts

i C amps

BJT Characteristics-Collector Circuit

0 μa

iB=50 μa

40 μa

30 μa

10 μa

Q POINT maICQ 5.2 20 μa

VCEQ = 5.9 V

Using these maximum and minimum values for the base current on the collect circuit load line, we find:

At Maximum Input Voltage: VCE = 5 V, iC = 2.7ma

At Minimum Input Voltage: VCE = 7 V, iC = 1.9ma

Recall: At Q-point: VCE = 5.9 V, iC = 2.5ma

Note that in addition to the asymmetry around the Q-point there is an inversion between the input voltage and the collector to emitter voltage

24 μa

15 μa


245

0.E+00

1.E-03

2.E-03

3.E-03

4.E-03

5.E-03

6.E-03

7.E-03

0 2 4 6 8 10

v CE volts

i C amps

BJT Characteristics-Collector Circuit

Q POINT maICQ 5.2

VCEQ = 5.9 V

0

0.005

0.01

0.015

0.02

0.025

0 10 20 3000.0000050.00001

0.0000150.000020.0000250.00003

0.0000350.00004

icib

00.5

11.5

2

2.53

3.54

0 10 20 3001

234

56

78

vinvce

0 μa

iB=50 μa

40 μa

30 μa

10 μa

20 μa24 μa

15 μa


246

BJT AC Analysis Amplifier Gains

• From the values calculated from the base and collector circuits we can calculate the amplifier gains:– β = 125– Current gain = Δic / Δib = (2.7 – 1.9) m / (24 -15) μ

= 0.8/9 ✕ 103 = 88.9– Voltage gain = Vo / Vi = ΔVCE / ΔVBE

= (5 – 7) / (.63 - .59) = -2/0.04 = - 50– Voltage gain = Vo / Vs = ΔVCE / ΔVin

= (5 – 7) / .4 = -2 / .4 = - 5


247

BJT AC Analysis Summary

• Once we complete DC analysis, we analyze the circuit from an AC point of view.

• AC analysis can be performed via a graphical processes– Find the maximum and minimum values of the input

parameters (e.g., base current for a BJT)– Use the transistor characteristics to calculate the output

parameters (e.g., collector current for a BJT).

• Calculate the gains for the amplifier


248

0.E+00

1.E-03

2.E-03

3.E-03

4.E-03

5.E-03

6.E-03

7.E-03

0 2 4 6 8 10

v CE volts

i C amps

BJT Characteristics - Distortion

Q POINT maICQ 5.2

VCEQ = 5.9 V

Note that even though the input (base current) is symmetrical about its Q-point, the output voltage is uneven.

This is due to the non-linear spacing of the characteristic curves and leads to signal distortion

0 μa

iB=50 μa

40 μa

30 μa

10 μa

20 μa

24 μa

15 μa


249

The pnp Transistor

• Basically, the pnp transistor is similar to the npnexcept the parameters have the opposite sign.– The collector and base currents flows out of the

transistor; while the emitter current flows into the transistor

– The base-emitter and collector-emitter voltages are negative

• Otherwise the analysis is identical to the npn


250

PNP Bipolar Junction Transistors• Two junctions

– Collector-Base and Emitter-Base• Biasing

– vBE Forward Biased – vCB Reverse Biased

o

o

o

Base

Emitter

pnp

Collector

iE

iCiB

B

E

C

vBE

vCE

++-

-p-type

p-type

n-type

Emitter

Base

Collector

- +

VCC10V

RC 2k

Vin+-

RB

50k

1.6V

-

+


251

Homework

• Probs. 4.4, 4.5, 4.8, 4.10, 4.14, 4.15, 4.19, 4.20, 4.21, 4.22,

lesson #8 chapter 4joelsd/electronics/classwork/... · transistor characteristics, the following...

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