Bjt Emitter Bias Circuit (Voltage Divider Circuit).

To ensure the best amplification mode, people often choose the static operating point at about the middle of the static load line.

I CQ  0.5I C (ngm) and U CEQ  0.5U CE (hm)

I C

U CC

R C

I CQ

static load line

U CE

U CC U CE

Figure 3.4: Graph of static load equation.

2. Polarization by voltage divider circuit

R 1

R C

R 2

R E

U CC

Figure 3.5: BJT bias circuit using emitter current (voltage divider circuit).

From the circuit in figure (3.5), we have:

U B 

R 2

1 2

R R

U CC

(3.9)

R R // R 

R 1 R 2

B 1 2 R R

and:

1 2

U B I B R B  U BE I E R E

 I B R B  1  R E  U B E

(3.10)

I B 

R

U B  U BE

BEIGE

  1  R

Collector current in the circuit:

I C =  .I B

We have: U CC = I C .R C + U CE + I E .R E

 U CC = I C .(R C + R E ) + U CE (see IE  I C )

 voltage between two poles C and E is:

U CE = U CC – I C .(R C + R E )

The limit points of the static load line are:

U CE(hm) = U CC

I C ( ngm ) 

U CC

R R

For example:

Given the circuit as shown in figure (3.5) with U CC = 18V, R1 = 48k  , R2 = 12k  , RC = 1.5k  and R E = 500  . BJT has coefficient  = 80. Find the static operating point?

Prize:

U B 

R 2

R R

U CC

 12 18  3.6  V 

48  12

We have:

1 2

R R 1 R 2  48 *12

 9.6  k  

B R  R 48  12

I B 

R

1 2

U B  U BE

BEIGE

  1  R

3.6  0.7

 9.6   1  80  0.5

 0.06  mA 

Collector current I C =  I B = 80*0.06 = 4.8 (mA)

 U CE = U CC – I C .(R C + R E ) = 8.4 (V)

So, the static operating point Q(0.06mA; 4.8mA; 8.4V)

U CC

3. Polarization by feedback voltage

R B

R C

Figure 3.6: BJT bias circuit using feedback voltage.

In figure (3.6), the resistor R B is connected directly between the collector terminal and the base terminal. The fundamental difference between the voltage feedback bias circuit and the fixed current bias circuit is that: in the voltage feedback bias circuit, the I B current mechanism is involved , sensing the voltage (or current) in the output circuit, while in the fixed current bias circuit, this is not the case.

We can illustrate as follows: assuming the coefficient  increases as the temperature increases, this causes I C to increase and therefore the voltage applied across R C is U C = I C .R C increases, the voltage U CE = U CC – I C .R Cdecreases and the voltage across R B decreases, thus reducing the current I B, causing I C to decrease and through a feedback loop the current I C is compensated in a phase-reversed manner and the operating point Q is thus better stabilized than with a fixed current bias. In the opposite case, assuming  decreases , the process leading to an increase in current I C compensates for the initial decrease in I C due to the decrease in  .

From the circuit in figure (3.6), we have:

U CC = (I B + I C )R C + I B .R B + U BE

= I B [(1+  )R C + R B ] + U BE

I U CC  U BE

B  1 R R

So the collector current I C =  .I B

From the expression: U CC = (I B + I C )R C + U CE

 the voltage applied across the two CE poles is:

U CE = U CC – (I B + I C )R C

4. Implementation sequence

4.1. Steps and how to do the job

4.1.1 Equipment, tools, materials

(Calculated for a practice session of 20 students)

Equipment Type	Quantity
1	VOM multimeter	10 pieces

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Resistors of all kinds, capacitors of all kinds, Diodes of all kinds, BJTs of all kinds	100 pieces
3	Each student prepares practice materials, test board, circuit board, and electrical wires.	10 sets
4	Workshop	1

4.1.2. Implementation process

4.1.2.1. General procedure

Name of work steps	Equipment, tools, materials	Work performance standards	Common errors and how to fix them
1	Experiment	Testboard, resistors of all kinds, capacitors of all kinds, BJTs of all kinds, diodes of all kinds, electric wires, VOM multimeter	Follow the specific procedure described in section 4.1.2.2.1.	-Incorrect experiment operation - Incorrect operation with VOM meter - Recorded incorrect results * Must strictly follow the procedures and regulations of the supervisor.
2	Record the experimental results	Practice materials, pens	Record experimental results accurately.
3	Submit collected and recorded documents to supervisor	Paper, pen, note taking materials.	Ensure full volume
4	Perform industrial cleaning	- Testboard, resistors, capacitors, diodes of all kinds, BJT, VOM meters, electric wires - Clean cloth	-Clean

4.1.2.2. Specific process

4.1.2.2.1. Experiment to determine the pins, check the good/bad properties of BJT and install the bias circuit for BJT

a. Check equipment and components

b. Conduct the experiment: Each group of 2 students, in which one student performs the measurement and assembles the circuit, one student reads the results and records the measurement results.

c. Record the experimental results

4.1.2.2.2. Submit collected documents and notes to the instructor.

4.1.2.2.3. Perform industrial cleaning

4.2. Practice exercises

4.2.1. Prepare equipment, tools and materials

4.2.2. Grouping

Each group of 2 students practice

4.2.3. Implement general and specific procedures

4.3. Requirements for evaluation of results

Target

Content	Point
Knowledge	- Present bias circuits for BJT	4
Skill	- Perform the experiment correctly. - Teamwork skills. - Recording and calculation skills.	4
Attitude	- Be careful, listen, take notes, be slow, and practice good industrial hygiene.	2
Total		10

REVIEW QUESTIONS AND EXERCISES

1. Describe the structure and symbol of npn BJT?

2. Describe the structure and symbol of pnp BJT?

3. What is the operating principle of npn BJT?

4. Describe the connection methods of BJT?

5. Describe how to bias NPN reverse BJT using voltage divider circuit

6. Describe how to bias a PNP forward BJT with a fixed current.

7. Given the circuit as shown, the BJT is made of Si with U CC = 12V, R B = 680k  , R C = 2.2k  and R E = 1K  . The BJT has a coefficient  = 100. Calculate the static operating point of the circuit.

8. Given the circuit as shown in the figure, the BJT is Ge type, U CC = 18V, R 1 = 47k  , R 2 = 12k  , R C = 1.5k  and R E = 560  . The BJT has a coefficient  = 80. Calculate the static operating point of the circuit.

9. Given the circuit as shown, the BJT is made of Si with U CC = 12V, R B = 560k  , R C = 3.3k  and R E = 1k  . The BJT has a coefficient  = 100. Calculate the static operating point of the circuit.

LESSON 4: OTHER SEMICONDUCTOR COMPONENTS

Target:

After completing this chapter, students will be able to:

- Correctly present the structure, conventional symbols and operating principles of other semiconductor components.

- Assemble and calibrate circuit types of other semiconductor components .

- Self-awareness, high discipline, sense of responsibility in work. Proactive and creative in learning

Main content:

- FET field transistor

- Thyristor

- Triac and Diac

I. FET FIELD TRANSISTOR

1. Structure, conventional symbols, operating principles

1. Structure

FET (Field Effect Transistor) conducts electricity by only one type of conductor, while BJT conducts electricity by both electrons and holes, so they are also called unipolar transistors. They have two different groups in structure: the group with a pn junction gate (JFET) and the group with an insulated gate (MOSFET).

Figure (4.1) describes the structure of n-channel JFET and p-channel JFET. Thus, there exists an electron conduction channel (n-type channel in figure (4.1a)) or a hole conduction channel (p-type channel in figure (4.1b)) connecting two poles named drain D (Drain) and source S (Source); the third pole taken at the semiconductor region opposite the conduction channel is named gate G (Gate), there can be one gate or two gates.

Gate(G1)

Drain (D)

trough

Gate(G2)

Gate(G1)

Drain (D)

trough

Gate(G2)

Door 1 pp

Door 2

Door 1 nn

Door 2

Source ( S) Source

(a) (b)

Figure 4.1: Structure of n-channel (a) and p-channel (b) JFET.

1.2. Conventional symbols

n-channel JFET p-channel JFET

1.3. Operating principle

1.3.1. JFET field-effect transistor

Consider an n-channel JFET with the D pole connected to the positive source and the S pole connected to the negative source as shown in figure (4.2). When the G pole is open (UGS = 0V):

At this time, the current will pass through the channel from the positive pole of the source to the D pole and out the S pole to return to the negative source of the UDD, the channel acts as a resistor (Figure 4.2).

I D

U DD

Figure 4.2: n-channel JFET when UGS = 0V and UDS > 0V.

If the UDS voltage is increased from 0V, the ID current increases rapidly, but then reaches a limit voltage, the ID current cannot increase anymore, which is called the saturation current IDSS. The UDS voltage with IDSS is called the Up cut-off voltage.

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