Amplifier Teaching Aid (DED Philippinen, 86 p.) Lesson 6 - Transistor Biasing II Lesson Plan (introduction...) Transistor biasing II VDB analysis Worksheet No. 6

Amplifier Teaching Aid (DED Philippinen, 86 p.)

Lesson 6 - Transistor Biasing II

Lesson Plan

Title: Transistor Biasing II

Objectives:

- Know the advantage of voltage divider bias
- Able to analyse VDB circuits

Figure

Transistor biasing II

Voltage Divider Bias

The most famous circuit based on -the prototype of emitter bias is called the voltage divider bias (VDB).

Recall the steps of analyzing the emitter bias circuit:

1. V_E
2. I_E
3. I_C
4. Voltage drop across R_C
5. V_C
6. V_CE

The three most important steps are:

1. I_E
2. V_C
3. V_CE

Fig. 6-1: Emitter biased circuit

Problem: Sometimes the voltage from the V_CC power supply is too large to apply directly at the base.

Solution:

- extra power supply for the base
- or ==> VDB

Fig. 6-2: VDB circuit

The voltage drop across R₂ is applied directly to the base, which means:

V₂ = V_B

1. step: find voltage drop across R₂
2. step: subtract 0.7V to get V_E

VDB analysis

Design errors of 5% or less are acceptable, because of resistor tolerances.

Fig. 6-3: VDB example circuit

Find the base voltage:

Assumption: Base current is so small that it has no effect on the voltage divider.

5% error - > base current is 20 times smaller than the divider current.

V_B = I * R₂ = 0.82 mA * 2.2KW = 1.8V

V_E = V_B - V_BE = 1.8V - 0.7V = 1.1V

V_C = V_CC -(R_C * I_C) = 10V - (3.6KW * 1.1 mA) = 6.04V

V_CE = V_C - V_E = 6.04V - 1.1V = 4.94V

Checking the assumption:

5% error -->

The current gain can vary from 30 to 300.

Even under the worst case condition the calculation is within the 5% limit, hence the assumption can be done.

Summary of Process and Formulas

Divider current
Base voltage	VB = I * R2
Emitter voltage	VE = VB - VBE
Emitter current
Collector voltage	VC = VCC - (IC * RC)
Coll.- emitter voltage	VCE = VC - VE

HO: What will change if the emitter resistor increases to 2KW? (unchanged voltage divider)

Fig. 6-4: VDB circuit

Solution:

I = 0.82 mA

V_B = 1.8V

V_E = 1.1V

V_C = V_CC - (R_C * I_C) = 8.02V

V_CE = V_C - V_E = 6.92V

VDB Load-Line and Q-Point

Fig. 6-5: VDB circuit

Saturation point:

Visualize short between collector and emitter

V_RC = V_CC - V_E = 10V - 1.1V = 8.9V

Cutoff point:

Visualize open between collector and emitter

- - > V_{CE (cut)} = V_CC - V_E = 8.9V

Q-point:

V_C = V_CC - (I_C * R_C) = 10V - (1.1 mA * 1KW) = 6.04V

V_CE = V_C - V_E = 6.04V - 1.1V = 4.94V

Now we plot these values and get the load line and the Q-point:

Fig. 6-6: Output curve with load line and Q-point

The values V_CC, R_C, R₁, and R₂ are controlling saturation current and cutoff voltage. To move the Q-point is possible by varying the emitter resistance (R_C).

Get the Q-point in the Middle of the Load Line

To set the Q-point is a important preparation as you will see later on.

Effect of R_E:

R_E too large -- > Q-point moves into cutoff
R_E too small --> Q-point moves into saturation

Q - point in the middle of the load line:

Half the value of I_{C (sat)} and redesign R_E

I_{C (sat)} = 2.47 mA ==> 1.23 mA

Look for the nearest standard value:

===> 910 W

Fig. 6-7: Output curve, Q-point in the middle

Worksheet No. 6

Figure

No. 1 What is -the emitter voltage? The collector voltage? Given: R₁= 10k, R₂= 2.2k, R_C = 3.6k, R_E = 1k, V_CC = 25V

Draw the load-line, plot the Q point!

No. 2 What is the emitter voltage? The collector voltage?

Given: R₁= 330k, R₂= 100k, R_C = 150k, R_E = 51k, V_CC = 10V

Draw the load-line, plot the Q point!

No. 3 What is the emitter voltage? The collector voltage? Given: R₁ = 10k, R₂ = 2.2k, R_C = 2.7k, R_E = 1k, V_CC = 10V

Draw the load-line, plot the Q point!

Redesign the circuit to get the Q-point in the middle of the loadline!