PHYS 162 - Chapter 5 Transistor Bias Circuits

Prepared By: Syed Muhammad Asad – Semester 102 Page 1

Figure 1 Example of linear and nonlinear operation

CHAPTER 5 TRANSISTOR BIAS CIRCUITS

5-1 THE DC OPERATING POINT - A transistor must be properly biased with a DC voltage to operate in the linear region.

- It ensures an amplified and accurate signal production at the output.

- The DC operating point is often referred as Q-point.

- The DC parameters that need to be found to determine the Q-point are collector current IC and

collector-emitter voltage VCE.

5.1.1 DC Bias

- If an amplifier is not biased with the correct DC voltages, it can go into saturation and cutoff.

- Figure 1(a) shows the correct

linear operation with amplified

output.

- Figure 1(b) shows nonlinear

operation where the amplifier is in

cutoff. The clipping in the positive

cycle is always due to cutoff.

- Figure 1(c) shows nonlinear

operation where the amplifier is in

saturation. The clipping in the

negative cycle is always due to

saturation.

5.1.1.1 Graphical Analysis

- In Figure 2, we chose three values

of IB and observe what happens to

IC and VCE.

o For 𝐼𝐵 = 200𝜇𝐴,

𝑉𝐶𝐸 = 5.6𝑉

o For 𝐼𝐵 = 300𝜇𝐴, 𝑉𝐶𝐸 = 3.4𝑉

o For 𝐼𝐵 = 400𝜇𝐴, 𝑉𝐶𝐸 = 1.2𝑉

- The corresponding Q-points can be seen on the graph.

PHYS 162 - Chapter 5 Transistor Bias Circuits

Prepared By: Syed Muhammad Asad – Semester 102 Page 2

5.1.1.2 DC Load Line

- The DC operation of a transistor circuit can be described graphically using a DC load line.

- It is a straight line connecting 𝐼𝐶 = 𝐼𝐶 𝑠𝑎𝑡 on the y-axis to 𝑉𝐶𝐸 = 𝑉𝐶𝐶 on the x-axis.

- At saturation 𝐼𝐶 𝑠𝑎𝑡 =𝑉𝐶𝐶−𝑉𝐶𝐸 𝑠𝑎𝑡

𝑅𝐶 and at cutoff 𝑉𝐶𝐸 = 𝑉𝐶𝐶 .

- Figure 3 shows the three Q-points.

Figure 2 Q-point adjustment

Figure 3 The Dc load line

PHYS 162 - Chapter 5 Transistor Bias Circuits

Prepared By: Syed Muhammad Asad – Semester 102 Page 3

5.1.1.3 Linear Operation

- All point along the DC load line between saturation and cutoff is the linear region of operation for a

transistor.

- Figure 4 is an example of linear operation.

- AC voltage Vin produces an AC base current 𝐼𝑏(𝑝𝑒𝑎𝑘 ) = 100𝜇𝐴 above and below the Q-point.

- This produces an AC collector current 𝐼𝑐(𝑝𝑒𝑎𝑘 ) = 10𝑚𝐴 above and below the Q-point.

- This change in the collector current changes the collector-emitter voltage 𝑉𝑐𝑒(𝑝𝑒𝑎𝑘 ) = 2.2𝑉.

- This changing Vce is the required voltage amplification at the output of the transistor.

NOTE: REFER EXAMPLE 5-1 PAGE 221

5-2 VOLTAGE-DIVIDER BIAS - Voltage-divider bias is one of the widely used biasing techniques for a

transistor.

- It uses a single power source and a voltage-divider to attain the voltage

base bias voltage.

- For circuit analysis, it is assumed that the base current IB is small enough

to be neglected.

- There are two types of voltage-dividers.

o Stiff voltage divider where

𝑉𝐵 = 𝑅2

𝑅1 + 𝑅2 𝑉𝐶𝐶

If 𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 ≥ 10𝑅2

o Non Stiff voltage divider where

𝑉𝐵 = 𝑅2||𝑅𝐼𝑁 𝐵𝐴𝑆𝐸

𝑅1 + 𝑅2||𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 𝑉𝐶𝐶

If 𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 < 10𝑅2

- 𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 = 𝛽𝐷𝐶𝑅𝐸

Figure 4 Variation in AC current and voltage

Figure 5 Voltage-divider bias

PHYS 162 - Chapter 5 Transistor Bias Circuits

Prepared By: Syed Muhammad Asad – Semester 102 Page 4

NOTE: REFER EXAMPLE 5-2 PAGE 224

5-3 OTHER BIAS METHODS - Other types of biasing methods are

o Emitter Bias

Excellent Q-point stability.

Uses two voltages sources instead of one.

o Base Bias

Mainly used for switching circuits.

Not suitable for linear amplifier because of poor Q-point stability.

o Emitter-Feedback Bias

Adding an RE in Base bias circuits gives emitter-feedback bias.

Better Q-point stability than the base bias but still not well enough for linear operation.

o Collector-Feedback Bias

Better Q-point stability than emitter-feedback bias.

Can be used in linear amplifier circuits.

- A summary of all the equations is given in Table 1.

Table 1 Transistor Bias Circuit Formula Sheet

Voltage-Divider Bias Emitter Bias Base Bias Emitter-Feedback Bias

Collector-Feedback Bias

Stiff voltage-divider

𝑉𝐵 = 𝑅2

𝑅1 + 𝑅2

𝑉𝐶𝐶

If 𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 ≥ 10𝑅2

Non Stiff voltage divider

𝑉𝐵 = 𝑅2||𝑅𝐼𝑁 𝐵𝐴𝑆𝐸

𝑅1 + 𝑅2||𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 𝑉𝐶𝐶

If 𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 < 10𝑅2

𝑉𝐵 = 𝑉𝐸 + 𝑉𝐵𝐸

𝑅𝐼𝑁 𝐵𝐴𝑆𝐸 = 𝛽𝐷𝐶𝑅𝐸

𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 𝑉𝐸 ≈ −1𝑉 (neglecting effect of 𝛽𝐷𝐶 ) 𝑉𝐸 = 𝑉𝐸𝐸 + 𝐼𝐸𝑅𝐸(taking 𝛽𝐷𝐶 into account)

𝐼𝐶 ≅ 𝐼𝐸 =𝑉𝐸𝑅𝐸

Without 𝛽𝐷𝐶

𝐼𝐶 ≅ 𝐼𝐸 =−𝑉𝐸𝐸 − 1𝑉

𝑅𝐸

With 𝛽𝐷𝐶

𝐼𝐶 ≅ 𝐼𝐸 =−𝑉𝐸𝐸 − 𝑉𝐵𝐸𝑅𝐸 + 𝑅𝐵/𝛽𝐷𝐶

𝐼𝐶 = 𝛽𝐷𝐶 𝑉𝐶𝐶 − 𝑉𝐵𝐸

𝑅𝐵

𝐼𝐶 ≅ 𝐼𝐸 =𝑉𝐶𝐶 − 𝑉𝐵𝐸

𝑅𝐸 + 𝑅𝐵/𝛽𝐷𝐶 𝐼𝐶 =

𝑉𝐶𝐶 − 𝑉𝐵𝐸𝑅𝐶 + 𝑅𝐵/𝛽𝐷𝐶

𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶

𝑉𝐶𝐸 = 𝑉𝐶 − 𝑉𝐸 𝑉𝐶𝐸 = 𝑉𝐶 − 𝑉𝐸 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 + 𝑅𝐸 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶

NOTE: REFER EXAMPLE 5-6, 5-7, 5-8, 5-9, 5-10 PAGE 230-236