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Ch05

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  • 1. 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 𝐼 𝐵 = 200𝜇𝐴, 𝑉𝐶𝐸 = 5.6𝑉 o For 𝐼 𝐵 = 300𝜇𝐴, 𝑉𝐶𝐸 = 3.4𝑉 o For 𝐼 𝐵 = 400𝜇𝐴, 𝑉𝐶𝐸 = 1.2𝑉 - The corresponding Q-points can be seen on the graph.
  • 2. 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 𝐼 𝐶 = 𝐼 𝐶 𝑠𝑎𝑡 on the y-axis to 𝑉𝐶𝐸 = 𝑉𝐶𝐶 on the x-axis. - At saturation 𝐼 𝐶 𝑠𝑎𝑡 = 𝑉 𝐶𝐶 −𝑉 𝐶𝐸 𝑠𝑎𝑡 𝑅 𝐶 and at cutoff 𝑉𝐶𝐸 = 𝑉𝐶𝐶. - Figure 3 shows the three Q-points. Figure 2 Q-point adjustment Figure 3 The Dc load line
  • 3. 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 = 𝛽 𝐷𝐶 𝑅 𝐸 Figure 4 Variation in AC current and voltage Figure 5 Voltage-divider bias
  • 4. 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 = 𝛽 𝐷𝐶 𝑅 𝐸 𝑉𝐸 = 𝑉𝐵 − 𝑉𝐵𝐸 𝑉𝐸 ≈ −1𝑉 (neglecting effect of 𝛽 𝐷𝐶 ) 𝑉𝐸 = 𝑉𝐸𝐸 + 𝐼 𝐸 𝑅 𝐸(taking 𝛽 𝐷𝐶 into account) 𝐼 𝐶 ≅ 𝐼 𝐸 = 𝑉𝐸 𝑅 𝐸 Without 𝛽 𝐷𝐶 𝐼 𝐶 ≅ 𝐼 𝐸 = −𝑉𝐸𝐸 − 1𝑉 𝑅 𝐸 With 𝛽 𝐷𝐶 𝐼 𝐶 ≅ 𝐼 𝐸 = −𝑉𝐸𝐸 − 𝑉𝐵𝐸 𝑅 𝐸 + 𝑅 𝐵/𝛽 𝐷𝐶 𝐼 𝐶 = 𝛽 𝐷𝐶 𝑉𝐶𝐶 − 𝑉𝐵𝐸 𝑅 𝐵 𝐼 𝐶 ≅ 𝐼 𝐸 = 𝑉𝐶𝐶 − 𝑉𝐵𝐸 𝑅 𝐸 + 𝑅 𝐵/𝛽 𝐷𝐶 𝐼 𝐶 = 𝑉𝐶𝐶 − 𝑉𝐵𝐸 𝑅 𝐶 + 𝑅 𝐵/𝛽 𝐷𝐶 𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼 𝐶 𝑅 𝐶 𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼 𝐶 𝑅 𝐶 𝑉𝐶𝐸 = 𝑉𝐶 − 𝑉𝐸 𝑉𝐶𝐸 = 𝑉𝐶 − 𝑉𝐸 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼 𝐶 𝑅 𝐶 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼 𝐶 𝑅 𝐶 + 𝑅 𝐸 𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼 𝐶 𝑅 𝐶 NOTE: REFER EXAMPLE 5-6, 5-7, 5-8, 5-9, 5-10 PAGE 230-236

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