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Midterm 1 Performance

Midterm 1 Performance. Concepts of electrons and holes in semiconductors. Forward and reverse bias in a p-n junction. Under forward bias the width of the depletion region decreases. Current increases exponentially.

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Midterm 1 Performance

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  1. Midterm 1 Performance

  2. Concepts of electrons and holes in semiconductors

  3. Forward and reverse bias in a p-n junction • Under forward bias the width of the depletion region decreases. Current increases exponentially. • Under reverse bias the width of the depletion region increases. Very low current flow (leakage current Is)

  4. Chapter 4: Bipolar Junction Transistor 4.1 Basic Operation of the npn Bipolar Junction Transistor Figure 4.1 The npn BJT Note: • The current flowing in a BJT is mostly due to electrons moving from the emitter through the base to the collector • We define the current gain of the transistor as β = IC/IB • The BJT can be considered as a current controlled current source. Input current is Ib and output current is Ic. Fig. 4.3

  5. Basic Operation in the Active Region – Cont’d Basic Operation in the Active Region: An npn transistor (CE configuration) with variable voltage sources operating in the active region: • VBE ≈ 0.6 V to forward bias the BE junction • VCE >VBE - the base collector junction is reverse biased First-Order Common-Emitter Characteristics

  6. Device Equations – Cont’d From Eq. 4.3 and 4.4 we have:

  7. Example Figure 4.4

  8. Load line analysis – Input circuit The slope of the load line is -1/RB Fig. 4.11 From Kirchoff’s voltage law applied to the input circuit: • Three major steps for analyzing BJT: • Solve the input circuit to determine the base current • Determine output characteristics corresponding to the base current • Solve the output circuit to determine the VCE and IC

  9. Load line analysis – Output circuit From Kirchoff’s voltage law applied to the output circuit: As IB changes with input voltage VB, the output current and voltage changes according to the transistor characteristics. Almost always an amplification can be obtained. Fig. 4.11 For each value of ib, there is specific ic vs. VCE characteristic. We have to draw the load line and find the equilibrium point for that load line to find the equilibrium ic and VCE

  10. Example First step: Find iB corresponding to the Q-point and the two extreme values of the applied voltage , i.e. vin = +0.4 V, and -0.4 V The values come out to be IBQ = 25 µA, IB, max = 35 and IB, min = 15 µA, respectively.

  11. Example Thus, from the Q-point of the o/p circuit, we have, ICQ = 2.5 mA, and VCEQ = 5V corresponding to iB = 25 µA . Second Step:Compute IC and corresponding VCEQ ICQ,max = 3.5 mA, and VCEQ, min = 3 V corresponding to iB, max = 25 µA . ICQ,min = 1.5 mA, and VCEQ, max = 7 V corresponding to iB, min = 15 µA . Thus, the peak-to-peak value of ac component of VCE = 4 V, and peak-to-peak value of vin = 0.8 V Thus the gain is 5, but with a negative sign, as seen from Fig. 4.13. Confirm that when the base voltage VB is higher the o/p voltage is lower.

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