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Midterm 1 Performance
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Concepts of electrons and holes in semiconductors
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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)
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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
Fig. 4.3
• 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.
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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
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Device Equations – Cont’d
From Eq. 4.3 and 4.4 we have:
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Example
Figure 4.4
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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:
1. Solve the input circuit to determine the base current
2. Determine output characteristics corresponding to the base current
3. Solve the output circuit to determine the VCE and IC
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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
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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.
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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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