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Bipolar Junction Transistor

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Bipolar Junction Transistor

  • 1
    Chapter Three The Bipolar Junction Transistor 3-1
  • 2
    Figure 3.10 3-2 Common-emitter circuits: (a) with an npn transistor, (b) with a pnp transistor, and (c) with a pnp transistor biased with a positive voltage source - Vic
  • 3
    Figure 3.13 3-3 Transistor current-voltage characteristics of the common-emitter circuit ic (rnA) 3 2 1 O Forward-active mode 2 4 6 iB 30 VA 25 20 15 10 5 8 10 12 VCE (V) npn or VEC (V) pnp
  • 4
    Figure 3.18 3-4 (a) Common-emitter circuit with an npn transistor and (b) dc equivalent circuit, with piecewise linear parameters CE VBE (on) cc cc
  • 5
    Figure 3.19 Circuit for Example 3.3 Vcc= 10 v Rc=2kQ Rc=2kQ RD = 220 k") BE 4-0.7 V 220 kQ 0.7 V CE 3-5 = 3 mA IE=IC+IB = 3.02 mA
  • 6
    Figure 3.21 Circuit for Example 3.4 vcc=5V VBB = = 580k') EC RB=580kQ +1.5 V (5-0.6) -IS 580 kQ 3-6 VCC = 5 v = 0.505 mA 0.6 v = 2.5 V VBB = 1.5 v 5-25 - 0.5 mA
  • 7
    Figure 3.22 3-7 (a) base-emitter junction characteristics and the input load line and (b) common- emitter transistor characteristics and the collector-emitter load line -2! = 18.2 B-E junction characteristics 0 Quiescent base current and B-E voltage IÅ)ad line ic (mA) 6 5 lc (sat) — 4 3 2 1 0 Forward-active mode Saturation Load line Q-point (on) = 07 V VBB- o 2 4 6 8 30 25 20 15 ILA 10 Cutoff 10 VCE (sat)
  • 8
    Figure 3.23 Circuit for Example 3.5 +10 V Rc=4kQ = 3.32 RB 220 + Rc-4kQ = 220k') ICE: --3.28 v I 0.7 V x 220kQ 33.2 /IA pcssible 3-8 10—02 4 RB-220kQ = VCE(sat) =o.2V 1-33.2 0.7 v EZlc+lB = 2.483 mA
  • 9
    3-9 Problem Solving Method: Bipolar DC Analysis Analyzing the dc response of a bipolar transistor circuit requires knowing the mode of operation of the transistor. In some cases, the mode of operation may not be obvious, which means that we have to guess the state of the transistor, then analyze the circuit to determine if we have a solution consistent with our initial guess. To do this, we can: L 2. 3. 4. Assume that the transistor is biased in the forward-active mode in which case V BE VBE(on), 1B > 0, and Ic FIB. Analyze the ' 'linear" circuit with this assumption. Evaluate the resulting state of the transistor. It the initial assumed para- meter values and V CE > VcE(sat) are true, then the initial assumption is correct. However, if 1B < 0, then the transistor is probably cut off, and if VCE < O, the transistor is likely biased in saturation. If the initial assumption is proven incorrect, then a new assumption must be made and the new "linear" circuit must be analyzed. Step 3 must 'then be repeated. McGraw-Hill Copyright 0 2001 by the McGraw-Hill Companies, Inc. All rights reserved.
  • 10
    Figure 3.26 Circuit for Example 3.6 VCC = 12 v = 0.4 k') VBB = 6 V 3-10 +12 V Ic =ßIB = 0.4 kQ = 5.63 mA = 12 CE BE = 0.6 0.7 v = 0.6k') = 6.32 v +ß)IB = 5.71 mA
  • 11
    Figure 3.33 3-11 Circuit for Example 3.8 RE=2k = 2.5 V RE=2kQ EC 0.6 v 190 k — = 20.5 5-2.5 - 1.25 mA = 1.23 mA (a)
  • 12
    Figure 3.44 3-12 (a) A bipolar inverter used as an amplifier; (b) the inverter voltage transfer characteristics Cutoff Rc=4kQ RB=100kQ (a) 0.2 o 0.7 Forward-active mode Saturation 1.9 5 VI(V)
  • 13
    Figure 3.45 (a) The inverter circuit with both a dc and an ac input signal; (b) the dc voltage transfer characteristics, Q-point, and sinusoidal input and output signals; (c) the transfer characteristics showing improper dc biasing 3-13 AVI R BB = 1.3 v Rc=4 kS2 vo 100 kS2 (a) vo(V) vo(V) Avo Tirne Q-point Avr Time Q-point AVI Tirne (b) Avo Time o (c)
  • 14
    Figure 3.53 (a) A common-emitter circuit with an emitter resistor and voltage divider bias circuit in the base; (b) the dc circuit with a Thevenin equivalent base circuit 3 — 14 cc cc (a) CEQ CEQ IEQ

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