Analytical exercises, Circuits, Devices, Networks and Microelectronics
CDNuE
Bipolar Junction Transistor (BJT) devices
version 2.5
Defaults: Resistances in k unless otherwise indicated.
9-1. For the npn BJT slice shown, the ‘uncovered
charge’ layer is assumed to be of approximate
thickness WCB = 0.8m, and approximately
constant (not exactly true but OK for rough
analysis). Determine the collector electric-field in
V/m that occurs for each of the transistors shown
in figures (A) & (B). Let hFE =  F = 100 and
assume VBE (default) = 0.7V.
9-2. Determine V1 and V2 for the following
circuits using the approximations. Assume VBE
(default) = 0.7V.
(1) hFE =  F → large (so that IB → 0)
(2) hFE =  F = 50
Note that you will need to find node voltages VE,
VB and VC for each circuit in order to determine Efield = ECB = VCB/WCB. And to find these
voltages you must first find IB and IC.!
Notice that if you have a current source it is the
same as designating one (if not all) of the currents
through the transistor.
(We often use  F → ∞ to get a rough indication of
the electrical facts of a BJT circuit.)
Answers: (a)(1) V1 = 5.7V (a)(2) V1 = 7.3V
9-3. The I-V characteristics when IB = .01 mA for
an npn transistor are as shown. From these output
characteristics determine
(a) output conductance gO (= slope in the
active regime) and its equivalent resistance rO.
(b) Early voltage VA (Note: r0 = VA/I)
Answers: (A) VB = 3.85V, ĒCB = 2.29 V/m
Answers: ( rO = 160k, VA = 80V)
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Analytical exercises, Circuits, Devices, Networks and Microelectronics
9-4. For the following circuits some
measurements have been made and are indicated.
Determine the other labeled voltages and currents
from this information using:
(1) hFE =  F = large (so that IB → 0)
(2) hFE =  F = 50
9-6. A modification of the super-beta Darlington
pair is shown by figure BJT-6, called the Sziklai
configuration. A pnp version is shown. The
forward current gain hFE is large and is of the
approximate form
hFE = IC2/IB1 = S = (1 2
if R = ∞
(We often use  F → ∞ to get a rough indication of
electrical facts for a BJT circuit.)
Answers: (a)(1) I1 = 0.5mA, V1 = 8.0V
Typically a resistance R is included as shown to
speed up the action of the transistor Q2 during
turn-off and is chosen such that only 50% of the
current from Q1 is available to drive Q2.
9-5. The BJT can be cascaded with one transistor
supplying current to another in cascade. This
configuration is called a Darlington pair or a
‘super-beta’ transistor and is often used in power
electronics applications. An npn version is shown.
The forward current gain is large and is of the
approximate form
Determine:
(a) Resulting VBE
(b) Resulting S = hFE if 1 = 30 and 2 = 25
(= discrete transistors)
hFE = S = 12 + (1 + 2)
Determine:
(a) The resulting VBE
(b) S if 1 = 100 and 2 = 20 (= integrated
Darlington component)
(c) S if 1 = 30 and 2 = 25 (= discrete BJTs)
Answers: (b) hFE = 2120
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