Industrial Motor Control
7th Edition
ISBN: 9781133691808
Author: Stephen Herman
Publisher: Cengage Learning
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Textbook Question
Chapter 58, Problem 1RQ
How is a zener diode connected in a circuit as compared to a common junction diode?
Expert Solution & Answer
To determine
The way in which the zener diode is connected in the circuit as compared to the junction diode.
Explanation of Solution
The PN junction is commonly known as the diode. The diode the electronic device that acts as a swtich and conducts in one direction only.
To forward bias the diode the P terminal is connected to the ANODE and the N type is connected to the cathode. Reverse of this is done to reverse bias the diode.
The symbol for the diode is shown below.
The diode operates like a swtich, when the diode is forward biased the current flows through it in one direction only that is from P terminal to N terminal. To reverse the diode the cathode is connected to P and anode to N, for ideal condition the diode do not conduct in reverse but there is a small amount of current through it that is called the leakage current.
The zener diode is the diode that can operate in the reverse biased and the region of operation in that case is called the zener region. The diode conducts when the diodes breaks down in the reverse direction known as avalanche.
Therefore, the zener diode is connected in the circuit in reverse polarity as compared to common junction diode.
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cutting
Instructions:
Do not copy the drawing.
Draw In third-angle orthographic projection, and to scale 1:1,
the following views of the hinge:
A sectional front view on A-A
A top view
⚫ A right view (Show all hidden detail)
Show the cutting plane in the top view
. Label the sectioned view
Note:
All views must comply with the SABS 0111 Code of Practice for
Engineering Drawing.
Galaxy A05s
Assessment criteria:
⚫ Sectional front view
026
12
042
66
[30]
11
10
1. Plot the moment (M), axial (N), and shear (S) diagrams as functions of z.
a)
b)
F₁ = 1250 N
F₁ = 600 N
M₁ = 350 000 N mm
F2 = 500 N
200 N
a = 600 mm
b=1000 mm
a=750 mm
b = 1000 mm
d)
M₁ = 350 000 N mm
F₁ = 600 N
F₂ =200 N
a = 600 mm
b = 1000 mm
M₁ 175 000 Nmm
F = 900 N
a-250 mm
b-1000 mm
-250 mm.
Figure 1: Schematics problem 1.
Given the following cross-sections (with units in mm):
b)
t=2
b=25
h=25
t = 1.5
b=20
b=25
t=2
I
t = 1.5
a=10
b=15
h-25
b=15
t=3
T
h=25
Figure 3: Cross-sections for problem 2.
1. For each of them, calculate the position of the centroid of area with respect to the given coordinate system
and report them in the table below.
2. For each of them, calculate the second moments of inertia I...
and I, around their respective centroid
of area and report them in the table below. Note: use the parallel axes theorem as much as possible to
minimize the need to solve integrals.
Centroid position
x
y
box
Moment of inertia
lyy
by
a)
b)
c)
d)
e)
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