Chapter 3 Problems: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
Line 12: Line 12:
A Diode at room temperature (298.15K) is at rest. If 100mA is induced and the voltage is 0.65V. What is the temperature of the diode if after several minutes the voltage is 0.45V after the diode heats up.
A Diode at room temperature (298.15K) is at rest. If 100mA is induced and the voltage is 0.65V. What is the temperature of the diode if after several minutes the voltage is 0.45V after the diode heats up.


"As the temperature increases, the knee voltage decreases by about 2mV/K."<ref><sup>Hambley, Allan. Electronics 2nd Edition. 2000. Page 133.</sup></ref> Therefore we had a voltage drop of 0.65V-0.45V=0.20V With 0.002V/K, we take <math>\frac {0.20V}{0.002V/K} = 100K </math>. An estimated tempterature change is 100 Kelvin.
"As the temperature increases, the knee voltage decreases by about 2mV/K."<ref><sup>Hambley, Allan. Electronics 2nd Edition. 2000. Page 133.</sup></ref>. Therefore we had a voltage drop of 0.65V-0.45V=0.20V With 0.002V/K, we take <math>\frac {0.20V}{0.002V/K} = 100K </math>. An estimated tempterature change is 100 Kelvin.


=== Problem 3.11===
=== Problem 3.11===

Revision as of 17:09, 19 January 2010

Problem 3.1

Typical diode (drawing redrawn by Ben Henry)

Show the graphical symbol and general design of a diode used in electrical schematic diagrams. Identify the anode and cathode to show the polarity of the diode.

Problem 3.3

Define a Zener diode, its other names, and how it is generally used. Show a graph of a volt/ampere properties for a 5.6-V Zener diode.

A Zener diode functions like a normal diode in the normal orientation. However, in the reverse polarity, that is with the band or cathode side to the positive input, the diode has what is called the Zener breakdown voltage. At this voltage the diode becomes conductive and allows current to flow through it at manufactured specified voltages.

Problem 3.7

A Diode at room temperature (298.15K) is at rest. If 100mA is induced and the voltage is 0.65V. What is the temperature of the diode if after several minutes the voltage is 0.45V after the diode heats up.

"As the temperature increases, the knee voltage decreases by about 2mV/K."<ref>Hambley, Allan. Electronics 2nd Edition. 2000. Page 133.</ref>. Therefore we had a voltage drop of 0.65V-0.45V=0.20V With 0.002V/K, we take . An estimated tempterature change is 100 Kelvin.

Problem 3.11