Problem 1: Determine the following: (d) Current gain (e) Power gain (f) Output resistance R₂ Wo 10 ΚΩ R₂ R₁ *100 ΚΩ 100 ΚΩ RE 2kQ2 Vcc=+20 V B= 200 VBEQ = 0.7 V RL IK
Protection System
A system that protects electrical systems from faults by isolating the problematic part from the remainder of the system, preventing power from being cut from healthy elements, improving system dependability and efficiency is the protection system. Protection devices are the equipment that are utilized to implement the protection system.
Predictive Maintenance System
Predictive maintenance technologies are designed to assist in determining the state of in-service equipment so that maintenance can be scheduled. Predictive maintenance is the application of information; proactive maintenance approaches examine the condition of equipment and anticipate when it should maintain. The purpose of predictive maintenance is to forecast when equipment will fail (depending on a variety of parameters), then prevent the failure through routine and corrective maintenance.Condition monitoring is the continual monitoring of machines during process conditions to maintain optimal machine use, which is necessary for predictive maintenance. There are three types of condition monitoring: online, periodic, and remote. Finally, remote condition monitoring allows the equipment observed from a small place and data supplied for analysis.
Preventive Maintenance System
To maintain the equipment and materials on a regular basis in order to maintain those running conditions and reduce unnecessary shutdowns due to unexpected equipment failure is called Preventive Maintenance (PM).
![Problem 1:
Determine the following:
(d) Current gain
(e) Power gain
(f) Output resistance
R₂
Wo
10 ΚΩ
R₂
R₁
*100 ΚΩ
100 ΚΩ
RE
2kQ2
Vcc=+20 V
B= 200
VBEQ = 0.7 V
RL
K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93e18a88-13ba-47d2-8c84-ec9d29397d2a%2F02b0e722-6617-49f5-b9b6-ca59519d5fd3%2Fd99g1g_processed.jpeg&w=3840&q=75)
![Vs
+
RS
ww
YOK-n
Step 1: open-ckt capacitors
IB= VH-VBE
gm
HE
remove AC ✓ source:
RTH = R₁11 R ₂ = 10011 100 = 50kn
VTH =
Vcc B₂
20 x 100 = IDV
R₁ + R₂
100 + 100
=
RI 100KA
R₂100kn
= 10-c.1
RTH + (1+B) RE 50+201×2
a) Small signal parameters:
re = UT
25mv
IE
4.221 MA
=
b) Voltage gain :
= 200
1.19×103
= 0.021 MA
I E = (1 + B) IB = 201 × 0.071 = 4.221 MA
IC IE-IB = 4.221 - 0.021
= 4.2mA
AV = = R₁
не
RE 2K-2 RL {IK_2
:
Av= - Rc 11 R₂
re
5.92
VCC
=0.168
= +20v
rπ = (1+ß) re = (1+200) 5.92 = 1.19kn
B
= -103
Ri
VTH
P₂{
Т
Since Rc=0,
= -168.91 v
B=200
VBEQ = 0.7
¥
2K
Vec 20v
RTH IB
HE
IE√ { 2K₁
Vec
5.92
c) Input Resistance :
Rin (Rill R₂)| VTT
Rin= (1001/100) 11.19 → Rin: 5011 1.19 → Rin= 50×1.19 = 1.16K1
50+1.19
✔](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93e18a88-13ba-47d2-8c84-ec9d29397d2a%2F02b0e722-6617-49f5-b9b6-ca59519d5fd3%2F6h05rxq_processed.jpeg&w=3840&q=75)
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