The circuit illustrated has a transistor with a VBEact = 0.7 V and a ß =100. It is assumed that is 0 V and VBIAS is 1 V. V CEsat 1. What is the minimum value for input voltage (V;) that will prevent the transistor from cut-off? Note the value should be negative 2. What is the maximum value for input voltage (V₁) that will prevent the transistor from saturation? Note the value should be positive 3. To achieve equivalent magnitudes of the answers in the solutions for questions 1 and 2, what adjustment should be made to the value of VBIAS. 4. To what extent/factors are variations in the input voltage (Vi) amplified at the output voltage (V)

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)
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Author:Russell E. Smith
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Chapter12: Electronic Control Devices
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The circuit illustrated has a transistor with a V
V CEsat is O V and VBIAS is 1 V.
BEact = 0.7 V and a ß =100. It is assumed that
1. What is the minimum value for input voltage (V₁) that will prevent the transistor
from cut-off? Note the value should be negative
2. What is the maximum value for input voltage (V;) that will prevent the transistor
from saturation? Note the value should be positive
3. To achieve equivalent magnitudes of the answers in the solutions for questions 1
and 2, what adjustment should be made to the value of V BIAS.
4.
To what extent/factors are variations in the input voltage (V;) amplified at the
output voltage (V)
Transcribed Image Text:The circuit illustrated has a transistor with a V V CEsat is O V and VBIAS is 1 V. BEact = 0.7 V and a ß =100. It is assumed that 1. What is the minimum value for input voltage (V₁) that will prevent the transistor from cut-off? Note the value should be negative 2. What is the maximum value for input voltage (V;) that will prevent the transistor from saturation? Note the value should be positive 3. To achieve equivalent magnitudes of the answers in the solutions for questions 1 and 2, what adjustment should be made to the value of V BIAS. 4. To what extent/factors are variations in the input voltage (V;) amplified at the output voltage (V)
Vi
V BIAS
+ 1
HilHi
10K
w
+10 V
1K
Vo
Transcribed Image Text:Vi V BIAS + 1 HilHi 10K w +10 V 1K Vo
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Hello! I just wanted to ask how did you get VT = 26m in the Emitter dynamic resistance in step 5? Also in step 6, is the KVL in Vwrong? since V= -(IBR10k + IBBre) which makes Vo/Vi = (1k)(100) / -(86700+867) = -1.14 
Av = -1.14
Hoping for your response. Thank you!

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