Electronic Devices - BJT Amplifiers - Discussion
Discussion Forum : BJT Amplifiers - General Questions (Q.No. 35)
35.
Refer to this figure. Determine the value of VC.
Discussion:
8 comments Page 1 of 1.
Effy said:
7 years ago
Vcc-IcRc-Vce-IeRe=0.
Ie = 6mA approx. = to Ic
20-(6mA)(2500)-Vce-(6mA)(500)=0.
Vce = 2V,
Vce = Vc - Ve.
Therefore, Vc = Vce + Ve.
Ve = 3.003 V approx. = to 3V,
Vc = 2V +3V = 5V.
Ie = 6mA approx. = to Ic
20-(6mA)(2500)-Vce-(6mA)(500)=0.
Vce = 2V,
Vce = Vc - Ve.
Therefore, Vc = Vce + Ve.
Ve = 3.003 V approx. = to 3V,
Vc = 2V +3V = 5V.
(1)
Manasa said:
7 years ago
Vbe = (5/27) * 20 = 3v
Ie = 3/500 = 6 mA
Ie = Ic; (α =100/101=~1).
Vc = Vcc-(Ie * Re) = 20-(6m * 2.5K) = 5V.
Ie = 3/500 = 6 mA
Ie = Ic; (α =100/101=~1).
Vc = Vcc-(Ie * Re) = 20-(6m * 2.5K) = 5V.
(1)
Albert said:
6 years ago
Vbe= (5/27)*20 = 3.7v - 0.7v = 3v.
Ie = 3/500 = 6mA =~ Ic.
Vc= Vcc - IcRc = 20 - (6mA)(2.5K) = 5v.
Vc = 5v.
Ie = 3/500 = 6mA =~ Ic.
Vc= Vcc - IcRc = 20 - (6mA)(2.5K) = 5v.
Vc = 5v.
(1)
Sandy said:
5 years ago
Where did 27 came from?
(1)
Daniel said:
5 years ago
@Sandy.
R1+R2=27.
R1+R2=27.
(1)
Sandy said:
4 years ago
@Effy:
Where did 3V came from? Explain.
Where did 3V came from? Explain.
(1)
Naresh said:
9 years ago
VCC-IcRc-Vceq-RE.IE = 0,
IE = 6ma,
Vc = VCEQ-IE.RE = 20-14.9 = 5.1
IE = 6ma,
Vc = VCEQ-IE.RE = 20-14.9 = 5.1
Hemanth said:
8 years ago
Anyone tell me the right way to find Vc? please.
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