Electronic Devices - Bipolar Junction Transistors - Discussion
Discussion Forum : Bipolar Junction Transistors - General Questions (Q.No. 1)
1.
Refer to this figure. Determine the minimum value of IB that will produce saturation.
0.25 mA
5.325 
A
A1.065 A
A10.425 A
ADiscussion:
25 comments Page 1 of 3.
Sabi said:
1 decade ago
Please give explaination.
Hariom Gupta said:
1 decade ago
Apply kirchoff's voltage law in the output side
10=4.7*10^3*Ic+0.2
on solving Ic=2.085 ma.
For common emmiter configuration we know Ic=b*Ib
where b=200
So, Ib = 2.085/200 = 10.425 uA .
10=4.7*10^3*Ic+0.2
on solving Ic=2.085 ma.
For common emmiter configuration we know Ic=b*Ib
where b=200
So, Ib = 2.085/200 = 10.425 uA .
Inshaf said:
1 decade ago
How 0.2 in the first equation?
Manikantan said:
1 decade ago
That is the constant saturation value of Vce?
Hapakhoa Riad said:
1 decade ago
Vc/Rc = 10/4700 ohms = 0.0021276595744681 A.
Betta = Ic/Ib.
Ib = Ic/Betta = 2.1276595744681/200 = 1.063829787234043 e-5 = 10.63829787234043 micro A (D).
Betta = Ic/Ib.
Ib = Ic/Betta = 2.1276595744681/200 = 1.063829787234043 e-5 = 10.63829787234043 micro A (D).
Anonymous said:
1 decade ago
Ic(sat) = Vcc/Rc,
Ib(min) = Ic(sat)/beta.
Ib(min) = Ic(sat)/beta.
Bem said:
1 decade ago
Given:
B = 200.
Rc = 4.7.
Rb = 1.2.
from KCL,
Vin=IbRb+Vbe+Vcb.
B = 200.
Rc = 4.7.
Rb = 1.2.
from KCL,
Vin=IbRb+Vbe+Vcb.
Eric Orellana Romero said:
1 decade ago
About of first answer, the equation ic=B*ib is not valid for saturation zone in a bjt transistor. In this zone, bjt behavior is not lineal.
Unknown said:
10 years ago
The Circuit is Fixed biased not emitter stabilized. It does not have resistor on the emitter (Re).
Ragavendar said:
9 years ago
Find ic first,
Ic = Vcc /Rc = 2.2mA / 4.7kohm.
Then find Ib from formula Bdc = Ic/Ib.
Ic = Vcc /Rc = 2.2mA / 4.7kohm.
Then find Ib from formula Bdc = Ic/Ib.
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