[SOLVED] NPN Transistor biasing question

[peeyushsigma]

hi folks,

In the attached transistor circuit i want to calculate V_CE ..



my analysis " transistor BE junction will be reverse biased and V_CE=5v ".

but answer says "Valid Answer range between: 4.26 and 4.27 ".

 

[ArticCynda]

How do you end up with a reverse biased transistor? When calculating the base voltage of the transistor, I get \[V_b = (2 - (-5)) * \frac{100}{100+15} - (-5) \approx 1.0\,V\].
A bipolar transistor will be conducting when its base emitter voltage is above ca. 650 mV, so in this circuit it's definitely forward biased. Of course, this is a "quick" calculation that neglects the base current of the transistor.

 

[peeyushsigma]

Thanks ArticCynda Got it ...

 

[arunkumarshanti]

How do you end up with a reverse biased transistor? When calculating the base voltage of the transistor, I get \[V_b = (2 - (-5)) * \frac{100}{100+15} - (-5) \approx 1.0\,V\].
A bipolar transistor will be conducting when its base emitter voltage is above ca. 650 mV, so in this circuit it's definitely forward biased. Of course, this is a "quick" calculation that neglects the base current of the transistor.

Hi,

Your above calculation will result in 11.08V which is wrong. The calculation goes like this https://obrazki.elektroda.pl/2800079700_1418056556.jpg. I hope attached solution explains everything.

regards
Arun

 

[chuckey]

Looking at the two bias resistors centre tap. Voc = (7 X 100/115) -5 ~ (6.086 ) -5 = 1.086V. I sc = 2/15 - 5/100 mA = .133 - .05 = .0833 mA. So we can replace the resistors and their voltages by a 1.086 V in series with a 1.086/.0833 resistor = 12.977 K
Frank