Vc vs Vin of NPN

Discussion in 'Homework Help' started by TsAmE, Oct 4, 2010.

Apr 19, 2010
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0
Calculate the collector voltage Vc as Vin increases from 0-5V and hence plot the transfer characteristic Vc vs Vin (consider many points where Vc is changing rapidly). Calculate Vin when Vc = 2V. An NPN with a gain of B = 100 is used.

Attempt:

When Vin = 0V and 1V Vc = 0.

Vin = 2V: Ib = 2 - 0.7 / 10 x 10^3
= 1.3 x 10^-4

Ic = IbB
= 1.3 x 10^-4 x 1000
= 0.013

Vc = 5 - 0.013 x 1 x 10^3
= -8V

I dont understand why I got a negative answer.

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2. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
In principle once you have enough base current to drive the collector voltage to 0V (actually it would be to Vce_sat) it can't be driven less than 0V - a condition called "saturation".

So for Vc to be zero then Ic would be 5mA. To just drive Ic to 5mA would require Ib=5mA/β=50uA.

To get Ib=50uA with Vbe assumed as 0.7V would require

Vin=10k*50uA+0.7=1.2V

An input voltage greater than this would lead to the saturation condition.

This isn't exactly what will happen in practice - but will suffice for the sake of this simple problem statement.

Last edited: Oct 4, 2010
3. Markd77 Senior Member

Sep 7, 2009
2,803
594
You are getting a negative answer because you have calculated a collector current that is greater than is possible to pass through the 1K resistor with the given voltage (5mA).

Apr 19, 2010
72
0
Howcome Vin = 1.2V would lead to saturation? Since the rule that the transistor will turn on if Vin - 0.7 >= 0.7V isnt fufilled? (with 1.2 - 0.7 = 0.5V)

5. Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
To open small signal BJT you need Vbe grater then 0.5V.

Your circuit will be in saturation if Ib is grater then Ib > ( Vcc / β*Rc ) = 5V/100KΩ = 50μA
This base current will cause voltage drop on RB resistor equal to :
Vb = Ib * RB = 0.5V
So for Vin grater then Vin > ( Vcc*Rb / β*Rc ) + Vbe BJT will be in saturation.

Of course we assume that β is constant, witch is not true .

6. Georacer Moderator

Nov 25, 2009
5,151
1,266
Saturation isn't a state before the transistor turn-on (or active region as it is called). It is actually achieved if you turn on your transistor "too-much".
In increasing order of ib (and Vbe consequently) the transistor goes from cut-off, to active, to saturation. In the last state it cannot increase any more the current of the collector and therefore the relation Ic=β*Ib isn't valid anymore.