transistor problem: sketching relationships

Thread Starter

ABoul

Joined Mar 30, 2009
15
the question can be found here:
http://www.MailFreeOnline.com/uploader/DEA77C24.jpg

V_in vs V_BE: i'm guessing the 2 are independant quantities, so i should see a flat line
V_in vs I_b: I_b is very close to 0 until V_in gets to 0.7, where it shoots up exponentially
V_in vs I_c: since I_c = BI_b, i expect this graph to have the same shape as the one for V_in vs I_b, but I_c flatlines when it reaches the saturation current of 10mA.
V_in vs V_L: no idea
V_in vs V_out: V_out = 10 - R_c*I_c, so i expect this to look the same as the reltionship between V_in and I_c but reflected along the y axis, scaled up and given an intercept of 10.


^ the above were ALL on-the-spot guesses. i have no idea if they are right. could anyone verify what i've done/correct me? thanks in advance.
 

t_n_k

Joined Mar 6, 2009
5,455
Remember Vin varies from a negative (-6V) to a positive (+6V) value.

Until the value of Vin equals the emitter base froward conduction value of 0.7V (nominally), then the only current that will flow in the negative range of Vin will be the the base-emitter junction reverse saturation current - effectively zero for the purpose of this exercise. Once Vi is equal to or greater than +0.7V, then the base-emitter junction will become forward biased and current will begin to flow into the base. This base current is limited by the 15kΩ series resistor, so it will increase according to the relationship

Ib=(Vin-Vbe)/15k

So Ib will not increase exponentially - rather more linearly in fact.

If you keep these things in mind you will hopefully be able to make another more successful attempt at the question.
:)
 

Thread Starter

ABoul

Joined Mar 30, 2009
15
Remember Vin varies from a negative (-6V) to a positive (+6V) value.

Until the value of Vin equals the emitter base froward conduction value of 0.7V (nominally), then the only current that will flow in the negative range of Vin will be the the base-emitter junction reverse saturation current - effectively zero for the purpose of this exercise. Once Vi is equal to or greater than +0.7V, then the base-emitter junction will become forward biased and current will begin to flow into the base. This base current is limited by the 15kΩ series resistor, so it will increase according to the relationship

Ib=(Vin-Vbe)/15k

So Ib will not increase exponentially - rather more linearly in fact.

If you keep these things in mind you will hopefully be able to make another more successful attempt at the question.
:)
thanks so much for the reply! one more question -- does I_b get limited to the satuation current divided by beta?
 

t_n_k

Joined Mar 6, 2009
5,455
No - once the transistor is in saturation the relationship no longer holds. In fact, Ib will continue to increase with increasing Vin, but since Ic cannot increase beyond the saturation value then increasing Ib has no effect. Since you are not given any data as to the saturation conditions - Vbe for instance - then you can probably assume the simple equation stated for Ib still holds true.
 
Top