NPN transistor numerical

Thread Starter

khatus

Joined Jul 2, 2018
95
Same problem with an additional base resistor

g


I think the steps will be




STEP-1:


Vx - VBE = IB×RE

STEP-2:

IB = Ira - Irb

STEP-3:

Vx = VIn×(RB/RB+RA)

Please tell me i am correct or not??
 

ebeowulf17

Joined Aug 12, 2014
3,307
I guess we can assume this is homework?

Start with something easy: How is Vb related to Vin?
This site is humbling sometimes. I can't come up with any way to determine Vin based on the information provided. If that's the easy part, I wouldn't stand a chance. Good thing this isn't my homework assignment!
 

wayneh

Joined Sep 9, 2010
17,498
Well maybe I’m all wet but I assume Vin is the unknown independent variable and can be any value. All voltages in the circuit will be expressed as a function of Vin.
 

ebeowulf17

Joined Aug 12, 2014
3,307
Well maybe I’m all wet but I assume Vin is the unknown independent variable and can be any value. All voltages in the circuit will be expressed as a function of Vin.
Ahhhh, I see! I was trying to come up with a number, not a formula. You're right - we can certainly build formulas as a function of Vin. I have no idea if that's what the thread starter needs, but it certainly helps my sanity a little!
 

wayneh

Joined Sep 9, 2010
17,498
Please tell me i am correct or not??
Looks good so far. Note that the equations are not continuous though as the base voltage reaches 0.7V. The base current is never negative.

And without Re, there is very little resistance base-to-emitter. It's essentially a short for most purposes.
 
Last edited:

WBahn

Joined Mar 31, 2012
30,082
Same problem with an additional base resistor

g


I think the steps will be




STEP-1:


Vx - VBE = IB×RE

STEP-2:

IB = Ira - Irb

STEP-3:

Vx = VIn×(RB/RB+RA)

Please tell me i am correct or not??
You are not correct.

In step three you make two common mistakes. The first is purely arithmetic in that you fail to recognize that

RB/RB+RA = 1 + RA

The second is that you are blindly applying the voltage divider formula without stopping to ask whether you have a voltage divider. You don't. The voltage divider formula is based on the assumption that the current flowing in both resistors is the same. In this case you have Ib which flows in Ra but not in Rb. If Ib is much, much smaller than the other currents, the voltage divider formula might give acceptable results, but you need some basis upon which to assert that claim before you go using it.
 
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