Simplifying a BJT circuit.

Thread Starter

Panterulez

Joined Mar 25, 2009
7
I am kind of weak in simplifying circuit using Thevenin equivalent. Can someone explain to me how I could go about simplifying the base portion of the transistor so that I could evaluate the circuit for it's Q points.


 

PRS

Joined Aug 24, 2008
989
Thevenin's theorem says you can simplify the analysis of a circuit by finding the equivalent resistance and voltage looking into the base. I'll call the resistor hooked to +15V R1 and the resistor hooked to -15 volts R2. Notice that with respect to the base they are in parallel. Therefore R(eq)=R1*R2/(R1+R2)=you do the math.

These resistors are looking at a total of 30 volts, so that's V(eq). I can't draw as I have no software for it, but think of a single Req looking into the base with Veq on the other end of it.

To find the Q point you need to find the voltage at the base and to do this you write a KVL loop around Veq, Req, Vbe and Re. If you're still confused I'll give you more help, but as this is homework, you're supposed to figure it out. :)
 

steveb

Joined Jul 3, 2008
2,436
The equivalent voltage needs to consider the voltage division with R1 and R2 under a no load condition.

i.e. 30V*(R1/(R1+R2))=15V relative to the -15V supply in this case.

So you have a 15 V supply with a 100K source resistance driving your base circuit, relative to the -15V as a ground.

A good way to think about the equivalent voltage is to ask what would a voltmeter read if no load (i.e. no transistor base) was hooked up to the circuit.
 

scythe

Joined Mar 23, 2009
49
To find the Q points, you need to know your gain (β or α), and at least one of your currents(Ie, Ib or Ic). To find those, do what PRS says. That should help you find Ib, which in turn will allow you to find Ie (remember that Ie is (β+1)Ib. and from there I think you got it.
 
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