To solve simulataneous equations, only managed to get one equation

[wanna_know123]

Homework Statement

Given beta = 100, Vce= 4V, Ic=1.25mA, Vbe =0.7V

Homework Equations

IbRth = Vth - Vbe - Ve

Rth= R1R2/(R1+R2)
Vth = R2*Vcc/(R1+R2)

The Attempt at a Solution

In order to find the two unknown (R1 and R2) using simultaneous equations, I need another equation right? But I have no idea how to get another one, hope you guys can let me know, thank you so much.

 

[tedbradly]

Homework Statement

Given beta = 100, Vce= 4V, Ic=1.25mA, Vbe =0.7V

Homework Equations

IbRth = Vth - Vbe - Ve

Rth= R1R2/(R1+R2)
Vth = R2*Vcc/(R1+R2)

The Attempt at a Solution

In order to find the two unknown (R1 and R2) using simultaneous equations, I need another equation right? But I have no idea how to get another one, hope you guys can let me know, thank you so much.

The ratio, not the individual resistor values, is mostly all you need. So choose a reasonable R_1 or R_2 and solve for the remainder.

Things to consider when selecting a 'reasonable' resistor: you don't want that DC voltage source needlessly wasting tons of power on those biasing resistors, so choose the biggest resistor value that works. What defines 'working'? The ability for enough current to reach the base does.

Ideally (beta = inf), the total current down that way would be v/(R_1 + R_2), and even for beta in the hundreds, the current down that branch is still roughly equal to that. So just make sure v_+/ (R_1+R_2) is much, much greater than what I_B you may expect to see. That way, you will definitely have enough current to fuel your base. Basically, don't use megaohm resistors, resulting in microamp currents, when the base wants to draw microamp currents itself.