I'm trying to determine the **Linear region** of the transistor by setting an upper and lower limit for the resistor on the lower left hand corner. I've already fixed the lower limit and tried to find a value for upper limit by using: $$I_B+I_C=I_E$$

$$R_EI_E+R_CI_C = 15$$ $$\frac{15R_2}{R_2+10^5}-B_{BE} – I_ER_E =0$$ Clearly these set of equations ain't enough. Is there any way out ? or i'm fundamentally wrong?

# Electrical – How to determine the value of the variable resistor to land on saturation region

amplifierdclinearresistorstransistors

## Best Answer

## added

Another explicit method is choose Vc= 15/2 then add Ve= 1.5 and add 0.5V for saturation margin at low current.

Thus Vc=8V and Ic= (15-8)V/2.21k=3.17mA and Ve=3.17m*221R = 700mV and thus Vb=1.4V from which you can compute R2 = 10.3k and raising R2 to compensate for the Re loading effect on input impedance at base.

Using hFE=100 we get Zin=100*221R = 22k //R2= 10.3k or (1/22)-(1/10.3)=1/R2

Thus

R2 = 19.3k//22k(Zin) = 10.3k for Vb=1.4V, Vc=8V, Ve=0.7V and 14Vpp swing max with 1.4Vpp in.(final answer)

For an ideal transistor say with the current gain at 100, and an H biased voltage amplifier with a gain of 10 you expect The resistors to drop 10+1 times the emitter resistor V(Re), value in voltage to match Vcc=15V with Vce=0. Then if we assume Vbe=0.7V we can choose Vb for linear symmetry.

We expect 15V supply with an AC gain of 10 will lose some range for Re thus a good bias point for R2 results in a t least a 13Vpp swing with a gain of 10 for an input of 1.3Vpp . So intuitively you can say Vb must be 0.7{Vbe}+1.3V/2 or about 1.4V thus R2 will be just under 10% of 100k.

This is how we do it on our heads in about a minute after years of practice and success.

But poking on this iPod takes forever!

Vb >1/(1+Av)*Vcc+Vbe = R2/(R1+R2)*Vcc.

So we want Vb such that Vce=0 but Vc will achieve a full swing on both peaks at the same amplitude for a gain of 10.

The following would DC bias Vce into saturation but you would not want this for full AC voltage swing, rather you want Vc to be mid range between Vcc and Ve.

15V/11+0.7=1.1V =R2/(100k+R2)*15V.

or inverting both sides.

100k/R2 +1 = 15V/1.1=13.6 =12.6+1

So R2=100k/12.6 = 8k approx.

The equations can easily be simplified by you to be a function of Rc/Re=Av and Vcc.

advanced —

In reality hFE drops rapidly to 10% max with Vce starting at Vce<=1V for low current and 2V for high current so we avoid Vce<2 for good linearity in sine waves such that Vp+=|Vp-| <<1% for low distortion.