Two Stage Amplifier

The amplifier

Find the values of the first stage components from the two stage amplifier for the conditions: Overall current amplification $A_i=-200$, overall voltage amplification $A_v=-20$, $R_{c1}=R_{in2}$ d) Q point at the middle of the work line. Sketch the dc and ac work lines for the stage two of this amplifier.

amplifier in two stages

The two transistors

The second transistor is emitter repeater (or equivalent common collector).

Common Collector:

$A_{2u} =1$ and $A_{2i} =\beta+1 =\beta$

input impedance of T2 is  $r(in) =\beta*Re =100*400 =40 K$  and therefore $Rc1 =40 K$

The first transistor is common emitter (with degenerated emitter)

Common Emitter

From the DC analysis we have $A2u = -R1c/R1e$

$Au =A1u *A2u$ therefore   $R1c/R1e =20$

$R1e =40 k/20 =2 k$

Taking $V c e(sat) = 0.2 V$ we draw the DC load line between $I c(max) = (10V -V c e(sat))/ 40 K =9.8/40 k=0.254 mA$   with a slope of $1/40 K$ up to $V c e(max) =10 V$

The Q point is in the middle this means $I c =0.254/2 =0.1225 mA$

(From AC analysis we have $v1(out) =ic1*(Rc1 || Rin2) = ic1*20 K$ so the the AC work line will pass through Q point above and have a slope of 1/20 K)

The amplification

The DC current amplification is $A1i =-\beta$, therefore current through the base divider is

$Ibb = Ic/A1i =0.1225 mA/100 =1.225 \mu A$

Base voltage is 0.7 V. Total DC source voltage is $V d c =10 V$

$R11/(R11+R21) =0.7/10$

$R11/R21 =0.7/9.3$   or equivalent $R21/R11 =13.28$

$V d c/(R11+R21) =I b b$

$R11 +R21 =10/(1.225 \mu) = 8.16 M\Omega$

$R11 =571 K$ and $R21 =7.59 M\Omega$

The load lines

We do the computations for the load lines of second transistor of this amplifier.

$V c e(sat) =0.2 V$

In DC, $Re2 = 400 ohm$ so the DC load line will have a slope of (1/400) , and will pass through $I max =(10-0.2)/400 =24.5 mA$    and $VCE max =10 V$.

The Q point is at the middle. $Q = (5 V, 12.25 mA)$

In AC, the emitter resistance is $R =Re|| R L =400/2 =200 ohm$ so the load line will pass through point Q and have a slope of 1/200 .

The working lines are drawn below. From the figure the maximum voltage swing is the intersection point of AC load line with $VCE$

$V_{swing}= 5+(1/200)*12.5 mA =7.45 V$

The work line