# 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.

#### The two transistors

The second transistor is emitter repeater (or equivalent 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)

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 $VCE$

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