Question

4

4. In figure 4, the ac generator has an internal resistance (Ra) of 40002. What is the output voltage (Vout) ifs-300? Draw the DC equivalent circuit. Fiure 4 О +10 V Rc R1 10 kQ out Ru 1 mV RE 1 ko 2.2 kQ

Answer 1

Dc analysis

DC equivalent circuit rightarrow Thevenin Equivalent AC equivalent circuit for AC analysis DC analysis This circuit can be changed to equivalent thevenin equivalent circuit shown in fig a Thevenin (open circuit) voltage = V_CC * R2/R1 + R2 = 10 * 2.2/10 + 2.2 = 1.8 V Thevenin resistance = R1 parallel R2 = 10 * 2.2/10 + 2.2 = 1.8 k ohm We know beta = Ic/Ib, Ic = 300lb And Ie = Ib + Ic Ie = 301 lb Applying kvl -1.80 + 1.80lb + 0.7 + 301lb = 0 lb = 3.64 uA Ie = 301 * 3.64 uA = 1.09 mA Using ac analysis, rbe = (hfe + 1) [0.025/Ie] = 301 * [0.025/1.09 * 10^-3] = 6863.1 ohm Ib = [Vin/rbe] = [10^-3/6863.1] = 0.145 uA Ic = hfe * Ib = 300 * 0.145 uA = 0.043 mA Output voltage vo = ic(Rc parallel Rl) Vo = [0.043/1000] * [3.6 * 10

This circuit can be changed to equivalent thevenin equivalent circuit shown in fig a

Thevenin (open circuit) voltage= $\frac{Vcc*R2}{R1+R2}$

= $\frac{10*2.2}{10+2.2}$

= 1.8 V

Thevenin resistance = R1|| R2

= $\frac{10*2.2}{10+2.2}$

= 1.8 k$\Omega$​​​​​​

We know  $\beta =\frac{Ic}{Ib}$ , Ic= 300Ib

And Ie=Ib+Ic

Ie= 301 Ib

Applying kvl

-1.80+1.80Ib+0.7+301Ib=0

Ib=3.64uA

Ie= 301*3.64uA=1.09mA

Using ac analysis,

rbe= (hfe+1)$[\frac{0.025}{Ie}]$

=301*$[\frac{0.025}{1.09*10^{-3}}]$

= 6863.1 ohm

Ib = $[\frac{Vin}{rbe}]$ = $[\frac{10^{-3}}{6863.1}]$ = 0.145uA

Ic= hfe*Ib= 300*0.145uA=0.043mA

Output voltage vo= ic(Rc||RL)

Vo= $[\frac{0.043}{1000}]$ *$[\frac{3.6*10}{3.6+10}]*1000$

Vo= 0.1157V (answer)