Question

For the differential input stage shown below, the differential input impedance is required to be 100k. Estimate the bias current in the differential amplifier to meet this specification. Hence calculate the value of R2 (in ko) required to set this bias current. The Early voltage of the NPN transistor is 150 V and that for the PNP is 50 V. Vcc (+9 V) 2N2907 Q5 2N2907 Vin (-) Q6 2N2907 2N2907 Q8 2N2222 09 2N2222 VEE (-9 V) assume lo = Iref (approx) (assumed in calculating the numerical answer) Assume for the NPN transistor, Bo = B = 255; rc = VA/Ic (VA is given in the questions where it is required). Use VBE=0.6V for all the transistors. Provide the numerical answers with THREE DECIMAL PLACES at least.

Answer 1

Solution:

Given that

$\beta _{0} =\beta=255$

$r_{ce}=\frac{V_A}{I_{C}},\, \, \, \, \, \, \, V_{BE}=0.6\,V$

differential input impedance =100 k$\Omega$

We have to find 1.Bias current

(Ibias)

2.$R_{2}$ to set this bias current

, Vec: (+90) &N 2907 ras aN2907 VOG. #N2907 2N2907 on y Nin (-) - R₂ (+) Vout 2N2222 OB ela Naleak VEE (9)

input resistance $=2(\beta +1)\left ( r_{e}+\frac{R_{CE}}{2} \right )$

$R_{CE}=\frac{V_{A}}{I_{C}}\, \, \, \, \, \, \, \, \, \,( given)$

$\alpha =r_{e}g_{m}\Rightarrow r_{e}=\frac{\alpha }{g_{m}}=\frac{\alpha V_{T}}{I_{C}}=\frac{\beta V_{T}}{(\beta +1)I_{C}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \left [ \because \left ( \alpha =\frac{\beta }{\beta +1} \right ) \right ]$

$R_{in}=2(\beta +1)\left [ \frac{\beta V_{T}}{(\beta +1)I_{C}}+\frac{V_A}{2I_{C}} \right ]$

$=\frac{2\beta V_{T}}{I_{C}}+\frac{2(\beta +1)V_{A}}{2I_{C}}$

$R_{in}=\frac{2\beta V_{T}}{I_{C}}+\frac{(\beta +1)V_{A}}{I_{C}}$

$I_{C}=\frac{2\beta V_{T}+\beta +1)V_{A}}{R_{in}}$

$=\frac{2(255)(0.025)+(256)(50)}{100\,k\Omega }$

$I_{C}=128.1275\,mA$

Now,

$2(I_{C})R_{2}=9-06+9\, \, \, \, \, \, \, \, \, \, \, \, \, \, \left [ \textrm{from Q4 transistor} \right ]$

$2R_{2}=\frac{17.4}{128.1275}\,mA=135.802\, \Omega$

$\therefore R_{2}=67.901\, \Omega \simeq 68\,\Omega$

NOTE : Hope you understand the solution, if you have any doubts comment below, kindly upvote. Thank you