CALCULATIONS :-

i) Consider the graph for INPUT CHARACTERISTICS

ri = [delta(VEB) / delta(IE)] at VCB = constant

a) ri = [delta(VEB) / delta(IE)] at VCB = 0 Volts
= (VEB corresponding to point E - VEB corresponding to point F) / (IE corresponding to point E - IE corresponding to point F)
= (0.65 - 0.6) Volts / (20 - 2.5) mA
= 0.05 / 17.5 x 10-3
= 2.857 ohms.

b) ri = [delta(VEB) / delta(IE)] at VCB = 4 Volts
= (VEB corresponding to point C - VEB corresponding to point D) / (IE corresponding to point C - IE corresponding to point D)
= (0.6 - 0.55) Volts / (11 - 1.75) mA
= 0.05 / 9.25 x 10-3
= 5.405 ohms.

c) ri = [delta(VEB) / delta(IE)] at VCB = 8 Volts
= (VEB corresponding to point A - VEB corresponding to point B) / (IE corresponding to point A - IE corresponding to point B)
= (0.6 - 0.55) Volts / (20 - 5) mA
= 0.05 / 15 x 10-3
= 3.333 ohms.

So, mean value of ri = 3.865 ohms.


ii) Consider the graph for OUTPUT CHARACTERISTICS

ro = [delta(VCB) / delta(IC)] at IE = constant

Here we see that for any change in VCB, the corresponding change in IC is zero.
So, ro = Finite number / zero = infinity
So, ro = infinite.


iii) & iv) Consider the graph for TRANSFER CHARACTERISTICS

a) alphaDC = (IC) / (IE)

Choose point B on the graph
So, alphaDC = (IC corresponding to point B) / (IE corresponding to point B)
= 10 mA / 10 mA = 1

b) alphaAC = [delta(IC) / delta(IE)] at VCB = constant

Choose points A and C on the graph
Then, alphaAC = (IC corresponding to point A - IC corresponding to point C) / (IE corresponding to point A - IE corresponding to point C)] at VCB = 50 mV
= (15 - 10) mA / (15 - 10) mA = 1