COMPUTING

Chapter 3 Bipolar Junction Transistorsi1 = 5.10e-007 ATherefore,v2 = -1.02e-001 Vi 1= 0.51 μA (3)v 2=– 102 mV(4)Next, we use (1) and (2) to find the new values of v 1 and i 2= 1.2 × 10 3× 0.51 × 10 – 6+ 2× 10 – 4× (–102 × 10 – 3) = 0.592 mi 2 = 50 × 0.51 × 10 – 6× 50 × 10 – 6× (–102 × 10 – 3) = 20.4 μAThe voltage gain isvG 2V = ---- = ------------------------–102 mV= – 172.30.592 mVv 1and the minus (−) sign indicates that the output voltage ininput.The current gain isiG 2I = --- =20.4--------------------μA= 400.51 μAand the output current is in phase with the input current.i 1180°out−of−phase with the3−114Electronic Devices and Amplifier Circuits with MATLAB Computing, Second EditionCopyright © Orchard Publications