1) In the following circuit, VBE = 0.7 volt and ẞ= 100. Determine Vs, V6, V7, and branch currents IE, IC, IB. 22 ΚΩ www +3 V 1.6 ΚΩ -OV 2.2 ΚΩ -3 V

Using superposition, determine the voltage across the capacitor C₂ for the circuit in figure 1. + ECIEVED ~ fez0kHz Note : Ri 2ksz C₁6081F Xe c It C₂=3.3nF w = сос zof 3.9k2 IcbmA/180" F-20kHz

Determine the approximate value of VGG of Figure 52. VDD T15V 1.5kQ Q1 V, = 9.5V 1MQ = 8mA DS V =- 4.5V EVGG Figure 52. 1.453 V -1.2 V -7.547 V O -3.281 V

Calculations and Discussion: 1. Perform the theoretical calculations to determine the Q-point for both circuits and compare them with the measured values. Vcc =+12V Quantity Measured Rc 1k2 VB 0-6 28 volt Rp 1 MQ Vc 07-82 Volt 0. 628 U Volt V BEQ BC107 0-0000||3A A 0-00118 A 0.782 UoIt IBQ Vc IcQ VB V CEQ Bac Transistor DC Biasing Circuits Vec =+12V Quantity Measured Re VB 4.459 V R, 849V 3-860Y 10k2 Vc VE вс107 Vc Y.459V 0:00 351 A V BEQ VB 5.6k2 IcQ Ve R Ika V CEQ 8.49

In the figure below, since Vcc=12V, Vbb= 5V, ß=100, Rc= 3 kΩ , Rb= 200 kΩ , how many Ib=? milliamperes (take Vbe=0.6 V)

What should be Rl and what is Pmax in order to transfer maximum power from the source to the load.

Calculate Voltage across R,L,CCalculate the Amper across R,L,C

o +20V 3.3 KO Given that the FET shown in figure 2 has an Ipss- 12mA and a Vp= -6V. Calculate the following parameters: 1 KO 2 MO a) Vout T2 b) le for T3 Vout T3 c) IDQ -2V T1 d) Veso e) VD 1 KO 4 KO 200 0 -5V -10V Figare 2 LAM

imput e, ou put Using the following subsystem S x[n] yEn] consider the following global system What are the poles of this global system ? O single pole 1 O single pole -1 double pole 1 double pole -1 O poles 1 and -1 poles 1 + /2 and 1-/2 O poles j and -j 1+j O poles and 2 2 O None of the above QUESTION 3 Click Save and Submit to save and suhmit Cliak S.

Determine the DC currents (B. Ic and le) and de juncion voltages (VBE, VCE and Vca) mA. Ic ma, E mA VBE= V. VCE Vand VCB Use 2 decimal places. RB Use the following values: VBB = 3V RB = 4 kQ Rc = 166 0 Vcc = 19 V BDC = 70 %3D %3D VCC RC VBB