Solutions of Homework problems

Resistive circuits Problem 1 Use KVL and Ohms law to compute voltages v a and v b v2v v1v1 From Ohms law: v 1 =8k  i 1 =8[V] v 2 =2k  i 2 =-2[V] Form KVL: v a =5[V]-v 2 =7[V] v b =15[V]-v 1 -v a =0[V]

Resistive circuits Problem 2 Write equations to compute voltages v 1 and v 2, next find the current value of i 1 From KCL: 50 mA=v 1 /40+(v 1 -v 2 )/40 and 100 mA=v 2 /80+(v 2 -v 1 )/40 Multiply first equation by 40: 2=v 1 +v 1 -v 2 =2v 1 -v 2 From second equation: 8=v 2 +2(v 2 -v 1 )=3v 2 -2v 1 add both sides: 10=2v 2 => v 2 =5 [V], v 1 =1+v 2 /2=3.5[V] i 1 = (v 1 -v 2 )/40=-1.5/40=37.5 [mA] 50 mA 100 mA i1i1 i1i1 v2v2 v1v1

Thevenin & Norton Problem 3: Find Thevenin and Norton equivalent circuit for the network shown. I1I1 N2N2 I2I2 vtvt N1N1 From KVL

Thevenin & Norton I1I1 N2N2 I2I2 I sc N1N1 From KVL

Thevenin & Norton Note: Negative v t indicates that the polarity is reversed and as a result this circuit has a negative resistance. + _ V t =-6 V R Th =-1.33Ω A B Thevenin Equivalent I n =4.5 A R Th =-1.33Ω A B Norton Equivalent R Th =v t /I sc =-1.33Ω

Problem 4: Find the current i and the voltage v across LED diode in the circuit shown on Fig. a) assuming that the diode characteristic is shown on Fig. b). Draw load line. Intersection of load line and diode characteristic is the i and v across LED diode: v ≈ 1.02 V and i ≈ 7.5 mA.

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V. (a) Diode is on for v > 0 and R=2kΩ. +v_+v_ 2kΩ i In a series connection voltages are added for each constant current

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V. (a) +v_+v_ 2kΩ i Resulting characteristics

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V. (b) Due to the presence of the 5V supply the diode conducts only for v > 5, R = 1kΩ +v_+v_ 1kΩ i +_+_ 5V First combine diode and resistance then add the voltage source

(c) Diode B is on for v > 0 and R=1kΩ. Diode A is on for v < 0 and R=2kΩ. +v_+v_ 2kΩ i 1kΩ AB Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

(d) Diode D is on for v > 0 and R=1kΩ. Diode C is on for v < 0 and R=0Ω. +v_+v_ i 1kΩ C D Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (v o versus v in ) for the circuit shown below. Case I: v in > 0 Both diodes are on, and act as short circuits. The equivalent circuit is shown here. v o = v in v in +_+_ +vo_+vo_ 1kΩ 3V v in +_+_ +vo_+vo_ 1kΩ

Case II: v in < 0 Both diodes are reverse biased and v o is the sum of the voltage drops across Zener diode and 1kΩ resistor. v in +_+_ +vo_+vo_ 1kΩ 3V v in +_+_ +vo_+vo_ 1kΩ 3V Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (v o versus v in ) for the circuit shown below.

Case II: v in < 0 Both diodes are reverse biased and v o is the sum of the voltage drops across Zener diode and 1kΩ resistor. v in +_+_ +vo_+vo_ 1kΩ 3V v in +_+_ +vo_+vo_ 1kΩ Case IIa: -3V < v in < 0 v o = v in, because the current through Zener diode is zero, all negative voltage drop is across the Zener diode. Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (v o versus v in ) for the circuit shown below.

v in +_+_ +vo_+vo_ 1kΩ Case IIb: v in < -3V Excess voltage below -3V is dropped across the two resistors (1k  and 1k  ), with v o = (1/2)*(v in +3)-3= v in /2-1.5 [V]. v in vovo V +_+_ Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (v o versus v in ) for the circuit shown below.

IaIa 4V - + 5V - + (a)

IaIa 4V - + 5V - + (a) S D G

IbIb 1V - + 3V - + (b) D S G

IcIc 4V + - 5V - + (c) G D S c

IdId 3V - + 1V - + (d) G S D