ECEN325SolutionHW4

ECEN325SolutionHW4 - ELEN 325: Electronics Homework #4 3.3:...

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Unformatted text preview: ELEN 325: Electronics Homework #4 3.3: 51 Answer: M V. OWWWWI a) Here we make plausible assumptions: ' + VD1 - F irst: Assuming that both D1 and D2 are conducting, this +3 V WMWMW y $0653 4 V I)”. : : + V32 7 2 m Resulting: VD1= —3 V; v.32: —1 V; The result is not possible to make both diodes conducting, original assumption is failed. Second: Assuming that only D1 conducting, this makes: V: 1V Resulting: VD2= 1 V; The result is possible to make D2 conducting, original “3V assumption is failed. 7/ 7 Third: Assuming that only D] conducting, which makes: V: 3 V Resulting: VD]: —2 V; the result is the same as we assumed. Using the result of assumption 3, we can have V is equal to 3 V, the current is: I _ 3V —(—3V) _ = 3mA L/ 2000 Ohm b) Using the same assumption as problem a), we can find that D1 is conducting and D2 is cutoff. The voltage at node V is equal to 1 V, the current is: 3V—1V 1=—————=1 A 2000 Ohm m ’/ 3.4 AnsWer: Assume the input signal as following wave form: ll The output wave form is shown as the above figures, in which onl the conducting voltage can be seen at the output node. If we neglect the voltage that drop across the diodes, negative peak value is 0 V and the y the voltage that is above the positive peak value is 10 V, the frequency is the same as input 1 KHz. b) If we neglect the voltage drop across the diodes, the positive peak value is O V, the n egative peak is the same as input 1 KHz. / / g voltage, D1 conducting and D2cutoff; when input signal is below the minus conducting voltage, D2 conducting and D1 cutoff. If we neglect the voltage drop across the diode, the positive peak value is 10 V, the negative peak value is -10 V and the frequency is the same as input 1 KHz When input signal is above the conducting voltage, D1 conducts current, which makes voltage at output equal to the voltage drop diodes, the positive peak value is 10 V, Th dame as input at l KHz. the across Dl. If we neglect the voltage drop across the e negative peak value is 0 V and the frequency is the When VI<0 V, D] is conducting and the circuit becomes a voltage divider. If we neglect the voltage drop across the diodes, the negative peak is: lKohm .(_10V) : _5V lKohm +1K0hm So the positive peak is 10 V, negative peak is -5 V and the output frequency is l KHz. v j) W“ When Vi>0 V, the output follows the input signal as D1 is conducting; when Vi<0V, D1 is cut off and the circuit becomes a voltage divider. The output is the same as the problem (g), positive peak is 10 V, negative is —5 V and the output frequency is l KHz. k) m. g” / When Vi>0 V, D] is cutoff and D2 is conducting. The output signal becomes: lkohm -1mA = 1V ; when Vi<0, D1 is conducting and D2 is cutoff. The output signal becomes Vo=Vi+lV. ‘ If we neglect the voltage drop across the diodes, the positive peak value is 1 V, The negative peak value is -9 V and the frequency is the dame as input at l KHz. 3.5 Answer: 9 f it,“ n ,I n, I I I I Vx 100 mA ---- Here, we assume that the node voltage between D1 and D2 is Vx. Also we make a assumption that there is a continuous voltage drop between the input node and output node, which means that when Vi is larger than Vo, we have Vi>Vx>Vo, when Vi is smaller than Vo we have Vi<Vx<Vo. Based on the assumption above, we can get: When Vi<4.5 , D1 conducts and D2 is cutoff, so i3 is zero; When Vi>4.5 V, D2 conducts and D1 is cutoff, thus disconnecting the input Vi and all the current flows through the battery. Assume the conduction angel is (9 , we get: 10Sinl9 = 4.5V I 4.5 +6=S‘4—— m (10) The conduction angel is 7: —2 6 The fraction of cycle that iB=100 mA is given by: Z = 7’ _ 26 = 0.35 2% We can get: . 1 . 1 13m, = —T- [13m =;(100 . 0.35 . T) = 35mA T If Vi is reduced by ten percent, the peak value of i3 remains the same i Bpeak =100mA But the fraction of the cycle for conduction changes It — 23in“ (12—5) _fl—26_ l Z 2% 2” 3 1 1 Th :' =—100-—-T =33.3mA us 13an T( 3 ) 3.54 Answer: Rs I \ 57; To find the small signal response, Vo, open the dc current source I and short the capacitors C1 and C2. Also replace the diode with its small signal resistance, we can get: _ nVT I rd So: nVT V” = rd .VS :4.“ :Vv_n_VT_ rd +Rs "VT ‘ nVT + [R + RS I The equation is proved. We set n=2, Vs=10 mV Rs=1K ohm and Vt=25 mV: l/U : . 0.05+1000[ So, V” = 0.476mV , when I=lmA V0 = 3.333mV , when I=0.1mA V = 9.804mV , when I=1uA r) For V0 =1VY = VY x—fl— 2 ' ‘ 0.05 + 10001 =50 uA *3.92 Answer: 73) / W a) For virtual ground, V- always equal to zero. Make plausible assumptions at first: F irst: Assuming that both D1 and D2 are conducting, this makes: Vo=-1.4 V Resulting: 1 —1.4 —1.4 + R R DR1+ DR2 . 2.4 — 1.4 ’ —-—) — : -————— R DR1+ DR2 The result is impossible, since the resistance of diodes can not have a negative value. Second: Assuming that only D1 conducting, this makes: VA: -0.7V; The voltage at output is 0V The result is impossible to make D2 conducting, original assumption is correct. Third: Assuming that only D2 conducting, the voltage at output is -1 V this make: VA=-1V+0.7V=-0.3 V, this can not make D1 conduct, original assumption is also correct. Fourth: Assuming that none of diodes conducting, there is no path to conduct the current. The assumption is incorrect. So we have Vo=0 V, VA=-0.7 V or Vo=-1V, VA=—0.3 V. . M b) Usmg the same assumption as problem a), we have D1 conducts and D2 cutoff or D2 conducts and D1 cutoff: so Vo=0 V VA=-0.7 V or Vo=—1V, VA=-1.3 V. c) Using assumption, we find that D2 conducting and D1 cutoff or D1 conducts and D2 cutoff so Vo=1 V, VA=1.7 V or Vo=0V, VA=-O.7 d) Using assumption, we find that D2 conducting and D1 cutoff or D1 conducts and D2 cutoff so Vo=2 V, VA=2.7 V or Vo=0V, VA=—0.7 ‘4" '"l. wo {Lisump ions are correct, I am not sure about the resuit. ~~ 3.105 Answer: a) 10 V L/ -10 V b) 20 V /* When Vi=—1O V, Vo=0V, The charge in capacitor is C-AV = C-(0——10)=10C 0 V When Vi=10 V, The charge in capacitor is unchanged, this make: C - (Va-10) = 10C, so the V0 is 20 V. c) 0 V / The same solution method as b) —20 V d) 0 V .7 «r: The same solution method as b) X 0" j —20 V \A \/ W L22 Wrzav/ e) 10 V -10V f) Here there are two different time constants involved. V t V1 During TI: V0 2 V1-e7E V1 When t=T1, two resistors are parallel we get: 'V2I . V2 “@ T : : E V12=V1-e 2'zV1‘(1——) ' T1 ' T2 ' RC __1_ During T]: M = |v2| . e When t=T2, only one resister in the bottom: Moreover, we have: T V12+ |V2| = 20 V1 - (1”— E) + |V2| = 20 -) |V22|+V1= 20 |V2|-(1+:—:)+V1=20 So we can get: |V2| =6.67 V and V1=13.33 V g) 18 V -2V h) The same solution way of (f). 13.33V .671 ...
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This note was uploaded on 04/29/2008 for the course ECEN 325 taught by Professor Silva during the Spring '08 term at Texas A&M.

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ECEN325SolutionHW4 - ELEN 325: Electronics Homework #4 3.3:...

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