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anonymous
 3 years ago
Rth helpp....Vin =27V , R 1 =5Ω , R 2 =7Ω , R 3 =12Ω , A=6 and B=0.1 .
anonymous
 3 years ago
Rth helpp....Vin =27V , R 1 =5Ω , R 2 =7Ω , R 3 =12Ω , A=6 and B=0.1 .

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0put Isc=0.25 in rth=vth/Isc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's just missing this one to me too .. Does anybody help me too ? Rth ... Vin=15V, R1=0.5Ω, R2=1.5Ω, R3=5Ω, A=2 and B=1.5.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cn u post formula of power in q2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Q2 is the same of HW3.. yeah.. for sure .. one minute !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0EXPLANATION: VS=5.0 VOH=4.5 VIH=4.0 VIL=1.5 VOL=1.0 The low noise margin is defined as VIL−VOL. For this problem, it is 0.5V. As an example, if an inverter outputs a valid voltage signal of 1.0V (VOL) to another inverter, then that signal can rise by 0.5V all the way to VIL, before it becomes an invalid logical 0 input to the second inverter. The high noise margin is defined as VOH−VIH. For this problem, it is 0.5V. Again, if an inverter outputs a valid voltage signal of 4.5V(VOH) to another inverter, then that signal can fall by 0.5V, all the way to VIH before it becomes an invalid logical 1 input to the second inverter. The width of the forbidden region is defined as VIH−VIL. For this problem, it is 2.5V. Valid inputs to our logic family are not allowed to fall between VIH and VIL. When VGS for the MOSFET is below VT, the MOSFET behaves like an open circuit, and iDS=0. That means no voltage drops over RPUI and the output voltage of our inverter is VS, so this case is fine with out static discipline, since the output voltage is above VOH. For VGS≥VT, the MOSFET is on and behaves like a resistor with resistance RON with our model. To find RPUI: VOL=VS(RON)RON+RPUI 1=5(5000.0)(5000.0+RPUI)→RPUI=20000.0Ω Similar to the inverter case, when either MOSFET is off, no current can flow through the pullup resistor because the MOSFETs and pullup resistor are in series. So VOUT for the NAND gate is simply VS when either MOSFET is off. When both MOSFETs are on, they behave like resistors with resistance RON. The two MOSFETs are in series, so we calculate: To find RPUA: VOL=VS(2RON)2RON+RPUA 1=5(10000.0)(10000.0+RPUA)→RPUA=40000.0Ω When both MOSFETs are off, VOUT for the NOR gate is simply VS because no current can flow through either MOSFET or RPUO. When one MOSFET is on, VOUT is: VS⋅(0.5)RON(0.5)RON+RPUO because the MOSFETs are in parallel with each other. For the cases with one MOSFET on or both MOSFETs on, VOUT must be at least as small as VOL. For the case with one MOSFET on, RPUO must be at least 20000.0 Ω. For the case with both MOSFETs on, RPUO must be at least 10000.0 Ω. So for our NOR gate to satisfy the static discipline, RPUO must be at least 20000.0 Ω. When the MOSFET in the inverter is off, no power is consumed because no current can flow as the MOSFET behaves like an open circuit. When it is on, VS must drop over RPUI and RON, so the power consumed is: (VS)2RPUI+RON Power consumed by inverter: 25V5000.0+20000.0Ω=0.001W When either MOSFET is off, no current flows through the MOSFETs or RPUA, so there is no power consumed. When both MOSFETs are on, VS must drop over RPUA and 2RON because the MOSFETs are in series with each other. The power consumed in this case is: (VS)2RPUA+2RON Power consumed by NAND: 25V2⋅5000.0+40000.0Ω=0.0005W When both MOSFETs are off, no curernt can flow, so no power is consumed. When one MOSFET is on, VS must drop over RPUO and RON, so the power consumed is: (VS)2RPUO+RON When both MOSFETs are on, VS must drop over RPUO and 12 RON because the MOSFETs are in parallel, so the power consumed is: (VS)2RPUO+0.5(RON) The maximum power is consumed by the NOR gate in the latter case: 25V(5000.0∥5000.0)+20000.0Ω=0.001111W

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i forget the formula..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The high noise margin is defined as VOH−VIH

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The width of the forbidden region is defined as VIH−VIL

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u plss solve this for me.. may b some calculation error vT=2, Ron =18000

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To find RPUI: VOL=VS(RON)/RON+RPUI

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah.. I'll solve .. one minute .. I don't know yet Thunda.. I am really want to know too

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0give me the values Vs etc etc ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey put Isc =0.75 in formula rth=vth/Isc

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and what about Vol ? what's value yours ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0short circuit that circuit then calculate the current in that circuit., now that current I=Isc+B*vb

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try this value 72000 ... because Vol=Vs*Ron/Ron*Rput then we have 1=5*18000/18000+Rput to be 1= 1 the variable Rput is 72000

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey for power of nand the formula is : Vs^2/(Rpul+2*Ron) but i got wrong answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0maybe your Rpua is not correct.. give me this value.. Rpua

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey i got.. thanx..:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ooh no problem.. did you get entire Q2 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you help with Rth in Q6 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Isc=0,75.. is it this value that you said ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's my last chance.. it's final check ..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so wait.. 1st i recheck

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0look my values are Vin=15V, R1=0.5Ω, R2=1.5Ω, R3=5Ω, A=2 and B=1.5.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah... THANK YOU SO MUCH !

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@kavi11: ur ans 1) 0.45 2)2.7 3)2.7 4)10.8

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@akash97 : how do u get that ? can you post the formula ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@xmark : I got using analysis der is no kind of formula dar I use directly

Qaisersatti
 3 years ago
Best ResponseYou've already chosen the best response.0Vin 27 v R1 5 ohm r2 = 7 Ohms R3 12 Ohms A=6 and B=0.1 QNO6 answers please please please ?

Qaisersatti
 3 years ago
Best ResponseYou've already chosen the best response.0time is running out for my friend....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0question 6 1)Vo* (R3(B*R1*R3)+R1+R2+A*R2)/(R3B*R1*R3)=VIN 2)i1=V0/(R3B*R1*R3) 3)vo = vth 4)Rth=Vth/Isc I=Isc+B*vb

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/508b8bc0e4b077c2ef2e9093
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