MIT 6.002 Circuits and Electronics, Spring 2007
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put Isc=0.25 in rth=vth/Isc
it'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.
cn u post formula of power in q2
Q2 is the same of HW3.. yeah.. for sure .. one minute !
EXPLANATION: 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
i forget the formula..
the first one
The high noise margin is defined as VOH−VIH
The width of the forbidden region is defined as VIH−VIL
can u plss solve this for me.. may b some calculation error vT=2, Ron =18000
To find RPUI: VOL=VS(RON)/RON+RPUI
Rth formula for Q6?
yeah.. I'll solve .. one minute .. I don't know yet Thunda.. I am really want to know too
I really want *
give me the values Vs etc etc ...
hey put Isc =0.75 in formula rth=vth/Isc
Thanks a bunch!!
and what about Vol ? what's value yours ?
short circuit that circuit then calculate the current in that circuit., now that current I=Isc+B*vb
try 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
hey for power of nand the formula is : Vs^2/(Rpul+2*Ron) but i got wrong answer
I'll help you.. 1 m
maybe your Rpua is not correct.. give me this value.. Rpua
hey i got.. thanx..:)
ooh no problem.. did you get entire Q2 ?
can you help with Rth in Q6 ?
Isc=0,75.. is it this value that you said ?
what is your vth?
vout= vth ... it's 3
it's my last chance.. it's final check ..
is it really 4 ?
so wait.. 1st i recheck
thank you so much
look my values are Vin=15V, R1=0.5Ω, R2=1.5Ω, R3=5Ω, A=2 and B=1.5.
yeah... THANK YOU SO MUCH !
@kavi11: ur ans 1) 0.45 2)2.7 3)2.7 4)10.8
@akash97 : how do u get that ? can you post the formula ?
@xmark : I got using analysis der is no kind of formula dar I use directly
Vin 27 v R1 5 ohm r2 = 7 Ohms R3 12 Ohms A=6 and B=0.1 QNO6 answers please please please ?
time is running out for my friend....
question 6 1)Vo* (R3-(B*R1*R3)+R1+R2+A*R2)/(R3-B*R1*R3)=VIN 2)i1=V0/(R3-B*R1*R3) 3)vo = vth 4)Rth=Vth/Isc I=Isc+B*vb