Need help!! electrical stuff -diodes
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diodes... argh D:
ty~
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qns 1) jus use the equation ID = IS ( e^(VD/VT) - 1 ) where VT is kT/q
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will there be any exceptions? or it's good?
ps: will check it in the afternoon.. got test soon D:
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bump >.<
the question keeps on friggin changing.. need the technique D:
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I presume Q1 has been done...
Q3)
a)
Vout = Vz = 8V
IR2 = 8/880 = 9.1 mA
Iz is just positive, can be assumed to be zero/neglible
therefore, current through Vs is 9.1 mA (Kirchoff's current law)
Then voltage drop across R1 = 0.0091 * 750 = 6.82V
Voltage drop across diode = 0.64 V
Thus, Vs = sum of voltage drops = 15.46 V (Kirchoff's voltage law)
b) Vz is still 8V
So Vout is still 8V, and IR2 = 9.1 mA
So voltage drop across R1 = 41 - 8V - 0.64 V = 32.36V (Kirchoff's voltage law)
so IR1 = 32.36 / 750 = 43.147 mA
Thus, Iz = IR1 - IR2 = 34 mA (Kirchoff's current law) -
Q2)
0 to 5 sec
Both off, so Vo = 0V
5 to 8 sec
Only V1 is on, so voltage drop of 0.6V, means Vo = 3.4V
8 to 13 sec
Can use law of superposition
From V1, it is 3.4V
From V2, it is 4.4V
So add up, you get 7.8V
13 to 15 sec
Both off, so Vo = 0V -
thanks, will go home and try it out cos the questions always change. 1 question for 1) though.. the negative values... i plugged that in the equation but i got a very small number. should it be 0?
i notice the two negative values give u the same answer
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edit: ignore above..
i managed to solve it on the second round :D
now i understand how to do them great....! however...
the 3rd part in the diode graph is wrong... i tried it twice n two different graphs and the superposition method doesnt apply... D:
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ahh how do i do this one D:
(using same time intervals)
they seem to cross.. so i guess average? :s
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no not superposition... mistake there...
I think the diode for the smaller voltage will not turn on at all. Means you consider only the bigger voltage source.
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thanks : D it works