Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Tue, 06 May 2008 08:04:14 +0800

WoAiMeiMei — Tue, 06 May 2008 08:04:14 +0800

thanks : D it works

Need help!! electrical stuff -diodes replied by eagle @ Mon, 05 May 2008 21:54:04 +0800

eagle — Mon, 05 May 2008 21:54:04 +0800

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.

Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Mon, 05 May 2008 20:38:08 +0800

WoAiMeiMei — Mon, 05 May 2008 20:38:08 +0800

ahh how do i do this one D:

(using same time intervals)

they seem to cross.. so i guess average? :s

Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Mon, 05 May 2008 20:30:02 +0800

WoAiMeiMei — Mon, 05 May 2008 20:30:02 +0800

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:

Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Mon, 05 May 2008 11:34:54 +0800

WoAiMeiMei — Mon, 05 May 2008 11:34:54 +0800

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

Need help!! electrical stuff -diodes replied by eagle @ Mon, 05 May 2008 11:15:12 +0800

eagle — Mon, 05 May 2008 11:15:12 +0800

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

Need help!! electrical stuff -diodes replied by eagle @ Mon, 05 May 2008 11:12:29 +0800

eagle — Mon, 05 May 2008 11:12:29 +0800

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)

Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Sun, 04 May 2008 17:10:20 +0800

WoAiMeiMei — Sun, 04 May 2008 17:10:20 +0800

bump >.<

the question keeps on friggin changing.. need the technique D:

Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Thu, 01 May 2008 06:24:28 +0800

WoAiMeiMei — Thu, 01 May 2008 06:24:28 +0800

will there be any exceptions? or it's good?

ps: will check it in the afternoon.. got test soon D:

Need help!! electrical stuff -diodes replied by justinfoo @ Thu, 01 May 2008 00:20:08 +0800

justinfoo — Thu, 01 May 2008 00:20:08 +0800

qns 1) jus use the equation ID = IS ( e^(VD/VT) - 1 ) where VT is kT/q

Need help!! electrical stuff -diodes replied by WoAiMeiMei @ Wed, 30 Apr 2008 18:40:52 +0800

WoAiMeiMei — Wed, 30 Apr 2008 18:40:52 +0800

diodes... argh D:

ty~

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