Electronics Questions

Join Date
Mar 2005
Location
Ontario
Posts
22
Hey guys, wolf here again,

I was wondering if someone could give me a hand and show me how to work out these answers, I have tried, but I cannot seem to figure them out, These question have nothing to do with any courses or school, this is something I like to do on the side!!

Anyhow I Hope someone will give help me,

Later

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Wolfy...

I did receive your PM. Since then I've been to Vegas for my son's wedding and then back home to find my computer in the classic "Crash & Burn" stance.

After getting my system restored I did begin working on your question... I then found myself distracted... sorry for neglecting you.

I also owe an explanation about a particular K-Map situation to BobO... I will post for both of you.

I will finish yours first. It will take a day or two to bring it together.

It will provide the algebraic solution (based on calculus) for any time=t.

Don't fret... it's easy!

I wonder if anyone understands the amount of effort required to post an explanation with graphics?
 
First I must sacrifice a pound of sweat to the exercise gods.

I will give someone else a chance to excel before I get back in about 3 hours. That is a challenge. This topic has been covered before. I still have the Mathcad worksheets from the last time we talked about this. Meanwhile I will let you guys struggle with this as I must struggle with my exercise.
 
Inductance and resistance
The time constant of an inductance L and a resistance R is equal to L / R, and represents the time to change the current in the inductance from zero to E / R at a constant rate of change of current E / L (which produces an induced voltage E across the inductance).

If a voltage E is applied to a series circuit comprising an inductance L and a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vL across the inductance and the magnetic linkage yL in the inductance are:
i = (E / R)(1 - e - tR / L)
vR = iR = E(1 - e - tR / L)
vL = E - vR = Ee - tR / L
yL = Li = (LE / R)(1 - e - tR / L)

If an inductance L carrying a current I is discharged through a resistance R, then after time t the current i, the voltage vR across the resistance, the voltage vL across the inductance and the magnetic linkage yL in the inductance are:
i = Ie - tR / L
vR = iR = IRe - tR / L
vL = vR = IRe - tR / L
yL = Li = LIe - tR / L

Rise Time and Fall Time
The rise time (or fall time) of a change is defined as the transition time between the 10% and 90% levels of the total change, so for an exponential rise (or fall) of time constant T, the rise time (or fall time) t10-90 is:
t10-90 = (ln0.9 - ln0.1)T » 2.2T

The half time of a change is defined as the transition time between the initial and 50% levels of the total change, so for an exponential change of time constant T, the half time t50 is :
t50 = (ln1.0 - ln0.5)T » 0.69T Note that for an exponential change of time constant T:
- over time interval T, a rise changes by a factor 1 - e -1 (» 0.63) of the remaining change,
- over time interval T, a fall changes by a factor e -1 (» 0.37) of the remaining change,
- after time interval 3T, less than 5% of the total change remains,
- after time interval 5T, less than 1% of the total change remains.

Courtesy of: http://www.bowest.com.au/library/formulae.html#13
 

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