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i need to work out the values for VGS and ID but both are in the same equation for each other and i dont know how to solve it
VGS = 2.2785 - (ID x 1.5k ohms)
ID = 8mA [ 1 - VGS / -4 ]squared
i dont know how to do it.. if anyone can enlighten me then please do..
my prof goes to do this
"As you can see there are two equations relating VGS and ID. Thus substitute from one equation preferentially for VGS in terms of ID and solve the simultaneous equations to get ID."
but i have no idea wat he means im running outta ideas to solve it..
just the main important thing is how one solved it.. so i can learn from it
Cheers Skye
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We know that VGS = 2.2785 - (ID x 1.5k ohms). A good way to see it is that VGS is "a name for" 2.2785 - (ID x 1.5k ohms).
So in ID = 8mA [ 1 - VGS / -4 ]squared, VGS is really just "a name for" 2.2785 - (ID x 1.5k ohms). So we can replace VGS in this equation with 2.2785 - (ID x 1.5k ohms), and get:
ID = 8mA [ 1 - (2.2785 - (ID x 1.5k ohms)) / -4 ]squared
Now we have an equation in the form of:
Where x is ID. So just use algebra to move things around, add, subtract, multiple, divide, and take roots. If you need any help with that part, don't hesitate to ask.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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cheers ricky my maths is poor i can understand what you've done by putting vgs into the ID equation but i cant solve x, im a network engineerer.. im not asking you to solve this equation but if you can provide step by step details on how you solved in this case id (x), then i can follow the steps you've taken and learn from you and then practice practice till i perfect.. the technique which you have showed me..
Jas
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The first thing to do is to simplify the inside:
Make everything into a decimal by dividing the fraction. Hope you have a calculator
Now multiply the ^2 part out:
Again, make sure you simplify the inside first. Now foil this out, and you come up with a quadractic equation. Subtract everything to one side so you get:
ax^2 + bx + c = 0
Where a, b, and c are numbers. Remember, they can be negative, so you may have minus signs instead of plus. Now use the quadratic equation to solve for x.
If you don't know foiling or the quadratic equation, or anything else for that matter, just ask.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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hm i think you got it wrong there, 1,5x/2. it should be 1,5x/4 right?
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i agree with kurre i think it should be 1.5/4
but the 1.5= 1.5x10^3 the 1.5kohms is 1500ohms
ricky u might think i am a dofus but i can do the basic quadratics but this is abit complex for me if you can show me how you to do it step by step because ive understood what you've done so far, but im lost on the simplifing do you basically mean divide all 2.2... by 4? and the 1.5x10^3x divided by 2 or 4 and then add the 1 and then finally multiply it all by the 0.008 (8mA = 0.008)
and the quadratics is way over my head..
Sorry about this ricky but im learning from all the posts youve replied to.
Skye
Yeah, true that, and the prefixes are important. Here's how I'd do it:
Split up the same way Ricky did and introduce a variable A for simplicity
Now we have:
From here, you can simplify the root (check it yourself):
Last edited by numen (2006-07-21 04:12:15)
Bang postponed. Not big enough. Reboot.
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