Thread: Solving -3 + (100)(29*10^{-12})(e^(V_D/(27*10^{-3})) = V_b

1. Solving -3 + (100)(29*10^{-12})(e^(V_D/(27*10^{-3})) = V_b

Can anyone one help me how i can solve this equation?

$-3\, +\, (100)\,(29\, \times\, (10^{-12})\, \left(e^{^{V_D}\big/{29\, \times\, 10^{-3}}}\, -\, 1\right)\, =\, V_b$

2. Originally Posted by zorro
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Can anyone one help me how i can solve this equation?
You have variables Vb and VD here. Whose value do you want to calculate? What is the value of the other one?

3. Originally Posted by Subhotosh Khan
You have variables Vb and VD here. Whose value do you want to calculate? What is the value of the other one?
both are VD:

$-3\, +\, (100)\,(29\, \times\, (10^{-12})\, \left(e^{^{V_D}\big/{29\, \times\, 10^{-3}}}\, -\, 1\right)\, =\, V_D$

4. Originally Posted by zorro
both are VD
You have a nonlinear equation.

What methods have you been taught to solve these types of equations?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

http://www.freemathhelp.com/forum/announcement.php?f=33

5. Originally Posted by Subhotosh Khan
You have a nonlinear equation.

What methods have you been taught to solve these types of equations?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

http://www.freemathhelp.com/forum/announcement.php?f=33
Capture.jpg
I'm not sure whether its true or not. But this is how i did and i'm stuck.

6. The given equation represents an equality between a transcendental function and an algebraic function. This type of equation cannot be solved in terms of elementary functions.

You can take a numerical approach (eg: Newton's Method), to approximate solution(s).

Exact solution(s) may be expressed, in terms of the LambertW function (generally not covered in undergraduate calculus courses).