Floating gate potential of a MOSFET

[divB]

Hi,

First of all, I posted a similar question in [1] but I think it is the wrong forum and wrong question (maybe the reason why I get no answer...).

I have to solve this (I think easy?) exercise:

http://img705.imageshack.us/img705/8241/floatingmosfet.png

I think (3.1) is easy and I solved it. I also solved (3.2) but my solution is so primitive that I think it is wrong.

In (3.2) I should express $$\phi_F$$ as a function of $$t$$ if I apply a ramp function (i.e. switch on with a high voltage $$V_{pp}$$).

If I look at it, the one parameter dependent on the gate voltage is $$V_g$$ (which is just replaced by $$V_{pp}$$) and $$Q_F$$.

The current which flows from/to the floating gate is given by the formula in the problem statement but I think it is NOT dependent on time ... What do you think?!

If this is the case I just integrate $$I$$ from $$0$$ to $$t$$ in order to get the total charge but of course this is only linear behavior which also grows without stopping!

So my total solution would be:

$$\phi_F(t) = \frac{C_1 V_{pp} + t \cdot A \cdot E^2 \cdot e^{-\frac{B}{E}}}{C_1 + C_2}$$

Do you think this is true or am I on the wrong way?divB[1] https://www.physicsforums.com/showthread.php?t=376713

 

[KataKoniK]

Hello divB,

Thank you for reaching out to us with your question. I am happy to help you with this exercise.

Firstly, I want to assure you that you are on the right track with your solution. The solution you have proposed for (3.2) is correct. The current flowing from the floating gate is indeed not dependent on time, but the potential on the floating gate is. This is because, as you correctly pointed out, the only parameter dependent on time is the gate voltage V_g (which is replaced by V_{pp} in this case).

Your solution for \phi_F(t) is correct and it takes into account the linear behavior of the potential on the floating gate. The total charge on the floating gate can be obtained by integrating the current from 0 to t, as you have proposed.

I hope this helps clarify your doubts. Keep up the good work and don't hesitate to reach out if you have any further questions.