Video:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/video-lectures/lecture-12/
Lecture Notes:- http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-002-circuits-and-electronics-spring-2007/lecture-notes/
From a pair of inverters, the output was not as expected. There was some delay in the final output.
Taking a look at the n-channel MOSFET - first, it is made up of silicon which is doped with a p-type material to become a semiconductor. Then put a thin layer of oxide (silicon dioxide) which is an insulator. .
Then another layer is put on the oxide. This layer can be many things. It could be a metal or something called polysilicon - a conductive form of silicon. So that layer is the gate. One can see that it is starting to resemble a capacitor-which has an insulator between two metal plates.
Next, two parts of the the p type are doped with n- type material. It is doped heavily with n type so it becomes a better conductor than the bulk which was reasonably doped.
A positive voltage is applied to the metal in relation to the bulk silicate. A set of positive charges appear at the boundary of the metal.
The positive voltage also attracts negative charges, electrons from the n type region which will migrate into the region close to the oxide, to form a channel. It can be seen that this resembles a capacitor.
The n-type regions are the drain and the source. The capacitor is commonly drawn between the gate and the source and it is called C_gs.
How does the MOSFET really work? If a positive voltage is applied to the drain with respect to the source, the electrons from the source are attracted to the drain. Current moves in the opposite direction to the electrons, so a positive current is flowing from the drain to source. The source produces electrons while the drain takes electrons.
The threshold voltage is where it takes a fair amount of voltage to form a conducting channel.
It is possible to build a capacitor. First, oxide is deposited on the silicon chip substrate. Then a plate is stuck on top of it. It could be a metal plate or polysilicon.
Of course, it is possible to build discrete capacitors by taking a pair of plates and having some insulator between them.
Capacitance is given by (e*A)/d where e is the permittivity of the dielectric, A is the area of the plates and d is the thickness of the dielectric.
When voltage v is applied to the terminals of the capacitor, there is a charge q on the capacitor.
q=Cv where C is the capacitance in farads. It can be considered to be a linear element and obeys the discrete matter discipline (DMD).
According to DMD, there is no net charge in my element but it seems that the capacitor does accumulate charge, so what is going on? If you put a box around the capacitor, it can be seen that the positive charge on one plate is exactly balanced by the negative charge on the other plate so there is no net charge.
Current is the rate of change of charge - i = dq/dt . If we substitute q by Cv, we get i=d(Cv)/dt. Assume that the capacitance is not a function of voltage or time - it has a fixed value and is constant.
So i=C(dv/dt).
v can be expressed in terms of i. v is the integral of i dt multiplied by 1/C with the limits between t and negative infinity.
Power can also be computed. P=vi= vC(dv/dt). From integration, E=(c*v^2)/2. This is an important equation as it proves that the capacitor is an energy storage device. This also makes it a memory device. It stores energy and that energy it stores, remembers what had happened before. The capacitor stores some state.
Capacitors are becoming important in energy storage but it can be harmful.
Capacitor c is connected to a the current source i(t). The voltage across the capacitor is v(t). The initial voltage of the capacitor is v(t_o).
We need to pick some current-a pulse which has a height of I for some amount of time T. We need to find v(t) for this input current.
To do that, we need to use the integral of i dt multiplied by 1/C with the limits between t and negative infinity. By splitting the integral into several parts, the graph of v(t) versus t can be drawn. The height during the time of charging is IT/C.
Circuits with one energy storage element are called first order circuits.
Analyzing RC circuits, it can be seen that the Thevenin's equivalent type of circuit could be used. Applying the node method and rearranging, we will get a first order differential equation RC (dv_c/dt) +v_c=v_I
The initial conditions of the capacitor is v_c(0)=V_0 and the input source v_I(t)= V_I. So now we will have a differential equation RC (dv_c/dt) +v_c=V_I
The method of homogeneous and particular solutions is used to solve the differential equation. The steps are :-
(1) Find the particular solution
(2) Find the homogeneous solution
(3) The total solution is the sum of the particular and homogeneous solution. Use the initial
conditions to solve for the remaining constants
So v_c(t)= particular + homogeneous solution.
The solution involves guesswork.
To find the particular solution, use trial and error on RC (dv_cp/dt) +v_cp=V_I where v_cp is a particular solution. Taking v_cp=V_I, then it seems to work.
The homogeneous solution is one where the drive is set to zero. Trial and error will also be used on RC (dv_ch/dt) +v_ch=0. Assume v_ch= Ae^(st) . Discarding the trivial A=0 solution, we will eventually get a characteristic equation of RCs+1=0. Therefore, s=-1/(RC) or v_ch=Ae^(-t/RC).
RC is a time constant.
From the initial conditions, it will be possible to find the value of A. A= V_0 - V_I
With that we can get the total final solution v_c=V_I+(V_0 - V_I)e^(-t/RC).
We can also get i_c which is equal to i=d(Cv_c)/dt. i_c= -(1/R)*(V_0 - V_I)e^(-t/RC).
To graph the solution, set t=0. It can be seen that v_c=V_0 when t=0. Then, when t is set to infinity, v_c= V_I. Assuming that V_I > V_0, then V_I can be marked on the graph. Between V_0 and V_I, there will be a rising curve.
One interesting aspect of this graph is that if we take the slope at the starting point, and look at where it intersects the final voltage value of the capacitor, that value is RC on the t-axis. Mathematically, this can be proven by differentiating v_c with respect to t at time t = 0.
Another point to note is that the equation v_c can be rewritten in a form of v_c(t) so that
v_c(t)=v_c(∞)+(v_c(0) - v_c(∞))e^(-t/RC). Any of the two forms can be used.
The last point to note is that if V_I < V_0, the curve would go in the opposite direction. If V_I=0, there would be an exponential discharge.
Finally, some examples that relate to the voltage outputs of the pair of inverters. Take the initial voltage v_c(0)=V_0=0V and v_I(t)=V_I=5V. This means that there is no voltage at the capacitor in the beginning and let's say at time t=0 the voltage switches suddenly from 0V to 5V. The curve starts out at 0 and after infinity time it reaches a voltage of 5V. The current across R decreases over time.
The equation for this curve is v_c= 5 - 5e^(-t/RC)
In the next scenario, v_c(0)=V_0=5V and v_I(t)=V_I=0V. If the input voltage V_I has been 5V for a long period of time, the voltage across the capacitor will be 5V. If at time t=0, V_I becomes 0 suddenly, then the capacitor will discharge. The curve starts out at 5V and after infinity time it reaches a voltage of 0 V. There will be an exponential decay. If we take the slope at time t=0 and see where it intersects the final voltage value, the value is time constant RC on the t-axis. The equation for the curve is v_c= 5e^(-t/RC)
If we combine the two curves, we will be able to see why the output of the pair of inverters behave the way they do.
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