I've already explained it!

Do this. Take in input signal and apply it to one side of a capacitor and tie the other side to ground. What affect does the input signal have on the voltage of the ground? Now tie tha other side to the positive side of a battery Ithat has the other side tied to ground). What affect does the input signal have on the voltage of the battery? Now swap out the battery with a DC voltage source called, oh, how about Vgs. What affect does the input signal have on the voltage of the source?

You really, really, REALLY need to spend some serious time and effort learning basic circuit fundamentals and analysis. You are going to be hindered, haunted, and handicapped until you do. This site has a decent E-book to help you along.

Take in input signal and apply it to one side of a capacitor and tie the other side to ground. What affect does the input signal have on the voltage of the ground?

For example, input signal is Vin then the voltage between gate and source of the transistor is Vin.

Now tie tha other side to the positive side of a battery Ithat has the other side tied to ground). What affect does the input signal have on the voltage of the battery?

The voltage between gate and source of the transistor now is Vin + Vbattery

Now swap out the battery with a DC voltage source called, oh, how about Vgs. What affect does the input signal have on the voltage of the source?

Vgs= Vin + Vdc

You really, really, REALLY need to spend some serious time and effort learning basic circuit fundamentals and analysis. You are going to be hindered, haunted, and handicapped until you do. This site has a decent E-book to help you along.

Thanks, what part do you recommed to start with?



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Answer my questions as stated. There's no transistor involved in them.

Place a capacitor between a signal Vfred and ground. Forget Vin as it appears in this circuit and if that's too confusing, call this new signal Vfred. What affect does Vfred have on the voltage on the ground node?

 

Answer my questions as stated. There's no transistor involved in them.

Place a capacitor between a signal Vfred and ground. Forget Vin as it appears in this circuit and if that's too confusing, call this new signal Vfred. What affect does Vfred have on the voltage on the ground node?

Sorry, I don't quite understand your questions.
Here is what I think. When a capacitor which has large capacitance is placed between Vfred and ground, the Vfred will see capacitor as a short wire and therefore one node of Vfred has 0V(ground).

 

Okay, but the question is, given a signal Vfred applied to one side of a capacitor in which the other side it tied to ground, what will the change be in the voltage on the grounded side of the capacitor?

 

Okay, but the question is, given a signal Vfred applied to one side of a capacitor in which the other side it tied to ground, what will the change be in the voltage on the grounded side of the capacitor?

If so I see that there are no difference in the voltage of the grounded plate of capacitor. The voltage in the grounded side is always 0V because it is always connected to ground.

 

If so I see that there are no difference in the voltage of the grounded plate of capacitor. The voltage in the grounded side is always 0V because it is always connected to ground.

EXACTLY!!!

Now, instead of tying the other side of the capacitor to ground, tie it to the output of an ideal voltage source and answer the same question -- by how much does the voltage on the side of the capacitor tied to the voltage source change as Vfred changes?

 

EXACTLY!!!

Now, instead of tying the other side of the capacitor to ground, tie it to the output of an ideal voltage source and answer the same question -- by how much does the voltage on the side of the capacitor tied to the voltage source change as Vfred changes?

The side of capacitor tied to the voltage source has a constant voltage that equals voltage source. Aha, I see what you mean now but in the case I consider
the capactior has a very large capacitance. Therefore, the voltage drop in it is 0V and what will happen if Vin and VGS is tied to each other and both are connected to the gate?

 

It doesn't matter how big the capacitor is. If Vin goes up by, say, 1mV then current will flow from Vin through the capacitor and down to Vgs until the voltage across the capacitor is 1mV.

As long as you are working with ideal voltage sources for Vin and Vgs, then that is what will happen. Now, if you want to assign output impedances to the two sources, then you can calculate how much of Vin will be seen at the gate. But that's really exactly what the inductor is -- an output impedance on Vgs that just happens to vary with frequency.

 

It doesn't matter how big the capacitor is. If Vin goes up by, say, 1mV then current will flow from Vin through the capacitor and down to Vgs until the voltage across the capacitor is 1mV.

I got this. \(i_{c}\left( t\right) =c\dfrac {\partial u_{c}\left( t\right) } {\partial t}\), where \(u_{c}= Vin - VGS\)

As long as you are working with ideal voltage sources for Vin and Vgs, then that is what will happen. Now, if you want to assign output impedances to the two sources, then you can calculate how much of Vin will be seen at the gate. But that's really exactly what the inductor is -- an output impedance on Vgs that just happens to vary with frequency.

I don't get this part. Could you explain what do you mean "output impedance" and do you mean two sources here is Vin and VGS?

 

In general and for most purposes, a source is generally modeled as an ideal source and an output impedance. For a voltage source, the output impedance is in series with the ideal voltage source and for a current source the output impedance is in parallel with the ideal current source.

That are very basic concepts. You really need to step back and get a handle on them. It is all but pointless to try to pick them up piecemeal like this. Sure, fill a hole there and smooth a rough edge there is one thing. But you need to spend some serious time learning these basic concepts and for that you need to delve into a book (such as the E-Book here) and get down to basics.

 

In general and for most purposes, a source is generally modeled as an ideal source and an output impedance. For a voltage source, the output impedance is in series with the ideal voltage source and for a current source the output impedance is in parallel with the ideal current source.

That are very basic concepts. You really need to step back and get a handle on them. It is all but pointless to try to pick them up piecemeal like this. Sure, fill a hole there and smooth a rough edge there is one thing. But you need to spend some serious time learning these basic concepts and for that you need to delve into a book (such as the E-Book here) and get down to basics.

Yes, I will start with the E-books here and then ask here about problems that I can't solve or understand. Can you suggest a method to learn these basic things. I have learnt most of basic concepts but maybe, I don't really understand it. I read many concepts but I don't use it effectively and I can't apply it. Do you think that I need to learn all things from beginning just like a newbie?

 

Yes, I will start with the E-books here and then ask here about problems that I can't solve or understand. Can you suggest a method to learn these basic things. I have learnt most of basic concepts but maybe, I don't really understand it. I read many concepts but I don't use it effectively and I can't apply it. Do you think that I need to learn all things from beginning just like a newbie?

The short answer is yes, I think you need to start at the beginning. But many things you will already know well enough to move through them quickly. The key is going to be making sure that you don't go to the next step until you are confident you know the present step.