MOSFET CS Amplifier (VTC)

Thread Starter

jegues

Joined Sep 13, 2010
733
See figure attached for the question and the figures given.

I'm having a little trouble for part a), I know at point A,

\(v_{GS} = V_{t}\) and the transistor is just turning on, and since \(v_{DS} > 0 \) we will enter saturation.

For point B,

\(v_{GS} - V_{t} = v_{DS}\)

but how do I determine Rd from this? I haven't been told anything about vGS or iD.

Hopefully once I get the ball rolling on part a), parts b), c) & d) should fall into place.

Thanks again!

EDIT:

a) Vgs = Vds + Vt, Vds = 0.6V

\(i_{D} = \frac{1}{2}k_{n}(v_{GS} - V_{t})^{2}\)

\(i_{D} = \frac{V_{DD} - V_{DS}}{R_{D}}\)

\(i_{D} = 1.25mA, \quad R_{D} = 1.6k \Omega\)

b)

\(v_{DS} = 0.5V \quad \rightarrow \quad v_{GS} = 1V\)

\(v_{DS} = 2.5V \quad \rightarrow \quad v_{GS} = 3V\)

I'm stuck on part c) now.
 

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Jony130

Joined Feb 17, 2009
5,155
In the first place tell me why you want to find Rd ??
Rd is given and its equal to Rd = 24kΩ

As for question a

The point A is equal to

\(V_{ds} = V_{t}\)

And point B

\(V_{ds} = V_{gs} - V_{t}\)

And we also know that:

\(V_{ds} = I_{d} * R_{d}\)

And

\(I_{D} = \frac{1}{2}K_{n}(V_{GS} - V_{T})^{2}\)

So you need to solve this equation

\(V_{dd} - (I_{d} * R_{d}) = V_{gs} - V_{t} = V_{dd} - \frac{1}{2}K_{n}(V_{GS} - V_{T})^{2}*Rd = (V_{gs} - V_{t})\)
 
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Thread Starter

jegues

Joined Sep 13, 2010
733
In the first place tell me why you want to find Rd ??
Rd is given and its equal to Rd = 24kΩ

As for question a

The point A is equal to

\(V_{ds} = V_{t}\)

And point B

\(V_{ds} = V_{gs} - V_{t}\)

And we also know that:

\(V_{ds} = I_{d} * R_{d}\)

And

\(I_{D} = \frac{1}{2}K_{n}(V_{GS} - V_{T})^{2}\)

So you need to solve this equation

\(I_{d} * R_{d} = V_{gs} - V_{t} = \frac{1}{2}K_{n}(V_{GS} - V_{T})^{2}*Rd = (V_{gs} - V_{t})\)
Sorry I had posted the wrong question by mistake, the correct one is up there now.
 

Jony130

Joined Feb 17, 2009
5,155
I must have done that incorrectly.

How do I get the value of Vgs for Vds = 2.5V

Is part a) okay?
The part a) looks okay except the last part Vgs=3V for Vds = 2.5V.
To get 2.5V as a Vds the drain currant must be equal to zero.
Then there will be no voltage drop across Rd resistor and Vds = Vdd.
 

Thread Starter

jegues

Joined Sep 13, 2010
733
The part a) looks okay except the last part Vgs=3V for Vds = 2.5V.
To get 2.5V as a Vds the drain currant must be equal to zero.
Then there will be no voltage drop across Rd resistor and Vds = Vdd.
So as a result, Vgs = Vt = 0.5V, correct?
 

Thread Starter

jegues

Joined Sep 13, 2010
733
For part C),

I solved the 2 equations for id for Vds and found that Vds = 0.077V, id = 1.51mA and rd = 50.85 ohms.

Is this correct?
 
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