MOSFET CS Amplifier (VTC)

jegues

Joined Sep 13, 2010
732
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.

Last edited:

Jony130

Joined Feb 17, 2009
4,931
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

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

Last edited:

jegues

Joined Sep 13, 2010
732
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
4,931
I also made the mistake

jegues

Joined Sep 13, 2010
732
I also made the mistake
It's okay, the correct question is up there now so hopefully things are more clear.

Jony130

Joined Feb 17, 2009
4,931
As for C

Simply find Id for Vgs = Vdd next find Vd

rds = Vd/Id

Jony130

Joined Feb 17, 2009
4,931
And can you tell me.
How can you get Vgs --> 3V for Vds = 2.5V for Vdd = 2.5V ?

jegues

Joined Sep 13, 2010
732
And can you tell me.
How can you get Vgs --> 3V for Vds = 2.5V for Vdd = 2.5V ?
I must have done that incorrectly.

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

Is part a) okay?

Jony130

Joined Feb 17, 2009
4,931
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.

jegues

Joined Sep 13, 2010
732
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?

Jony130

Joined Feb 17, 2009
4,931
So as a result, Vgs = Vt = 0.5V, correct?
Yes, for Vgs = Vt----> Id=0A

By the way what "VTC" means ?

jegues

Joined Sep 13, 2010
732
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?

jegues

Joined Sep 13, 2010
732
Yes, for Vgs = Vt----> Id=0A

By the way what "VTC" means ?
Voltage Transfer Characteristic

jegues

Joined Sep 13, 2010
732
Here's my work up to Part C), I'm unsure if the Vds->Id->rd I got for part C) is correct or not.

Jony130

Joined Feb 17, 2009
4,931
Can you telly me what value for Kn you use in your calculations?

jegues

Joined Sep 13, 2010
732
Can you telly me what value for Kn you use in your calculations?
$k_{n} = k_{n}^{'}(\frac{W}{L})$

Where the values of kn' and W/L are given in my attempt.

Jony130

Joined Feb 17, 2009
4,931
So Kn is 10m right ?