# MOSFET CS Amplifier (VTC)

Discussion in 'Homework Help' started by jegues, Sep 27, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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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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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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: Sep 28, 2011
3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Sorry I had posted the wrong question by mistake, the correct one is up there now.

Feb 17, 2009
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5. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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It's okay, the correct question is up there now so hopefully things are more clear.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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As for C

Simply find Id for Vgs = Vdd next find Vd

rds = Vd/Id

7. ### Jony130 AAC Fanatic!

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

8. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I must have done that incorrectly.

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

Is part a) okay?

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.

10. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
So as a result, Vgs = Vt = 0.5V, correct?

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, for Vgs = Vt----> Id=0A

By the way what "VTC" means ?

12. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
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?

13. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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Voltage Transfer Characteristic

14. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
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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.

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15. ### Jony130 AAC Fanatic!

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

16. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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$k_{n} = k_{n}^{'}(\frac{W}{L})$

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

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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So Kn is 10m right ?