# sketch Vin- Vout for the circuit

Discussion in 'Homework Help' started by screen1988, Apr 14, 2013.

1. ### screen1988 Thread Starter Member

Mar 7, 2013
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Hello, I need help again with sketch Vout versus Vin as Vin varies from 0 to Vdd.

The solution gives the characteristic Vin - Vout but I can't figure out how to get that.
For example:
Why Vout decreases when Vin increases?
How can you limit the number of case can be occur(M1 off, M2 off, M3 deep triode...)?
With this kind of exercise I really don't know how to analyse it properly.

• ###### MOS.JPG
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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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3. ### screen1988 Thread Starter Member

Mar 7, 2013
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Oh, thanks Jony!
This circuit in part c (video) is much simpler than mine but I think it is similar to the circuit above. I will try this method
Ah, in the video at 02:06 when PMOS is not cut off but null current he concludes that the PMOS is in triode. I don't get this because in the Ids-Vds characteristic I can't find a point in triode where i = 0 but Vgs ≠ 0.

4. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
Here is an example of Ids-Vds characteristic of PMOS.

In the video, Vsg = 5V and ids =0 A but as in the picture bellow with Vsg = 5V there is only a point at origin that ids=0.
But if so I think we have to call it is in cut off not triode?

PS.
I think I have just figured it out.
With PMOS in triode region:
Ids = -kn(Vsg -|Vt| - Vds/2)Vds (1)
With PMOS in saturation:
Ids = -1/2kn (Vsg - |Vt|)^2 (2)
As for the case in the video, Vsg =5V > Vt => PMOS is in triode or cut off.
Now because Vsg - Vt≠ 0 => Ids ≠ 0 => This case don't happen.
=> PMOS is in triode
Now I will try to analyse my circuit and post it here with hope anyone can look and correct it.

• ###### Pmos.JPG
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5. ### screen1988 Thread Starter Member

Mar 7, 2013
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Now my confusion in other thread has been clear. I will try to analyse the circuit.
With Vin < Vth1 => M1 is OFF and Id1 = 0 A
=> Vds1 = 0 V
With M2: Vgs2 = Vg2 - Vs2 = Vb1 - 0 = Vb1. I don't know what the value of Vb1 is but in the solution it says that M2 is OFF therefore I Vb1 < Vth2???
I get stuck here. With Id = 0 why I can't apply Ohm's law for M1, M2 and M3?
Does it because transistor is not a linear device? If so how can I calculate the voltage drop at M1, M2 and M3?

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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From which book you have this questions?

Also keep in mind that if Vin < Vth1 The M1 transistor is OFF.
So how can M2 transistor be ON when there is no physical path between M2 source to ground. Don't forget that VB1 is connect between M2 gate and GND.

7. ### screen1988 Thread Starter Member

Mar 7, 2013
310
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Hello Jony,
Design of Analog CMOS Integrated Circuits (Behzad Razavi). It is an exercise in chapter 3 and above is the solution.

If so(M2 is OFF) then M3 also OFF because it has no physical path between D of M3 and ground. But as in the solution above M3 is in deep triode.
Therefore, I am total confused.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
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But M3 is a PMOS not NMOS.

9. ### screen1988 Thread Starter Member

Mar 7, 2013
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Do you mean that PMOS can be ON if Vsg > Vth and the drain D is felt open?

10. ### Jony130 AAC Fanatic!

Feb 17, 2009
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To open a MOS transistor, the voltage between gate and source must be larger than the threshold voltage. Vgs > Vt .
So in M2 case when M1 is OFF. The M2 source is floating. So to OPEN M2 we need to concert a voltage source between the gate and source. But in our case VB1 is connect between gate and gnd.

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11. ### screen1988 Thread Starter Member

Mar 7, 2013
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And now assume that Vin > Vth then M1 is ON. Could you explain why at first M1, M2 are in saturation and M3 is in triode?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Because Vin is only slightly higher from Vth1 so M1 and M2 current is small.
But on the other end, we have the opposite situation. M3 Vgs voltage is much larger the Vth. So M3 want to deliver a large amount of current, but M1 and M2 allowed only the small current to flow.
So from M3 perspective. M2 and M2 behave just like a large resistor. But from M1 and M2 perspective. Thw M3 behave just like a small resistor.

Last edited: Apr 16, 2013
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13. ### screen1988 Thread Starter Member

Mar 7, 2013
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With Vin small (above Vth) then from NMOS characteristic I can know that Id is small. But even with Id small M1 can be in triode or saturation. I can't find the voltage in the drain D of M1 and therefore I don't know what value Vds is and can say if M1 is in triode or saturation.
And do you think that this kind of exercise is useful? If not I am going to skip it and move on.
I'm in over my head! There are many exercises of this kind with various configurations.

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The load resistance determines whether MOS is in triode or saturation.
See this two circuit

Solve for Id and Vout and determine if MOS work in triode or saturation region. If we assume Kn = 2.5m and Vt = 0.5V.

Well, I think that your main problem is that you really don't understand how the basic transistor circuit work (CS amplifier or CE for the BJT). So you really should return to basics transistor circuit.
And for IC designer this exercises are very useful, but for application engineer not so very useful. Because in 95% of a cases we use MOSFET as a switch.

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15. ### screen1988 Thread Starter Member

Mar 7, 2013
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Oh, I think this exercise is no problem because it has only a transistor and a load. My problem is the case there are many transistors in a circuit such as the above.
With the first circuit:
Vgs = 1.5 V > Vt => NMOS is ON
Because Rd is relatively small, therefore I guess it is in satuaration. Now I will use the assumption and solve Vout to check if the guess is right or not.
(I think that you mean W/L = 1 and Kn = 2.5mA/V^2 )
Id = 1/2Kn (Vgs - Vt)^2 = 1/2*2.5(1.5 - 0.5)^2 = 1.25 mA
Vout = Vdd - Id*Rd = 5 - 1.25mA* 1KΩ = 3.75V
=> Vgs - Vt =1 V < Vds = 3.75 V and therefore the assumption is right!
With the second circuit:
Vgs = 1.5 V > Vt => NMOS is ON
Because Rd is relatively big, therefore I guess it is in triode. Now I will use the assumption and solve Vout to check if the guess is right or not.
Id = Kn (Vgs - Vt - Vds/2)Vds = 2.5(1.5 - 0.5 - Vout) Vout = 2.5*(1- Vout)Vout mA
Vout = Vdd - Id*Rd = 5 - 2.5*(1- Vout)Vout mA * 10 KΩ
=>25Vout ^2 -26Vout + 5 = 0 => Vout = 0.04 V or Vout = 1 V(this value don't satisfy)
Then Vgs - Vt = 1V > Vds = 0.04 => NMOS is in triode and assumption is right.

16. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Well but we have a very similar situation in the circuit from post 1.
Simply, for Vin slightly larger than Vth1, M3 act just like 1K resistor in my first circuit.

As for the solution.

Id = Kn (Vgs - Vt - Vds/2)Vds = 0.0025*(1 - Vds/2) * Vds

And

Vds = Vdd - Id*Rd = 5V - ( 0.0025*(1 - Vds/2) * Vds )*10000

I get Vout = 0.214V
http://www.wolframalpha.com/input/?i=x+=+5+-+(+0.0025*(1+-+x/2)+*+x+)*10000

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17. ### screen1988 Thread Starter Member

Mar 7, 2013
310
4
Hi, thanks! You are right. I am mistaken in the final equation.