I want to ask about CS amplifier with active load.

Does this circuit always operate in triode region as long as Vin > Vth?
Here is my understanding:
1) Assume that the transistor is in saturation:
Id = 1/2*Kn* (Vgs - Vth)^2 = 1/2*Kn* (Vin - Vth)^2
If Vin increases => Id increases. This is impossible because Id = constant
2) Assume the transistor is in triode:
Id = 1/2* Kn * ((Vgs - Vth)*Vds - Vds^2) = 1/2* Kn * ((Vin - Vth)*Vds - Vds^2)
If Id = constant and Vin increase, then the only way to make the equation hold is Vds change.
=> In CS amplifier with active load the transistor always operate in triode region, right?

 

For your simplified model, you are correct.
Be aware that real world MOSFETs have varying degrees of channel length modulation in saturation, wherein the drain current is weakly dependent on Vds. See the attachment, which comes from here.
In this case, the voltage gain will not be infinite, as it is when your simplified model is in saturation, because, for a given drain current, changing vgs will change Vds.

 

As model in the page:
Triode mode or linear region (also known as the ohmic mode)
When VGS > Vth and VDS < ( VGS – Vth )

Saturation or active mode
When VGS > Vth and VDS ≥ ( VGS – Vth )

Now how can I know which region the transistor operate?
The problem here is that I can't calculate VDS to check if VDS < ( VGS – Vth ) or not.
As what I did in the first post, I checked the equations of Id in each region and see there is only one region possible but now both equations are dependent on VDS and this method doesn't work.

 

As model in the page:
Triode mode or linear region (also known as the ohmic mode)
When VGS > Vth and VDS < ( VGS – Vth )

Saturation or active mode
When VGS > Vth and VDS ≥ ( VGS – Vth )

Now how can I know which region the transistor operate?
The problem here is that I can't calculate VDS to check if VDS < ( VGS – Vth ) or not.
As what I did in the first post, I checked the equations of Id in each region and see there is only one region possible but now both equations are dependent on VDS and this method doesn't work.

If your original equations came from your textbook, use them. You already figured out the correct answer.

I'm sorry that I confused you. I was just pointing out what you will encounter when you use a real MOSFET.

 

If your original equations came from your textbook, use them. You already figured out the correct answer.

I'm sorry that I confused you. I was just pointing out what you will encounter when you use a real MOSFET.

Thanks, I am willing to know new things, too.

 

If you talking about a amplifier, first you need to set a proper bias point (Q point). So I add a bias two resistors. And I set ID ≈ 100mA and I use 100mA current source as a load.
I pick IRFZ44 and I know from LTspice that Vt = 4V
Also from LTspice I now that for Vgs = 5V and Vds = 10V drain current is equal to Id = 9.5A

The MOSFET K factor is equal

K = 9.5A/(5V - 4V)^2 = 9.5

So to get ID = 0.1A I need Vgs = 4V + √(0.1/9.5) = 4.1V.

So for Vdd = 20V I need Vds = 10V

R1/R2 = (10/4.1 - 1) = 1.43

So I choose R1 = 470K and R2 = 330K.

As you can see I build CS amplifier with a constant current load.

And additional this circuit work in saturation region.
VDS ≥ ( VGS – Vth ) = 10V ≥ (5V - 4V)

But first lets try verify this. So first we need to find a Q-pint for this circuit.
I use this to solve for Vgs and assume saturation region.

\(K*(Vgs - Vt)^2 = I1 - \frac{Vgs * (1 + \frac{R1}{R2})}{R1+R2}\)

Vgs = 3.89741, Vgs -> 4.10259V

And

ID = 0.0999847A = 99.98mA

And Vds = Vgs * (1 + R1/R2) = 9.94567V

Now I check VDS ≥ ( VGS – Vth ) and see it right.

And now I hope that you see why I connect R1 to drain instead Vdd.
Because we have two current sources in series in this circuit.
So how could I determine the Vds voltage without R1 and R2?
Without these two resistor Vds is undetermined.

And finally to answer your question

"In CS amplifier with active load the transistor always operate in triode region, right? "

No, as you can see we can build amplifier with active load which works in saturation region

 

Pretty cool! Jony
I though this configuration can't be an amplifier.
One more question.
At first you chose Vds =10V, meaning that the operating point Q in the between of the load line.
From that you chose Id = 0.1A and infer Vgs = 4.1V.
But when R1 and R2 is added, the current Id is no longer equal to 0.1A, it decreases a little bit. But at the end the trans still operates in saturation region.

I'd like to know if I can do it as follows:
Vdd =20V => choose Vds = 10V
The current of the source is 0.1A => the current through trans Id = I1 - IR1
Then I choose a certain Id and from that calculate Vgs and R1, R2.
Now the question is what are criteria for me to choose Id?
Or it is OK as long as 0< Id < I1.

 

If you talking about a amplifier, first you need to set a proper bias point (Q point). So I add a bias two resistors. And I set ID ≈ 100mA and I use 100mA current source as a load.
I pick IRFZ44 and I know from LTspice that Vt = 4V
Also from LTspice I now that for Vgs = 5V and Vds = 10V drain current is equal to Id = 9.5A

The MOSFET K factor is equal

K = 9.5A/(5V - 4V)^2 = 9.5

So to get ID = 0.1A I need Vgs = 4V + √(0.1/9.5) = 4.1V.

So for Vdd = 20V I need Vds = 10V

R1/R2 = (10/4.1 - 1) = 1.43

So I choose R1 = 470K and R2 = 330K.

As you can see I build CS amplifier with a constant current load.

And additional this circuit work in saturation region.
VDS ≥ ( VGS – Vth ) = 10V ≥ (5V - 4V)

But first lets try verify this. So first we need to find a Q-pint for this circuit.
I use this to solve for Vgs and assume saturation region.

\(K*(Vgs - Vt)^2 = I1 - \frac{Vgs * (1 + \frac{R1}{R2})}{R1+R2}\)

Vgs = 3.89741, Vgs -> 4.10259V

And

ID = 0.0999847A = 99.98mA

And Vds = Vgs * (1 + R1/R2) = 9.94567V

Now I check VDS ≥ ( VGS – Vth ) and see it right.

And now I hope that you see why I connect R1 to drain instead Vdd.
Because we have two current sources in series in this circuit.
So how could I determine the Vds voltage without R1 and R2?
Without these two resistor Vds is undetermined.

And finally to answer your question
No, as you can see we can build amplifier with active load which works in saturation region

What your circuit points out is that, with the original saturation mode equation, you MUST have feedback to establish an operating point.

 

What your circuit points out is that, with the original saturation mode equation, you MUST have feedback to establish an operating point.

Exactly what I was trying to say.

 

Pretty cool! Jony
I though this configuration can't be an amplifier.
One more question.
At first you chose Vds =10V, meaning that the operating point Q in the between of the load line.
From that you chose Id = 0.1A and infer Vgs = 4.1V.
But when R1 and R2 is added, the current Id is no longer equal to 0.1A, it decreases a little bit. But at the end the trans still operates in saturation region.

I'd like to know if I can do it as follows:
Vdd =20V => choose Vds = 10V
The current of the source is 0.1A => the current through trans Id = I1 - IR1
Then I choose a certain Id and from that calculate Vgs and R1, R2.
Now the question is what are criteria for me to choose Id?
Or it is OK as long as 0< Id < I1.

Why you want Id<<I1 ? Why you want to waste a power?

 

Why you want Id<<I1 ? Why you want to waste a power?

OK, I see it now. Now about the choosing R1 and R2. Because Vds = 10V
=> Power wasted on these resistors is 100/(R1+R2)
Meaning that the larger R1 + R2 the less wasted energy.
But is there any limit for R1 and R2?

 

It all depends on the specific application requirement. What is your goal that you try to achieve e.g. low noise, low power.

 

It all depends on the specific application requirement. What is your goal that you try to achieve e.g. low noise, low power.

I think the larger the current I1 (the current through R1 and R2), the better ability immune noise. Is this right?