Biasing circuit with mosfet: am I missing information or misreading it?

I have the following circuit with MOSFET (cascode amplifier):

Information given: $$V_t=1V$$ $$I_1=20\mu A$$ $$k=10 \mu AV^{-2}$$

Knowing that $$\V_b\$$ is 5V determine the minimum value of the output voltage so that both transistors are saturated.

I'm kinda confused on how to proceed on this.

First of all, don't we need VDD? How to relate VB with this? I started by assuming both transistors had the current I_1 (I assume this is correct).

Then I used the MOSFET equation for saturation $$i_D=k(v_ {GS}-Vt)^2$$ to determine $$\v_{GS}\$$ in both transistors. obtaining 2.4 V. Then I applied the boundary condition for saturation $$v_{DS}=v_{GS}-V_t$$ and so obtained for both transistors $$\v_{DS}=1.4V\$$ So the minimum value of $$\v_O\$$ is 2.8V.

This is incorrect, the answer should be 4V. I'm really not understanding how to proceed... Can someone give me an hint on how to do this? Thanks!

• The output impedance very high so a load RC must defined including stray then specs for desired GBW. Coss is a significant parameter as well as Vds max. Feb 28 '18 at 2:16
• for DC biasing and gain, I think you need to know Lambda and channel length allaboutcircuits.com/technical-articles/… Feb 28 '18 at 2:22

The current through both the MOSFETs is the same. Also it is given that both the MOSFETs are in saturation region. Therefore we can write: $$I_{D1} = I_{D2}= I_{D} = k(V_{GS} - V_t)^2 = 20 \mu A$$ on solving we get $$\V_{GS}\$$ should be $$\2.4V\$$ for both the MOSFETs, if $$\ 20 \mu A\$$ has to flow through both of them.

Now for M2, $$V_{GS} = 2.4V$$ $$\implies V_B - V_A = 2.4 V$$ Given $$\V_B\$$ is 5V.

$$\implies 5 - V_A = 2.4 V$$ $$\implies V_A = 2.6V$$

For M2 to enter saturation, $$\ V_{DS}\$$ of M2 should be atleast $$\ V_{GS} - V_t\$$ $$V_{DS_{min}}= 2.4 - 1 = 1.4 V$$ At this point, $$\ V_{DS} \$$ of M1 will be: $$V_{DS(M1)} = V_A = 2.6V > 1.4V$$

So M1 will also be in saturation, well saturated over the minimum value of $$\1.4V\$$.

Now from the circuit we can express $$\V_{DS}\$$ of M2 as $$V_{DS} = V_o - V_A$$ $$\implies V_o = V_A + 1.4 = 4 V$$

If you look closely, the node voltage $$\ V_A \$$ decides the minimum output voltage. When M2 entered the saturation region (Boundary condition), M1 had already been saturated way before. So your assumption of having boundary conditions in both the MOSFETs at the same time, went wrong. And hence the calculations too.

• Thank you so much! The fact that you identified that node changed everything and thanks for such a detailed explanation! Feb 28 '18 at 20:30

If Vds of M1 is 1.4V and Vgs of M2 is 2.4V, how can Vb be 5V?

Vdd is not needed because I1 is an ideal current source.

• Oh so that's why VDD is not needed. I know it doesn't make sense but I'm really confused on how to start the problem. Can you help me? Feb 28 '18 at 2:30
• @GrangerObliviate The mistake you made was "vDS=vGS−Vt" Since Vgs for M2 is 2.4V and Vb is 5V, the source of M2 must be 5V - 2.4V
– τεκ
Feb 28 '18 at 2:38
• But that's a known equation for MOSFET. In the boundary of saturation-triode, vDS=vGS-Vt. Feb 28 '18 at 10:48
• @GrangerObliviate M1 and M2 don't necessarily have the same Vds. M2 may be in triode while M1 is still in saturation.
– τεκ
Feb 28 '18 at 13:47
• No, the question says that they both need to be in saturation. That's a requirement of the exercise. Feb 28 '18 at 16:36