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I'm a little bit confused about applying KVL to circuits containing MOSFETs. I hope someone can provide some clarity. I'll be using the figure below to ask my questions.

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First Question

If we want M1 to be ON, we need \$V_{\text{in}, \min} > V_{\text{TH}1}\$.

For M2 to be ON, we need \$V_{b, \min} > V_{\text{GS}2} + V_{X}\$. This is shown in the figure with red arrows.

  • In the case of \$V_{b, \min}\$, it was straight-forward to write the KVL. However in the case of Vin,min, I had to use my knowledge of transistors to write the condition.
  • Let's say I want to write KVL for M1. I'll get \$V_{\text{in}, \min} > V_{\text{GS}1} + 0\$. This makes no sense since \$V_{\text{GS}1}\$ is \$V_{\text{in}}\$. Why was I able to derive the condition for M2 using KVL, whereas I wasn't able to do the same for M1?

Second Question

For M2 to be in saturation, we need \$V_{b, \max} < V_{\text{TH}2} + V_{\text{out}}\$. Again, I used my knowledge of transistors to get this condition. How can we get this condition from KVL? If I use the same technique that I used to derive \$V_{b, \min}\$, I'll get \$V_{b, \max} < V_{\text{GD}2} + V_{\text{out}}\$. But \$V_{\text{GD}2}\$ is not \$V_{\text{TH}2}\$.

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2 Answers 2

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First, KVL does not involve inequalities. It says that the sum of the voltages around a closed pathequals zero.

First Question

If we want M1 to be ON, we need \$V_{in,min}>V_{TH1}\$

This is correct but not KVL. It is a requirement of the FET.

For M2 to be ON, we need \$V_{b,min}>V_{GS2}+V_X\$

This is incorrect:

KVL says \$V_{b,min}=V_{GS2}+V_X\$. This is always true no matter what is going on.

\$V_{GS2}\$ is not the same as \$V_{TH2}\$.

\$V_{TH2}\$ is a particular value of \$V_{GS2}\$. So, from transistor theory, \$V_{GS2}>V_{TH2}\$ for M2 to conduct.

So then: \$V_{b,min}=V_{GS2}+V_X>V_{TH2}+V_X\$ for M2 to conduct.

I want to write KVL for M1. I'll get \$V_{in,min}>V_{GS1}+0\$

This is also incorrect. By KVL, \$V_{in}=V_{GS1}\$ always, whether M1 is on or off. Applying transistor theory, \$V_{in,min}>V_{TH1}\$ for M1 to turn on.

Second Question

For M2 to be in saturation, we need \$V_{b,max}<V_{TH2}+V_{out}\$

This is correct.

Again, I used my knowledge of transistors to get this condition. How can we get this condition from KVL?

You can't get device defining relations usising KVL.

transistor theory

According to the Wikipedia article on MOSFETs. the condition for saturation is

$$V_{GS2}>V_{TH2}\text{ and }V_{DS2}>V_{GS2}-V_{TH2}$$

Combining these shows that: $$V_{TH2}<V_{GS2}<V_{DS2}+V_{TH2}\tag{Equ 1}$$ revealing that $$V_{GS2min}=V_{TH2}\text{ and }V_{GS2max}=V_{DS2}+V_{TH2}$$

Applying KVL: $$V_{DS2}=V_{OUT}-V_{X}$$
and:$$V_{GS2}=V_b-V_{X}$$

Substituting into (Equ 1) and rearranging reveals:$$V_{X}+V_{TH2}<V_{b}<V_{out}+V_{TH2}$$

So: $$V_{b,min}=V_X+V_{TH2}\text{ and }V_{b,max}=V_{out}+V_{TH2}$$

But \$V_{GD2}\$ is not \$V_{TH2}\$.

Of course it is not. Proper application of KVL and substitution reveals that $$V_{X}+V_{TH2}-V_{out}<V_{GD}<V_{TH2}$$

There are two issues with your analysis:

  1. Incorrect application of KVL. KVL is a moment captured in time of the voltages in a closed path. There are no inequalities or maximums or minimums or threshholds.
  2. Confusion among variable names. \$V_{GS}\$ is not \$V_{TH}\$.
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  • \$\begingroup\$ Regarding your first answer, you had \$V_{b, \min} > V_{\text{TH} 2} + V_X\$. In the video that I was watching before making this post, the author used \$V_{b, \min} > (V_{\text{OV} 2} + V_{\text{TH} 2}) + V_X\$. Since \$V_{b, \min}\$ is the minimum \$V_b\$ required for the transitor to be ON, I didn't understand the presence of \$V_{\text{OV} 2}\$. Here's a screenshot of the video: i.imgur.com/qZy1BVz.png \$\endgroup\$ Commented May 14, 2023 at 10:37
  • \$\begingroup\$ Regarding your second answer. I'm taking about the condition required for the M2 to be in saturation. A transitor is in saturation when its gate is not higher than its drain by more than \$V_{\text{TH}}\$. This is what sets the maximum gate voltage that can be used. From KVL: \$V_b = V_{\text{GD}} + V_{\text{out}}\$. Since \$V_{\text{GD}} < V_{\text{TH}}\$, we get \$V_b < V_{\text{TH}} + V_{\text{out}}\$. This is what I meant. \$\endgroup\$ Commented May 14, 2023 at 10:44
  • \$\begingroup\$ This changes the question. Answers cannot be accurate when you change the question after receiving an answer. I can guess that \$V_{O2}\$ is the excess \$V_{GS2}\$ required to make \$R_{DS2}\$ insignificant. \$\endgroup\$ Commented May 14, 2023 at 13:16
  • \$\begingroup\$ The secon part: I see what you mean now, I will edit my answer \$\endgroup\$ Commented May 14, 2023 at 13:41
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0V is a perfectly valid value for a KVL expression. \$V_{GS1}=V_{IN} - 0V\$ is an application of KVL, and is as valid and true as \$V_{GS2} = V_B - V_X\$. Neither of those expressions is relying on transistor characteristics.

However, the statement \$V_{IN} - 0V > V_{TH1}\$ differs in two ways; firstly it is an inequality which has nothing to do with KVL, and secondly it refers to \$V_{TH2}\$ which is a transistor characteristic that KVL knows nothing about.

KVL is always true, and simply relates potentials around a loop. Inequalities describe constraints to certain values, or behaviour of certain elements under different conditions, neither of which is accounted for in KVL, or KCL.

All KVL says is "this plus that minus the other is such and such", and never makes any claims about why, or how. It doesn't account for impedance, or dynamic or conditional behaviour. It is not concerned with greater-thans or less-thans, its only claim is that potentials around a loop sum to zero.

The condition you describe for M2 to be on is incorrect or ambiguous. You stated \$V_{b,min} > V_{GS2} + V_X\$. It would be better to say that \$V_{b,min} > V_{GS(ON)2} + V_X\$, where \$V_{GS(ON)2}\$ is a constant (perhaps from a datasheet), independent of any actual measured potentials. The value \$V_{GS2}\$ is not a constant, it's the potential difference between the gate and source of M2; \$V_{GS2} = V_B - V_X\$.

If from the datasheet you find that \$V_{GS(ON)}=2.0V\$ for M2, then the statement \$V_{b,min} > 2.0V + V_X\$ is meaningful and correct, and describes the condition necessary for M2 to be on. This is very different from \$V_{b,min} > V_{GS2} + V_X\$, which is equivalent to \$V_{b,min} > (V_B - V_X) + V_X\$, making no sense at all.

If by \$V_{TH}\$ you mean \$V_{GS(ON)}\$, then the following conditions describe the MOSFETs being on.
For M1:

$$ V_{GS1} > V_{TH1} $$ $$ V_{IN} - 0V > V_{TH1} $$

For M2:

$$ V_{GS2} > V_{TH2} $$ $$ V_B - V_X > V_{TH2} $$

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  • \$\begingroup\$ Thanks for your answer. I was watching a video before making this post. The author used \$V_b > (V_{\text{OV} 2} + V_{\text{TH} 2}) + V_X\$ as the condition necessary for M2 to be on. I didn't understand why \$V_{\text{OV} 2}\$ was included. From your answer, it seems that I can interprete \$V_{\text{OV} 2} + V_{\text{TH} 2}\$ as \$V_{\text{GS} 2 (\text{ON})}\$. But from transistor theory I know that for a a transistor to be in saturation \$V_{\text{OV}} > V_{\text{GS}} - V_{\text{TH}}\$ ... \$\endgroup\$ Commented May 14, 2023 at 10:57
  • \$\begingroup\$ So, it seems that \$V_b > (V_{\text{OV} 2} + V_{\text{TH} 2}) + V_X\$ includes the requirement for M2 to be in saturation. What does M2 being in saturation have to do with M2 being ON? Here's a screenshot of the video: i.imgur.com/qZy1BVz.png. \$\endgroup\$ Commented May 14, 2023 at 10:57
  • \$\begingroup\$ @LeonhardEuler \$V_{OV}=V_{GS}-V_{TH}\$ , so you can see that when adding the threshold term again, everything drops out and you're left with Vgs + Vx \$\endgroup\$ Commented May 14, 2023 at 11:56

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