How is source-drain voltage defined?

[mzh]

Dear Physics Forum users
Is it true that conventionally, $V_{SD}$ is defined as the difference in potential between the drain and the source. Meaning, $V_{SD} = V_S - V_D$ with the voltage on the source as the reference.
As an example: if $V_{SD}$ is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source. Right?

Thanks for hints.

 

[NascentOxygen]

Is it true that conventionally, $V_{SD}$ is defined as the difference in potential between the drain and the source. Meaning, $V_{SD} = V_S - V_D$ with the voltage on the source as the reference.

To adhere to engineering conventions, V_SD is the voltage of point S with reference to point D,
i.e., V_SD ≡ V_S - V_D

As an example: if $V_{SD}$ is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source. Right?

http://img441.imageshack.us/img441/3331/nooo1.gif

The difference $V_{SD} = V_S - V_D$ reveals how many volts S is above D. Suppose S was at 6v and D was at 2v, V_S - V_D = +4, and you can see that's the voltage at S with reference to D.

Thanks for hints.

Thanks for asking.

 

[mzh]

The difference $V_{SD} = V_S - V_D$ reveals how many volts S is above D. Suppose S was at 6v and D was at 2v, V_S - V_D = +4, and you can see that's the voltage at S with reference to D.

Hi Oxy
Thanks for your outline. But please explain to me, why my statement is different from yours:

me: "If V_SD is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source"
you: "V_SD=V_S − V_D reveals how many volts S is above D."

I mean, it's the same to ask how many meters Paris is above sea level or how many meters below Paris is the sea level. No?

 

[NascentOxygen]

me: "If V_SD is given as -5V (note minus), then this means the drain is at a potential of 5 Volts lower than the source"
you: "V_SD=V_S − V_D reveals how many volts S is above D."

I think you might be confusing yourself with multiple negatives here.

If V_SD is -5v, then S is -5v with reference to D. So S has the lower potential, or, equivalently, you can say D is at the higher potential. So it is equivalent to saying D is +5v with reference to S.

 

[mzh]

I think you might be confusing yourself with multiple negatives here.

If V_SD is -5v, then S is -5v with reference to D. So S has the lower potential, or, equivalently, you can say D is at the higher potential. So it is equivalent to saying D is +5v with reference to S.

yes, i think i see it now. If, conventionally and for example, $V_{SD} = V_S - V_D = -5$, then this means $V_S < V_D$, which is what you say.

 

[NascentOxygen]

It follows that V_SD = –V_DS

 

[mzh]

still i could not find this conventional definition explicitly nowhere.

 

[gulu]

Can anyone help me for preparing mcq of electronics??

 

[jim hardy]

still i could not find this conventional definition explicitly nowhere.

GE Transistor Manual , General Electric Company, revised seventh edition 1969, pp 524 & 529:

V_KJ Circuit voltage between terminals K and J .

K, k Unspecified (general) measurement electrode. Also degrees Kelvin.
J, j Reference electrode

https://www.amazon.com/dp/B000U3J62U/?tag=pfamazon01-20

the way i was taught is "Your meter's red lead goes to first letter, black lead to second letter".
But that was before autoranging digital meters, so we often had to swap our meter leads to get an upscale reading..

 

[yungman]

I could swear $V_{DS}=V_D-V_S$ ! Say for a N channel MOSFET, the $V_{DS}$ is always positive. For N FET, Drain is always higher than source. For P-Channel FET, Drain is always lower than the source, so you have $V_{DS}$ is always negative.

 

[jim hardy]

I could swear $V_{DS}=V_D-V_S$ ! Say for a N channel MOSFET, the $V_{DS}$ is always positive. For N FET, Drain is always higher than source. For P-Channel FET, Drain is always lower than the source, so you have $V_{DS}$ is always negative.

for that N channel in a proper circuit:
red lead on Drain
black lead on Source
meter will show V_DS positive;

and for a P channel in a proper circuit, V_DS negative.

 

[NascentOxygen]

Can anyone help me for preparing mcq of electronics??

Hi gulu. For help with exam and homework questions go to the homework forum here: https://www.physicsforums.com/forumdisplay.php?f=152 Start a new thread, include details of one question that you are having difficulty with, and show a genuine attempt at answering it. Someone is sure to come by and help you. Guaranteed.

 

[yungman]

for that N channel in a proper circuit:
red lead on Drain
black lead on Source
meter will show V_DS positive;

and for a P channel in a proper circuit, V_DS negative.

Exactly.

 

[gnurf]

Say for a N channel MOSFET, the $V_{DS}$ is always positive. For N FET, Drain is always higher than source.

Except when it's not, e.g. when a NMOS is used as a synchronous rectifier. A MOSFET can conduct in either direction.

 

[NascentOxygen]

It's not unknown for an OP to ask a question different from that intended, or at least to be seeming to. If that's the situation here, then mzh should feel free to come back and ask again.

 

[yungman]

Except when it's not, e.g. when a NMOS is used as a synchronous rectifier. A MOSFET can conduct in either direction.

I am talking in general. I know you can even reverse the C and E of a BJT and it will work somewhat also.

 

[mzh]

It's not unknown for an OP to ask a question different from that intended, or at least to be seeming to. If that's the situation here, then mzh should feel free to come back and ask again.

Thanks for the feedback on my question.
No, getting the answer that in V_{SD}, the drain is considered the reference electrode, such that V_{SD} = V_S - V_D was most supportive.

 

[yungman]

Thanks for the feedback on my question.
No, getting the answer that in V_{SD}, the drain is considered the reference electrode, such that V_{SD} = V_S - V_D was most supportive.

This is the first time I heard of drain being used as reference voltage! BJT, MOSFET and JFET "usually" act as transconductance devices where the collector/drain is the output of the current source. Voltage is set by the load impedance and the power supply voltage. Control is usually done on $V_{GS}$.

I put "usually" because there are ways to use transistor in special ways, I am talking the general amplifier working in lineal situation. I know all about that you can even turn the FET around and use the source as drain and it'll still kind of work.

 

[mzh]

Control is usually done on $V_{GS}$.

Ok, so is it $V_{GS} = V_G - V_S$? My question is really only about the notational convention. Otherwise, correct sign will always be random.

 

[yungman]

Ok, so is it $V_{GS} = V_G - V_S$? My question is really only about the notational convention. Otherwise, correct sign will always be random.

Yes.

Sign is not random. For NMOSFET, $V_{DS}$ is "usually" positive. That is,. the Drain is higher than the Source. For PMOSFET, Drain is negative so $V_{DS}$ is negative.

 

[mzh]

Yes.

Sign is not random. For NMOSFET, $V_{DS}$ is "usually" positive. That is,. the Drain is higher than the Source. For PMOSFET, Drain is negative so $V_{DS}$ is negative.

but why then was it problematic for you to read $V_{SD} = V_S - V_D$? If V_SD = 5V, it means V_S is 5V higher than V_D.

 

[yungman]

but why then was it problematic for you to read $V_{SD} = V_S - V_D$? If V_SD = 5V, it means V_S is 5V higher than V_D.

Normally we call it $V_{DS}$ not $V_{SD}$.

 

[mzh]

Normally we call it $V_{DS}$ not $V_{SD}$.

check out this figure (which was the reason for my post):
[URL=http://picturepush.com/public/8993278][PLAIN]http://www5.picturepush.com/photo/a/8993278/640/8993278.jpg[/URL][/PLAIN]

from
http://www.nature.com/nature/journal/v445/n7127/abs/nature05498.html

btw., can anyone tell me if it is permitted to post such a figure on the web?

 

[yungman]

First time I ever seen it like this, check out all the data sheets of MOSFET and textbooks, it's always $V_{DS}$.