what is the purpose of the circuit

[matura713]

probably, very simple question for someone that has even basic knowledge of electronics (contrary to me)...

so, if we have circuit as on the attached picture, what is supposed to be the voltage on the point marked with "? ? ?". of course under the assumption that the 2 resistors and the 2 n-channel mosfet transistors are working properly.

it would good if the answer contains how the calculations to obtain the value are done and help me learn something new. however to understand if that's the point of failure the voltage value is enough. thanks!

[matura713]

A poorly implemented level translator.

would you care to elaborate a little, bare in mind i have no idea of electronics, i am just trying to pin point what potentially is the fault using digital multi-meter. so, considering +3.3 and 5V comes directly from the power supply what voltage is supposed to be on point "? ? ?"

[ataradov]

And just in case, here is a good version:

[ataradov]

It is not a switch, it does not switch anything, you can't draw any power from the output. It is a level shiftier, noting more.

EDIT: This was in response to coromonadalix, but the comment got removed while I was typing.

[matura713]

Guys, you're the expect, but let me tell you what confuses me - that MOSFET on the left, it's 3 pins are all connected to fixed volatges, as i marked on the schematic respectively +3.3V, +5V and GND directly from the power supply. So, that means nothing is really switched, if the input doesn't change, right? maybe it's done like that in case +3.3V or +5V rail disappeared, i.e. in case of fault in the power supply?

in any case I still cannot understand from your replies assuming by "output" you mean point "? ? ?" on the schematic what voltage should be there?

[ataradov]

They are not fixed voltages. +3.3V indicates a side with low voltage levels (0 or 3.3V). 5 V indicated high voltage level (0 or 5 V).

The circuit translates 0 on one side to 0 on the other, and 3.3V on one side to 5 V on the other.

[coromonadalix]

if the 3.3 volt is not there as a command,  you will have 5 volts after the 1 k resistor,  if you have the 3.3 v ,  the ? ? ? section will be at zero volt,  since it will pulled down or shorted to ground

[matura713]

If this circuit and its power supply rails all work, the "? ? ?" point should be ~5V.

hmm, so if that's the case, i just can bypass the whole circuit and connect +5V from the power supply to point "? ? ?", strange... what about those 1K and 100K resistors - they are their to divide the voltage or to reduce the current?

[ataradov]

hmm, so if that's the case, i just can bypass the whole circuit and connect +5V from the power supply to point "? ? ?", strange... what about those 1K and 100K resistors - they are their to divide the voltage or to reduce the current?

This is not a power supply thing, all signals here are logic levels.

This translates 0 V on the gate of the first transistor to 0 V on ? ? ?, and 3.3 V to 5 V on ? ? ?. But you can't power anything  from this, signal must go to another logic input, that expects 5 V. That's how digital logic circuits work.

[matura713]

This is not a power supply thing, all signals here are logic levels.

i understand, but my point then is that the Gate of the MOSFET on left is hardwired to +3.3V (i can check that on the board) and "? ? ?" in such case is always +5V, which no matter if you think of it as logic '0' or '1' it's still fixed to +5V and the whole circuit looks useless to me, because it can be fixed to +5V just with hard-wiring to the +5V rails instead pass via that circuit.

[ataradov]

If it is hard wired, then yes, the while thing can be replaced by a top 1K resistor. And depending on where ? ? ? goes, you can omit the resistor as well.

[matura713]

If it is hard wired, then yes, the while thing can be replaced by a top 1K resistor. And depending on where ? ? ? goes, you can omit the resistor as well.

ok, it's very weird, but i will try something like that.

[ebclr]

This circuit intend to do something like this

[matura713]

At least with my limited level of understanding what makes most sense to me is that the whole circuit is made for some unknown to me reason to set point "? ? ?" to 0V when +3.3V power rail disappears, because when +3.3V power rail is there it seems to do nothing and it can be bypassed.

[ataradov]

Well, yes, it can be a 5V level logic indicator of the presence of the +3.3 V rail. That makes sense.

It can also be a simple AND for 5 V and 3.3 V indicator, basically showing that both rails are present.

[matura713]

It can also be a simple AND for 5 V and 3.3 V indicator, basically showing that both rails are present.

do you mean if +3.3V is there, but +5V disappears then "? ? ?" goes to 0V as well?

[ataradov]

do you mean if +3.3V is there, but +5V disappears then "? ? ?" goes to 0V as well?

Correct.

[matura713]

Correct.

thanks, i believe i fully understand it now and i actually learned something from the whole discussion

[Decapitator]

Looks like a "power good" circuit. The output is only high, ~5v, if both voltages are present.