Yes, there is an equation for it. It is called "simple addition and subtraction".
On your diagram, mark the drain and source of each FET. Take a closer look at the voltages marked on the diagram and use the "equation" that I quoted.
Next time, try using your brain before you panic and post.

 

Yes, there is an equation for it. It is called "simple addition and subtraction".
On your diagram, mark the drain and source of each FET. Take a closer look at the voltages marked on the diagram and use the "equation" that I quoted.
Next time, try using your brain before you panic and post.

I think you misunderstood my question. I meant how do i calcuate them by myself without using the simulator xD
I mean the picture was taken from a software that calcuated them. I want to calcuate by myself

 

What type of devices do you think you're using? They're not standard symbols JFETs or MOSFETs. What do the colors on the symbols mean?

Your schematic drawing style is weird. Why do you have so many unnecessary wire jogs?

 

What type of devices do you think you're using? They're not standard symbols JFETs or MOSFETs. What do the colors on the symbols mean?

Your schematic drawing style is weird. Why do you have so many unnecessary wire jogs?

These are two MOSFETS NMOS (the lower one) and PMOS (the upper one)
The colored triangle just means that they are in saturation
My question is: Lets suppose we have a random main Voltage and the two transistors are identical (same dimensions ..etc) and they have different Gate Source voltages. How can i determine the Drain Source voltage of each one? Is there an equation?

 

These are two MOSFETS NMOS (the lower one) and PMOS (the upper one)

No they aren't. The arrow points in on N channel and out on P channel. When the vertical line representing the channel is solid, that indicates a depletion mode device. The line is dashed for enhancement mode devices.

 

Your Mosfets have no part number then you cannot see the wide range of conduction they have in their datasheet.
A simulation simply guesses that your Mosfet has "typical" specs. When you buy one then it might be very sensitive or have very low specs.

There are many opposing Mosfet symbols that are used:

 

Your Mosfets have no part number then you cannot see the wide range of conduction they have in their datasheet.
A simulation simply guesses that your Mosfet has "typical" specs. When you buy one then it might be very sensitive or have very low specs.

There are many opposing Mosfet symbols that are used:

If we assume that both MOSFETs are identical, how can i know how much voltage would drop on each of them? (We know Vgs1, Vgs2 and V of the voltage source (current is not known). How would you calculate the voltage in this case?

 

If we assume that both MOSFETs are identical, how can i know how much voltage would drop on each of them? (We know Vgs1, Vgs2 and V of the voltage source (current is not known). How would you calculate the voltage in this case?

If you were designing a MOS FET circuit, you would refer to the data sheet for the devices you plan on using.
In your diagram, there is not enough gate to source voltage on either device to allow any apreciable source to drain current to flow.

 

Here is the data sheet.

 

View attachment 269894
View attachment 269895
Here is the data sheet.

Maybe the simulation is wrong but i am asking a general question about how to calculate the drain source voltage of two transistors connected as above

 

Both mosfets are in saturation and the same current is flowing and the parameters where chosen exactly the same
SO why is the voltage drop on both mosfets different?
shouldnt it be 1.5V on each one?

 

Maybe the simulation is wrong but i am asking a general question about how to calculate the drain source voltage of two transistors connected as above

I am not familiar with the simulator that you are using. The "data sheets" that you show must be the device definitions that the simulator uses. They are meaningless to me. You will have to study how the simulator works to get the answer to your question.
In the real world, we use a data sheet like this to get the information on specific devices:

http://www.datasheet-pdf.com/PDF/IRFZ40-Datasheet-Vishay-966217

 

it's the upper P-channel one ... if you double the lower mosfet and add the current mirror to the upper mosfet - it likely brings the DS drops much closer.

 

You are simply guessing that the Mosfets have identical threshold voltages and temperatures.
Maybe they are different because they are so far apart.

 

You are simply guessing that the Mosfets have identical threshold voltages and temperatures.
Maybe they are different because they are so far apart.

well i used same parameteres in the software (of course i mean the absolute values)
so everything is identical..

 

The hand calculation will be a pain in the ass. Because we have two current sources connected in series. Thus, the transistor which tries to "make the lowest current" will set the current in the circuit.
https://forum.allaboutcircuits.com/...tant-current-source-load.147146/#post-1253377
In the first iteration, we can assume that both transistors are in the saturation region.
Thus:
Id1 = Id2
So we need to solve this for Vds1.
W/L *Kp/2 * (Vgs1 - Vt)^2 (1 + λVds1) = W/L *Kp/2 * (Vgs2 - Vt)^2 (1 + λ(Vdd - Vds1))
Next, we need to check if our assumption about saturation was wrong or not. And if it turns out that one of the transistors is not saturated we need to repeat the calculations. Again assuming Id1 = Id2 but we need to plug the triode region equation for the one that is not in saturation.

 

The hand calculation will be a pain in the ass. Because we have two current sources connected in series. Thus, the transistor which tries to "make the lowest current" will set the current in the circuit.
https://forum.allaboutcircuits.com/...tant-current-source-load.147146/#post-1253377
In the first iteration, we can assume that both transistors are in the saturation region.
Thus:
Id1 = Id2
So we need to solve this for Vds1.
W/L *Kp/2 * (Vgs1 - Vt)^2 (1 + λVds1) = W/L *Kp/2 * (Vgs2 - Vt)^2 (1 + λ(Vdd - Vds1))
Next, we need to check if our assumption about saturation was wrong or not. And if it turns out that one of the transistors is not saturated we need to repeat the calculations. Again assuming Id1 = Id2 but we need to plug the triode region equation for the one that is not in saturation.

Thank you that was a helpful answer

 

I'm glad that I could help. Any additional questions?

 

I'm glad that I could help. Any additional questions?

Thank you Jony for now i dont have further questions. I just think that calculating UDS somtimes can be quite frustrating