Hi, I have to find Vo and I did it, but I can not prove that M4 and M5 are in saturation(active mode), this is what I did

 

Are you sure that M5 isn't supposed to be diode connected?

 

yes, that is the circuit, everything is fine

 

yes, that is the circuit, everything is fine

Then how is the gate voltage on M5 and M6 being established? That node is floating. Have you tried simulating that circuit as drawn?

 

I think that in this case you don't need to establish a voltage directly in the gate because the purpose is to make a current mirror between M5 and M6 ..... Yes I did a simulation but the results are absurd

250MV ...

 

Perhaps the results are absurd because the node connecting the gates of M5 and M6 are floating and therefore undefined. The fact that the simulation even converged is a bit surprising.

How many mirror circuits have you seen in which gate nodes are floating?

Try diode-connecting M5 (just to humor me, if nothing else) and see what happens.

While you're at it, look at how you are connecting the body taps on your PFETs. Remember, you want the channel-body diode junctions to be reverse biased at all times.

Also, note that you don't need six 9 V supplies, just two.

 

But look that M5 and M6 have the same source gate voltage, so if you suppose that they are in active mode, then they form a current mirror

 

Note that I didn't ask you to replace M5 with a diode, I asked you to diode connect it. Very different thing -- especially when dealing with FETs. Even so, look at how your simulation values are MUCH more reasonable. Just like the other diode-connected FETs in that circuit, connect the gate of M5 to the drain of M5.

The node that connects the gates of M5 and M6 is floating. There is no way to establish a gate-source voltage based on the current in one that can then be transferred to the gate of the other -- that is how a mirror works!

You still have not corrected the body connection on M6. The symbols you are using indicate that the model internally connects the body to the source. This makes the FET asymmetric and it matters which way you connect the source and the drain. For a PFET you want the source to be at the higher potential (specifically so that you don't forward-bias the channel-body junction).

 

1. I don't understand what you mean with diode-connecting M5 (You are referring to connect the gate to the drain?)
2. Ok
3. In DroitTesla(The simulator) there isn't way to disconnect the body

 

 

1. I don't understand what you mean with diode-connecting M5 (You are referring to connect the gate to the drain?)
2. Ok
3. In DroitTesla(The simulator) there isn't way to disconnect the body

1) Yes - A diode-connected BJT has the base connected to the collector while a diode-connected FET has the gate connected to the drain. In both cases you end up with a two-terminal device that has a diode-like IV characteristic.

2) Don't know what this is referring to.

3) I'm not asking you to disconnect the body. As I said, the symbol you are using indicates that the body is internally connected to the source. But that means that it is YOUR responsibility to properly connect the source and the drain -- they are NOT interchangeable. Your PFETs are upside down! Look at the body diode in the symbol -- that diode needs to be reverse biased, which for the PFET means that the pin it is connected to (which defines it as the source pin) has to be connected to the higher potential (i.e., the +9V supply in your schematic).

 

Ok summarizing the circuit does not work because the gate node of M5 and M6 is floating... I don't understand why a PhD teacher put this exercise in class

 

I don't understand why a PhD teacher put this exercise in class

Are you sure that you "copy" a circuit in the right way ? I'm sure that your teacher give you this circuit

 

I don't understand why a PhD teacher put this exercise in class

You are far more likely to find morons and caoon artists among those highly educated than among the common people (aka rednecks).

 

Vds(m1)=Vds(m2)=Vds(m3), id(m3)=id(m7) ==> Vds(m3)=Vds(m7)=(9V-(-9V))/3=6V ==>Vo=-9V+6V=-3V