How does wilson current find its DC working point?

[wwwww12345]

For a normal current source , we first have a current , then the current flows through a dioded connected mos and generates a voltage vx, so vx makes M0 generate a current.

But for a wilson current mirror, how does it generates vy and vz ? Only formula
forces these two nodes to be the correct voltage? Can anyone tell me ? thank you in advance.

 

[gszczesz]

The Wilson Current mirror doesn't provide much advantage when using MOS. It's biggest strength is eliminating base current error when using BJT's.

However, here's how it functions in your configuration:

Assuming initially all devices are off, the following sequence of events will happen during start-up:

1) I4 will start increasing voltage vz.
2) When vz > VT, M5 will start increasing voltage vy.
3) When vy > VT, M3 will start to conduct current. It will keep increasing until it sinks all of I4.

The net voltage will be:
vy = vgs of M3 necessary to pass all of I4 current.
vz = vy + vgs of M5 necessary to pass all of ratioed current in M4. So if M4 = M3, it will be the vgs necessary to pass I4 current.

 

[FvM]

The Wilson Current mirror doesn't provide much advantage when uesing MOS

It also has a cascode enhanced output resistance, may be a srious advantage in some applications.

 

[wwwww12345]

The Wilson Current mirror doesn't provide much advantage when using MOS. It's biggest strength is eliminating base current error when using BJT's.

However, here's how it functions in your configuration:

Assuming initially all devices are off, the following sequence of events will happen during start-up:

1) I4 will start increasing voltage vz.
2) When vz > VT, M5 will start increasing voltage vy.
3) When vy > VT, M3 will start to conduct current. It will keep increasing until it sinks all of I4.

The net voltage will be:
vy = vgs of M3 necessary to pass all of I4 current.
vz = vy + vgs of M5 necessary to pass all of ratioed current in M4. So if M4 = M3, it will be the vgs necessary to pass I4 current.

Thank you , gszczesz! I got it .

 

[leohart]

it is a shunt series feekback,with the M5 as amplifier and current mirror formed by M3 and M4 as its feedback network.The beta is set by M3 and M4