[SOLVED] Question about Class A CMOS output stage

[petelee]

Hi,

I'm not quite understanding this text book description. :<

Regarding the attached snapshot, it says "A source follower is shown, biased at 0.1mA. A low resistor of 50ohm is connected to it. It's is clear that the maximum output voltage swing can only be 5mV. For higher voltages, we would need higher biasing current as well.".

To get 5mV, it seems like that they multiply 0.1mA and 50ohm. I don't understand why it should be. Output voltage swing is also determined by "In" input swing?

 

[rsashwinkumar]

First of all, be clear that a MOS or any transistor CAN NEVER deliver power of their own. They only convert the DC power to AC power. So, the maximum current that could be supplied to the load is the DC current itself and so it follows that the max output voltage can be 0.1mA*50ohms which is 5mV.

 

[yuvan]

Hi petlee,
Vout will follow vin as long as the transistor is in proper operating region & body effect; because as far as I can see all the equations are for large signal; The load res will be supplied with current from supply and not from IB for DC operation.

 

[petelee]

Thanks all for your comments. I think I got it.

 

[steveelliott]

How about?
When Vin goes low, at some point the FET turns off. When the FET turns off, the only current flowing is the IB current. That gives you the the 0.1mA times 50ohms = 5mV that you properly noted in your original post.

- - - Updated - - -

How about?
When Vin goes low, at some point the FET turns off. When the FET turns off, the only current flowing is the IB current. That gives you the the (-)0.1mA times 50ohms = -5mV that you properly noted in your original post.

 

[yuvan]

When FET turns off what is the possibility for IB in first place?

 

[petelee]

Thanks steveelliott, It's more clear explaining that way. When FET is off, the ideal current source IB pulls down the output voltage by IB*RL.