What does VDD represent in a BJT-MOSFET circuit with current source?

[etf]

Hi!
Here is my task:
a) Calculate and sketch Iout=f(Vc2), for 0<Vc2<1.5V. (neglect Early effect)
b) calculate output resistance for Vc2=Vc1.
BJT's Q1 and Q2 are in forward active mode, have identical characteristics and same temperature. MOSFET M3 is in saturation. Iin=10μA, Vt=25mV, Vc1=1.25V, VA=50V.

My question is, what does VDD represent? There are few examples in my book where VDD is connected to some current source like this...

 

[CWatters]

VDD will be a +ve DC supply voltage. Your text refers to VA =50V. My guess is this is a typo and VDD is meant to be VA?

 

[etf]

I don't think so. VA is Early voltage.

 

[cpscdave]

Generally with questions like this VDD is the biasing voltage source that puts the amplifer into the active region. My prof always used VCC for it :)

 

[CWatters]

I don't think so. VA is Early voltage.

I think you're right.

The actual value of VDD isn't really important in this circuit as there is a current source in series.

 

[donpacino]

In real life that current source will be a transistor network. It it frequently represented as a current source to simplify analysis.