firsttimedesigning
Full Member level 1
When Vout2 is much bigger than Vout1, what happens to the circuit? I think Ix will increase because Ix = (Vout2 - Vout1) / (R1+R2).
However, the books says that Ix will decrease and so is the drain current of transistor 7 (Id7) "I1 must sink both Ix and Id7"
but I dont see why...
The book also says that if R1+R2 or I1 is not large enough, then Id7 will drop to 0.
So I think that if R1+R2 is small then Ix will increase. Assume I1 stays constant, Id7 will drop 0. Is that right?
Also, when I1 is small, since Ix is large due to the fact that Vout2 is bigger than Vout1, Id7 decreases to 0. But I am not sure if that is right...
However, the books says that Ix will decrease and so is the drain current of transistor 7 (Id7) "I1 must sink both Ix and Id7"
but I dont see why...
The book also says that if R1+R2 or I1 is not large enough, then Id7 will drop to 0.
So I think that if R1+R2 is small then Ix will increase. Assume I1 stays constant, Id7 will drop 0. Is that right?
Also, when I1 is small, since Ix is large due to the fact that Vout2 is bigger than Vout1, Id7 decreases to 0. But I am not sure if that is right...