- #1

Quincy

- 228

- 0

There's a question in my Microelectronic Circuits book involving this circuit:

where it states that

voltages at the drain of Q_{2}can be as high as +1.3 V and voltages at the drain of Q_{5}can be as low as -1.3 V.

Part of the solution involves this equation:

V

_{OV4}= V

_{D4}- V

_{S4}(equation for overdrive voltage of Q

_{4})

V

_{OV4}= -1.5 - (-1.3)

My question is: Why is V

_{D4}equal to -1.5 V and why is V

_{S4}equal to -1.3 V? -1.3 V is the voltage at the drain of Q

_{5}, how did it become the voltage at the source of Q

_{4}?