This question is testing whether you remember which MOSFET capacitance dominates once the transistor is in saturation, and why. Let's check each option against how the channel behaves in saturation.
- \(C_{GS} > C_{GD} > C_{DS}\): In saturation, the channel pinches off near the drain, so almost the whole inversion layer sits under the gate on the source side. That makes \(C_{GS}\) by far the biggest term. \(C_{GD}\) is left as just a small gate-drain overlap capacitance, bigger than \(C_{DS}\), which has no gate-oxide contribution at all and comes only from the drain-body junction. This ranking matches the physics.
- \(C_{GD} > C_{DS} > C_{GS}\): This puts \(C_{GS}\) last, but \(C_{GS}\) carries almost the entire channel charge in saturation, so it cannot be the smallest term. Wrong order.
- \(C_{DS} > C_{GD} > C_{GS}\): This makes the junction-only capacitance \(C_{DS}\) the largest, but a junction capacitance with no gate-oxide contribution is normally the smallest of the three. Wrong order.
- \(C_{GD} > C_{GS} > C_{DS}\): This ranks the small drain-overlap capacitance above \(C_{GS}\), but \(C_{GS}\) always dominates in saturation since it carries the bulk of the channel charge. Wrong order.
Only the first option puts \(C_{GS}\) on top, \(C_{GD}\) in the middle, and \(C_{DS}\) at the bottom, which is exactly how these capacitances behave once the MOSFET enters saturation.
Let's summarize:
- \(C_{GS}\) is largest because the pinched-off channel's charge collects almost entirely near the source in saturation.
- \(C_{GD}\) is a small overlap-only term since the channel does not reach the drain in saturation.
- \(C_{DS}\) is smallest since it is a pure junction capacitance with no gate-oxide coupling at all.
So the correct ranking is \(C_{GS} > C_{GD} > C_{DS}\), option (A).