Enhancement vs depletion mode MOSFETs — Hardware notes

WHAT are we even talking about?

The word tells you the action:

Enhancement = you add carriers to make it conduct.
Depletion = you remove carriers to make it stop.

WHY does a channel form (or not)? — from first principles

Take an n-channel MOSFET (source & drain are n+ regions in a p-type body).

HOW the field works:

The gate sits on the oxide above the p-type body.
Apply a positive $V_{GS}$ . The gate becomes positively charged.
This field repels holes (majority carriers in p-body) away from the surface and attracts electrons toward it.
When enough electrons gather, a thin n-type inversion layer connects source to drain — a conducting channel.

The key distinction:

	Enhancement n-MOS	Depletion n-MOS
Channel at $V_{GS}=0$	none	pre-doped, exists
$V_{th}$ sign (n-type)	positive	negative
To turn ON	$V_{GS} > V_{th}$	already ON
To turn OFF	$V_{GS} \le V_{th}$	$V_{GS} < V_{th}$ (negative)
Default state	OFF	ON

For the p-channel versions, flip every sign: enhancement p-MOS has $V_{th}<0$ (turn ON with negative gate), depletion p-MOS has $V_{th}>0$ .

Figure — Enhancement vs depletion mode MOSFETs

WHERE $V_{th}$ sits — the transfer curve

Both modes obey the same drain current equation in saturation; only $V_{th}$ moves.

WHY squared? One factor of $(V_{GS}-V_{th})$ comes from how many carriers (charge), the second from how fast they're pushed (field). Two effects both scale linearly → product is quadratic.

The only thing that distinguishes the modes on this curve is where $V_{th}$ lands:

E-mode n-MOS: curve starts at a positive $V_{th}$ .
D-mode n-MOS: curve is shifted left, crossing a nonzero $I_D$ at $V_{GS}=0$ (its $V_{th}$ is negative).

Common mistakes (Steel-manned)

Active recall

Recall Test yourself before reading the answers

Is an enhancement MOSFET normally ON or OFF?
What sign is $V_{th}$ for a depletion n-MOS?
Why is $I_D$ a square law and not linear?
How do you turn a depletion n-MOS OFF?

Enhancement MOSFET default state (no gate voltage)

OFF — no channel exists until

V_{GS}

exceeds

V_{th}

Depletion MOSFET default state (no gate voltage)

ON — a pre-doped channel conducts at

V_{GS}=0

Sign of

V_{th}

for a depletion-mode n-channel MOSFET

Negative.

Sign of

V_{th}

for an enhancement-mode n-channel MOSFET

Positive.

What does "enhancement" physically mean?

You add/attract carriers with the gate to build a channel.

What does "depletion" physically mean?

You remove/repel carriers with the gate to empty the existing channel.

Saturation drain-current equation

I_D=\tfrac12 k (V_{GS}-V_{th})^2

for

V_{GS}>V_{th}

, else 0.

Why is drain current quadratic in overdrive?

One factor from channel charge (number of carriers), one from field-driven velocity — both linear in

(V_{GS}-V_{th})

Are mode (E/D) and channel type (n/p) related?

No — independent; all four combinations exist.

I_{DSS}

of a depletion device means

Drain current at

V_{GS}=0

(its "on" current with gate grounded).

Recall Feynman: explain to a 12-year-old

Imagine a water pipe with a squeeze-clamp. In an enhancement MOSFET the pipe is squeezed shut by default — you have to push the gate button to open it and let water flow. In a depletion MOSFET the pipe is open by default — water already flows, and you push the gate button to pinch it closed. Same pipe, same water rule; the only difference is whether it starts open or closed.

Connections

MOSFET structure and operation — the oxide/inversion physics behind both modes
Threshold voltage Vth — the single number that sets the mode
BJT vs MOSFET — voltage-controlled vs current-controlled
CMOS logic — uses complementary enhancement n- and p-MOS
MOSFET biasing and load lines — where the transfer curve meets circuit constraints
JFET — inherently depletion-mode (compare)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, MOSFET basically ek voltage-controlled switch hai — gate pe voltage lagao aur drain-source ke beech current control karo. Enhancement aur depletion mode ka farak sirf ek sawaal ka jawab hai: jab gate voltage zero ho, tab channel pehle se bana hua hai ya nahi? Enhancement mode me channel hota hi nahi — usko banana padta hai gate voltage se (isliye "enhance"), isliye ye normally OFF hai. Depletion mode me channel factory se hi doped rehta hai, current already flow karta hai zero gate pe, aur tumhe usko hataana padta hai (deplete) band karne ke liye — ye normally ON hai.

Sabse important cheez: n-channel me enhancement ka $V_{th}$ positive hota hai aur depletion ka negative. Yaad rakho — "E is Empty, D is Doped". Threshold ka sign hi mode bata deta hai. Bahut students galti karte hain ki depletion ko p-channel se jod dete hain — nahi bhai, mode (E/D) aur channel type (n/p) do alag baatein hain, chaaron combination possible hain.

Current ka formula dono ke liye same: $I_D=\frac12 k (V_{GS}-V_{th})^2$ . Square kyun? Kyunki ek $(V_{GS}-V_{th})$ carriers ki sankhya se aata hai aur doosra unki speed (field) se — dono linear, isliye product quadratic. Bas ek rule: pehle check karo device ON hai ya nahi — agar $V_{GS} \le V_{th}$ to $I_D=0$ , formula lagane se pehle region confirm karo, warna galat answer aayega.

Yeh concept CMOS, amplifiers aur real circuit design me daily kaam aata hai — normally-ON depletion device switch ki tarah, normally-OFF enhancement device logic gates me. Samajh gaye to aadha transistor chapter clear ho jata hai.