Depletion MOSFET

A depletion MOSFET conducts current even with zero gate voltage, unlike its enhancement-mode cousin. This makes it the go-to choice for constant-current sources and self-biased amplifier stages without any external bias network.

Core Concept

A depletion MOSFET has a physical channel already diffused between drain and source during fabrication. At VGS = 0V, this channel is wide open and drain current flows. You do not need to apply any gate voltage to turn it on.

Applying a negative VGS (for N-channel) depletes carriers from this channel, narrowing it. At a sufficiently negative voltage called the pinch-off voltage Vp, the channel pinches off completely and ID drops to zero. The device can also be operated with positive VGS, enhancing the channel and increasing current beyond IDSS.

A common N-channel depletion MOSFET used in labs is the 2N3797. Its dual-mode operation, depletion for negative VGS and enhancement for positive VGS, makes it unique among FETs. This allows zero-bias operation: tie gate to source, and the device still passes its rated IDSS.

Key Equations

The drain current in the saturation region follows Shockley's equation:

Drain current: I_D = I_DSS * (1 - V_GS/V_p)^2

Where I_DSS is drain current with VGS = 0 (device fully on), V_GS is gate-source voltage in volts, and V_p is pinch-off voltage (negative for N-channel, e.g. -4V).

Transconductance: g_m = (2*I_DSS / |V_p|) * (1 - V_GS/V_p)

Maximum transconductance at VGS = 0: g_m0 = 2*I_DSS / |V_p|

Saturation condition: V_DS >= V_GS - V_p

Given:
  I_DSS = 8 mA
  V_p   = -4 V
  V_GS  = -1 V

Why this formula:
  Device is in saturation region, so Shockley equation applies.

Formula:
  I_D = I_DSS * (1 - V_GS / V_p)^2

Substitution:
  I_D = 8 mA * (1 - (-1) / (-4))^2
      = 8 mA * (1 - 0.25)^2
      = 8 mA * (0.75)^2

Calculation:
  (0.75)^2 = 0.5625
  I_D = 8 mA * 0.5625

Final Answer:
  I_D = 4.5 mA

Exam Tip: GATE frequently asks you to find VGS for a given ID, which requires solving a quadratic from Shockley's equation. Remember Vp is negative for N-channel. Students often plug in |Vp| and get the sign wrong. Also, IDSS is NOT the maximum possible current for a depletion MOSFET; positive VGS can push ID above IDSS. Confusing IDSS with absolute maximum ID is a common error.

Key Properties

• A depletion MOSFET conducts at VGS = 0V because a physical channel exists in the device structure from fabrication.
• Pinch-off voltage Vp is typically -2V to -8V for N-channel devices like the 2N3797.
• IDSS (drain-to-source current, gate shorted) is the drain current when VGS = 0V and VDS is in saturation.
• The device can operate in both depletion mode (negative VGS) and enhancement mode (positive VGS), unlike a JFET.
• Input impedance at the gate is extremely high, typically greater than 10^10 ohms, because the gate is insulated from the channel by a thin SiO2 layer.
• Zero-bias operation (VGS = 0) simplifies biasing circuits, removing the need for a voltage divider network.
• Transconductance gm varies with the operating point and is maximum at VGS = 0, equal to 2*IDSS / |Vp|.

Quick Revision

• Depletion MOSFET: ON at VGS = 0V due to pre-formed channel.
• N-channel: negative VGS depletes channel. Positive VGS enhances it.
• Shockley equation: I_D = I_DSS * (1 - VGS/Vp)^2 applies in saturation.
• Vp is negative for N-channel (e.g. -4V). Channel pinches off at VGS = Vp.
• gm0 = 2*IDSS / |Vp| is the maximum transconductance at VGS = 0.
• Gate input impedance exceeds 10^10 ohms due to SiO2 insulation.
• Common use: constant-current source, zero-bias amplifier stage.
• Exam trap: Students treat Vp as positive in the Shockley equation for N-channel, getting a current higher than IDSS, which is physically wrong for depletion-mode operation at negative VGS.