JFET: The Pinch-off Voltage

The Pinch‒off Voltage V_P

• Fig. 5.6.1 shows a single‒ended‒geometry junction FET. Here, the substrate is of p‒type material onto which an n‒type channel is epitaxially grown. A p‒type gate is then diffused into the n‒type channel. In this device, a slab of n‒type semiconductor is sandwiched between two layers of p‒type material, forming two p‒n junctions.

• Let us assume that the p‒type region is doped with N_A acceptors per cubic meter, n‒type region is doped with N_D donors per cubic meter and that the junction formed is abrupt.

• We know that, if N_A >> N_D we have W_P << W_n and for the space‒charge width, W_n(x) = W(x) at a distance x along the channel in Fig. 5.6.2.

              …………... (5.6.1)

where

ε : Dielectric constant of channel material

q : Magnitude of electric charge

V_O : Junction contact potential at x

V(x): Applied potential across space‒charge region at x and is a negative number for an applied reverse bias

 a ‒ b(x): Penetration W(x) of depletion region into channel at a point x along channel

• When the drain current is zero, b(x) and V(x) are independent of x and b(x) = b. By substituting b(x) = b = 0 we have

     ………… (5.6.2)

• Now solving for V with assumption |V₀| << |V| we obtain the pinch‒off voltage V_P (the diode reverse voltage that removes all the free charge from the channel) as

     ………… (5.6.3)

• If we substitute V_GS for V_O ‒ V(x) and b(x) = b in equation (5.6.1) we get,

     ………… (5.6.4)

• From equation (5.6.3) we have

     ………… (5.6.5)

Substituting equation (5.6.5) in equation (5.6.4) we get,

...(5.6.6)

• This equation shows that the voltage V_GS (the reverse bias across the gate junction) is independent of distance along the channel if I_D = 0.

Ex. 5.6.1: For an n‒channel silicon FET, find the pinch‒off voltage and the channel half‒width.

Assume: a = 3 × 10^‒6 m, N_D = 10²¹ electrons/m², V_GS= 1/2 V_P, I_D = 0 and relative dielectric constant of silicon = 12.

Solution:

Since relative dielectric constant of silicon is 12 we have ε = 12ε₀. Using values of q and ε₀ we have

V_P = [ 1.6×10^‒19×10²¹×(3×10^‒6)² ] / [2×12×(8.85×10^‒12)] = 6.8 V

From equation (5.6.6) we have

 V_GS = (1 ‒ (b/a))² V_P

(1 ‒ (b/a))² = V_GS / V_P = 1/2

 1 ‒ (b/a) = 0.707

 b = (1 ‒ 0.707)a

= (0.293) (3 × 10⁻⁶) = 0.879 m

This result shows that the channel width is reduced to about one‒third its value for

V_GS = ½ V_P

Ex. 5.6.2: Assume that the p+n junction of a uniformly doped silicon n channel JFET at T = 300 K has doping concentrations of N_a =10¹⁸ cm and N_d = 10¹⁶ cm⁻³. Assume that the metallurgical channel thickness a is 0.7 μm Calculate the pinch off voltage. Also calculate pinch‒off voltage if the metallurgical channel thickness a is 0.75 μm.

Solution:

Here all quantities are in centimeters instead of meters. Channel thickness a = 0.7 μm = 0.7×10^‒4 cm. The internal pinch off voltage is given by

Review Question

1. Derive the expresion for pinch‒off voltage.