11/20/15

1

ECE 535 Notes for Lecture # 26

Class Outline: • Ballistic Field-Effect Transistor

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 1 In this lecture, we examine the ballistic field-effect transistor…

•  Assume that the source is grounded.

•  A threshold voltage Vt is needed to create a 2DEG.

Once created, the 2DEG forms the lower plate of a parallel plate capacitor with the gate metal.

Changes in charge density: True when turned on only.

•  Quantum capacitance arises because DOS of the semiconductor band is lower than the metal forcing the finite voltage drop in the semiconductor to hold charge.

•  It may be pictured as a spread in the 2DEG where the centroid is not at the surface.

11/20/15

2

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 2 When the 2DEG is present, the electrons are carrying current and the physics is best understood in k-space.

•  When VGS > VT, we create a quantum well resulting in ground state energy, Enz.

Total energy per subband:

The F-D distribution dictates the carrier distribution of the 2DEG in k-space:

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 3 Since the Fermi level is controlled by the gate alone when VDS = 0, we can rewrite things as a function of VGS.

Find the area of the occupied states:

How do we solve this equation?

substitutions

Obtained by integrating over theta in the above expression.

11/20/15

3

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 4

We can identify some different quantities:

Zeroth order F-D with j = 0.

VGS tunes the Fermi level of the 2DEG as:

Therefore:

Fermi level changes linearly with gate voltage at high gate overdrive.

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 5 If we want the subthreshold characteristics of a ballistic MOSFET, then we need to modify some things.

The RHS is always positive, but the LHS is not when VGS < VT. Let’s rewrite the division of voltage drops.

VT absorbs the surface barrier height, conduction band offset between the barrier and the semiconductor, and energy quantization.

Relation between ns and VGS then becomes:

11/20/15

4

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 6

Within this, we can find two characteristic densities that simplify the equations.

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 7

11/20/15

5

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 8 Now let’s focus on the behavior of the on-state of the transistor. Let’s begin by calculating the left and right going contributions to the current at VDS = 0.

We have an expression for the current:

We have assumed that the transmission probability is unity and d = 2 for the 2DEG.

•  Clearly evaluating this integral will lead to a null result.

•  Let’s evaluate the current moving in +k.

This means we need to restrict the plane:

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 9

•  The prefactor has units of A/m.

•  We can also think of the right going carriers as being in equilibrium with the source.

•  The left going carriers are in equilibrium with the drain.

Net current is zero

Net current is zero

The balance between left and right going carriers is broken via the application of an applied voltage.

11/20/15

6

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 10 When we apply a drain voltage, the energy band diagram now becomes position dependent:

•  The maximum in potential is referred to as the “top of the barrier”.

•  We will focus on the plane that

contains the maxima in the x-direction.

Therefore,

Where we have used the fol lowing substitutions:

and

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 11 With these new definitions, we can define the new total current:

Right going left going

•  In order to evaluate the current, we need to understand the dependence of ηS on the gate and drain voltages.

•  Let’s assume that the 2DEG density is completely controlled by the gate capacitance.

•  This means that the 2DEG density in the TOB plane has not changed from VDS = 0.

•  This further means that we are ignoring short-channel effects in our analysis.

11/20/15

7

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 12 Just like for the current, we split the carrier distribution equation into right and left going carriers.

Original

Where we may identify:

Making the substitutions, we get:

Which is a quadratic equation, so we solve for ηS:

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 13 Our expressions for the charge and the current provide us with the complete on-state output characteristics at any temperature.

•  If we want the off-state, then we need to solve the charge self-consistently using:

So the total expression for the current, both on and off state, is:

Where we have replaced:

11/20/15

8

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 14

Subthreshold slope: When VDS >> kT:

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 15

•  Current is higher in silicon because it takes advantage of the multiple valleys.

•  At high-drain bias, the on current is larger for materials with smaller effective mass due to the higher velocity.

•  This effect, however, leads to the need for larger voltages to achieve saturation.

•  The linearity of the ID-VDS curves is due to the limited number of k-states that are available for transport.

11/20/15

9

M.J. Gilbert ECE 535 – Lecture 26

Ballistic Field-Effect Transistor - 1