384 CHAPTER 8. FIELD EFFECT TRANSISTORS
- The electric field in the buffer (or bulk) is zero.
- The voltages in the system are specified
The first condition allows voltages in the system to adjust and the second allows charges and
hence fields to adjust. Both conditions should not be applied simultaneously. We can now use
the band diagram in figure 8.15c to calculate the charge in the 2DEG as a function of gate bias
VG. The methodology is to follow the energy bands from the Fermi level in the metal to that in
the GaAs and set the difference equal to the gate biasVG. After dividing by the electron charge
e, we get the following equation:
−VG+φb−V 1 +V 2 −ΔEc
e+Vdi−=0 (8.5.14)V 1 andV 2 are found by solving Poisson’s equation and are given by
V 1 =
enm(d−ds)
(8.5.15)
V 2 =
ensds
(8.5.16)
Substituting the relationships from equation 8.5.6 and equation 8.5.7 into equation 8.5.5 and
rearranging terms, we get
ens(d+Δd)
−
eNd+(d−ds)
−[VG−(φb−ΔEc/e)] = 0 (8.5.17)From figure 8.15a, we see thatd−ds=dδis the distance between the gate and theδ-doped
layer, andd+Δd=Dis the distance between the gate and the 2DEG. Solving fornsgives us
ns(VG)=eNd+dδ+[VG−(φb−ΔEc/e)]
eD(8.5.18)
The termNd+(dδ/D)in our expression fornsdepicts what is known as theLeverRule for
charge sharing. To illustrate its impact, consider the special case whereφb−VG=ΔEc/e.
When theδ-doped layer is half way between the gate and the channel(dδ=D/2), the charge
is shared equally between the gate metal and the 2DEG(ns=nm=Nd+/2). When theδ-
doped layer is brought closer to the metal, more of the charge is imaged on the gate; asdδ→ 0 ,
nm→Nd+. Similarly, as theδ-doped layer is brought closer to the channel, more of the charge
is imaged in the 2DEG. The change in the charge in the 2DEG can be related to the gate voltage
as
Δns=ns(Vg=0)−ns(Vg)
=Nddz−/eΔ(0)
D−
Nddz−/eΔ(Vg)
D
or
eΔns=
D
·[Δ (Vg)−Δ(0)]