Download Saturated Load Noise Margin Analysis

Supplement S6.1 - Noise Margins for the Saturated Load Inverter
We will now calculate the values of the VIL, VOH, VIH and VOL for the inverter with a
saturated load device. Remember that VIH and VIL are defined by the points in the
transfer function at which the slope is equal to -1. In Fig. S6.1.1 the slope of the transfer
function abruptly changes as MS begins to conduct at the point where vI = VTNS. This
point defines VIL:
VIL = VTNS = 1 V for
VOH = VH = VDD - VTNL = 3.39 V.
5.0V
NMOS Inverter with Saturated Load
4.0V
O
VOH = VH
u
t
p
u 3.0V
t
V
o
l
t
a 2.0V
g
e
1.0V
VOL
VL
0V
0V
VIL
1.0V
VIH
2.0V
v
3.0V
VH
4.0V
5.0V
6.0V
I
Figure S6.1.1 - PSPICE simulation of the voltage transfer function for the NMOS inverter with
saturated load
S6.1-1
Next let us find VIH. In order to find a relationship between vI and vO, we
observe that the drain currents in the switching and load devices must be equal. At vI =
VIH, the input will be at a relatively high voltage, and the output will be at a relatively
low voltage. Thus, we can guess that MS will be in the linear region, and we already
know that the circuit connection forces ML to operate in the saturation region. Equating
drain currents in the switching and load transistors:
iDS = iDL
Ê
v ˆ
K
2
K S Á v I - VTNS - O ˜vO = L (VDD - vO - VTNL )
Ë
2¯
2
with
(S6.1.1)
ÊW ˆ
K S = K n' Á ˜
Ë L ¯S
†
and
ÊW ˆ
K L = K n' Á ˜
Ë L ¯L
∂v
The point of interest is O = -1, but solving for the value of vO can be quite tedious.
∂v I
†
Since the derivatives are smooth, continuous and non-zero, we will assume that
-1
∂vO Ê ∂v I ˆ
=Á
˜ ,†and will solve for vI in terms of vO:
∂v I Ë ∂vO ¯
v I = VTNS +
†
or
†
vO
1
K
2
+
(VDD - vO - VTNL ) where K R = S
2 2K R vO
KL
2
˘
vO
1 È (VDD - VTNL )
Í
v I = VTNS +
+
- 2(VDD - VTNL ) + vO ˙
2 2K R ÍÎ
vO
˙˚
and
(S6.1.2)
2
˘
∂v I 1
1 È (VDD - VTNL )
Í˙
@ +
+
1
∂vO 2 2K R ÍÎ
vO2
˙˚
In this last expression, the dependence of VTNL on vO has been neglected for simplicity.
(This approximation
will be justified shortly.)
†
Setting the derivative equal to -1 at vI = VIH yields
S6.1-2
2
˘
1
1 È (VDD - VTNL )
Í˙
-1 = +
+
1
2 2K R ÍÎ
vO2
˙˚
and solving for VOL = vO for vI = VIH yields:
VOL
†V - VTNL
= DD
1+ 3K R
and
V
(V - VOL - VTNL )
+ OL + DD
2
2K RVOL
VIH = VTNS
2
(S6.1.3)
For†the inverter design of Fig. 6.26 with VTNL = 1 V, we find
VOL =
VDD - VTNL
=
1+ 3K R
VIH = 1+
VDD - VTNL
1+ 3
(W /L) S
(W /L) L
=
(5 -1)V
= 0.66V
1+ 3( 3.53)( 3.39)
0.66V
1
1
2
+
(5 - 0.66 -1) V 2 = 2.04V
2
2( 3.53)(3.39) 0.66V
With these values we can check our assumption of the operating region of MS:
VGS - VTNS = 2.04 - 1 = 1.04 V and VDS = 0.66 V .
†
Since VDS < VGS - VTN, the linear region assumption was correct.
In Fig. S6.1.1, it can be seen that the calculated values of VIH and VIL agree well
with SPICE simulation results. Thus, the approximation of neglecting the voltage
dependence of the threshold voltage of the load device does not introduce significant
error.
However, if we so desire, we can use an iterative procedure to improve on the
estimates above. The calculated value of VOL can be used to determine a better estimate
for VTNL, and this new value of VTNL can then be used to improve the estimate of VIH.
For VOL = 0.66V,
VTNL = 1V + 0.5 V
(
(0.66 + 0.6)V
)
- 0.6V = 1.17 V
and the new values of VOL and VIH from Eqn. (S6.1.3) are equal to
†
VOL = 0.63 V
and
S6.1-3
VIH = 1.99 V .
This process can be continued, but it has already converged as indicated in Table 7.2
below. (Note that this procedure is another example of an algorithm that is very easy to
implement using MATLAB or a spreadsheet.)
Table S6.1.1 - Iterative Update of VOL and VIH
Iteration #
VOL
VTNL
V'OL
VIH
0
---
1.00 V
0.66 V
2.04 V
1
0.66 V
1.17 V
0.63 V
1.99 V
2
0.63 V
1.17 V
0.63 V
1.99 V
NML = VIL - VOL = 1 - 0.63 = 0.37 V
NMH = VOH - VIH = 3.39 – 1.99 = 1.40 V
S6.1-4