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Unformatted text preview: 5.1 Design a resistiveload inverter with R = 1kg such that VOL = 0.6 V when an
enhancementtype nMOS driver transistor has the following parameters:
 VDD = 5.0 V
 Vm = 1.0 V
 'y: 0.2 V”2
' l = 0.0 V"
' k' = 22.0 W2
(a). Determine the required aspect ratio, W/L
(b). Determine V“ and V”
(0). Determine noise margins NM L and NMHr SOLUTION:
(a) When Vm = VOL. Vin = VD”: 5.0V, the driver transistor operates in linear region. Using
Eq.(5.12) . v —v ' w
[a = 4,91% = ETiZWOII ' Vro)V0L ‘ VOLZ] 5—0.6 22x10'6W 2
—=——2s1 .6—0.
1000 2 L[( )0 6]
SolveforWL E=90.1
L (b)When Vin = V1], driver transistor Operates in saturation region. Using Eq.(5.19) Va ‘ V95; =kyﬂ (th ‘me RL L 2 (iv Taking derivative with respect to Vin and set (Wmr —l v“ uvm I de W l
‘—' =k—(VIL'VT0)=_“
RL dig” mm L Rt
Thus.
1
V =V +———=1.0+—————=L5V
"— 7" R JEE 100022x10*.90.1 [ 1
L L
Vin = VI“, driver transistor operates in linear region. Using Eq.(5.24) repeated
here:
VDD'”V  k. W 2
TﬂL_?Z[2(Hn_VT0)meVw ]
. . . . . . dvour —_1
Dlﬁerenualmg both Sides wuh respect to Vm and set Wm ' ,
Vk=Vm
1 GM! , k' W (W (W
—— 0“ =—— 2(VmVm M +2vw2vm M
RL dun Vu‘ﬁu 2 dun Eur“H 0 a db?” Vh=ﬁﬂ é=k¥l<wrwen+2vm1 1
W
Ric'—
L L Plug in back to Eq.(5.24) repeated ane to solve for Vm “H = Vro +2Vm — VDD_VM=£E 1 RL 2 L VTﬂ+zvm_m_VT0 Vow—Vow:
L
V I = 2.122.—
“ Vau‘vm 3 W
RL'k': s 5 1
=L0+J————————=109V
3100022x10‘690J 100022x10'690J [ l (C) NML= vIL  V0515 015 = 0.9 [V]
NMH= 11W  v,” = 5.0  3.09 =1.91 [V] ...
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This note was uploaded on 04/10/2008 for the course ECE 1238 taught by Professor Kim during the Spring '08 term at Pittsburgh.
 Spring '08
 Kim

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