Consider the MOS transistor shown in the figure below-Both transistors M1 and M2 have VTN-0-8 V and -nCox-30 - A-V2 - If the value of V0-2-91 V- then the ratio of WL1WL2 is equal to

Ans : (a) Given, V_GS2=V_0=2.91 V ∴ V_GS1=5 2.91=2.09 V ∴ I_D1=I_D2 thus, μ_nC_ox2(WL)_1(V_GS1 V_T)^2=μ_nC_ox2(WL)_2(V_GS2 V_T)^2 ∴(WL)_1(WL)_2=(2.91 0.8)^2(2. ...

You visited us 7 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Direct Current vs. Alternating Current

Consider the ...

Question

Consider the MOS transistor shown in the figure below:

$Both transistors M_{1} and M_{2} have V_{T N} = 0.8 V and μ_{n} C_{o x} = 30 μ A / V^{2} . If the value of V_{0} = 2.91 V, then the ratio of \frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} is equal to$

\frac{1}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{14}{9}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{8}{3}$
Ans : (a)

$Given, V_{G S 2} = V_{0} = 2.91 V$
$∴ V_{G S 1} = 5 - 2.91 = 2.09 V$
$∴ I_{D 1} = I_{D 2}$
thus,
$\frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{1} (V_{G S 1} - V_{T})^{2} = \frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{2} (V_{G S 2} - V_{T})^{2}$
$∴ \frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} = \frac{(2.91 - 0.8)^{2}}{(2.09 - 0.8)^{2}}$

$\frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} = \frac{8}{3}$

Suggest Corrections

Similar questions

Q. Consider the MOS transistor circuit shown in the figure below:

Assuming both the MOS transistor are identical with

μ_{n} C_{o x} = 2.5 μ A / V^{2}, V_{T h} = 1 V a n d \frac{W}{L} = 3

then the value of current

I_{D}

is equal to _____

μ A

Q. Consider the MOS transistor circuit shown in the figure below

The two N-MOS transistors are identical with threshold voltage

V_{T} = 1 V

and assuming the value of

λ = 0

\frac{μ_{n} C_{o x} W}{L} = 2 m A / V^{2}

, the value of drain to source voltage for transistor

T_{2} (V_{D S 2})

is equal to ____V

Q. Consider the circuit shown in the figure below:

For the transistor

T_{1}

, the value of

μ_{n} C_{o x} = 50 \frac{m A}{V^{2}} a n d V_{t} = 1 V

.

Then the value of

(\frac{W}{L})

for the transistor is
(Assuming, there is no channel length modulation)

Q. Consider the circuit shown in the figure below.

The transistors M_{1} and M_{2} are in saturation have V_{T} = 1 V, and the process transconductors parameter

k_{a}^{'} = μ_{n} C_{o} x = 100 μ A / V^{2} . If {(\frac{W}{L})}_{1} = 1.5 {(\frac{W}{L})}_{2} = 20 and λ = 0, then the value of voltage V is equal to

c o n s i d e r t h e c i r c u i t s h o w n i n t h e f i g u r e b e l o w . t h e v a l u e o f C_{1}, C_{2} a n d C_{c} i s v e r y l a r g e a n d t h e t r a n s i s t o r i s b i a s e d s u c h t h a t t h e v a l u e o f g_{m} = 20 m A / V λ = 0.01 V^{- 1} I f t h e M O S t r a n s i s t o r h a s t h e n t h e v a l u e o f v o l t a g e g a i n \frac{V_{0}}{V_{s}} i s e q u a l t o . . . . . V / V .

Join BYJU'S Learning Program

Consider the MOS transistor shown in the figure below: Both transistors M1 and M2 have VTN=0.8 V and μnCox=30μA/V2 . If the value of V0=2.91V, then the ratio of (WL)1(WL)2 is equal to

The correct option is C 83Ans : (a) Given,VGS2=V0=2.91V ∴VGS1=5−2.91=2.09V ∴ID1=ID2 thus, μnCox2(WL)1(VGS1−VT)2=μnCox2(WL)2(VGS2−VT)2 ∴(WL)1(WL)2=(2.91−0.8)2(2.09−0.8)2 (WL)1(WL)2=83

Consider the MOS transistor shown in the figure below:

$Both transistors M_{1} and M_{2} have V_{T N} = 0.8 V and μ_{n} C_{o x} = 30 μ A / V^{2} . If the value of V_{0} = 2.91 V, then the ratio of \frac{{(\frac{W}{L})}_{1}}{{(\frac{W}{L})}_{2}} is equal to$