Consider the circuit shown in the figure below-If WL 1 - 1-75 and WL 3 - 27-8- then the value of WL 2 - Assume -nCox - 86 - A-V2 and VTN - 1 V for all the transistors

All the transistors are in saturation region, thus I_D = μ_n C_ox2 ( WL )_3 (V_GS3 – V_TN)^2 =362 × 10^ 6× (27.8) × (2 1)^2 I_D = 0.5 mA Now, since all the tran ...

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Standard XII

Physics

Transistor: Working

Consider the ...

Question

Consider the circuit shown in the figure below:

If ${(\frac{W}{L})}_{1} = 1.75$ and ${(\frac{W}{L})}_{3} = 27.8$ , then the value of ${(\frac{W}{L})}_{2} =$ ____. (Assume $μ_{n} C_{o x} = 86 μ A / V^{2} a n d V_{T N} = 1 V$ for all the transistors)

6.83

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Solution

The correct option is A 6.83
All the transistors are in saturation region, thus
$I_{D} = \frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{3} (V_{G S 3} - V_{T N})^{2}$
$= \frac{36}{2} \times 10^{- 6} \times (27.8) \times (2 - 1)^{2}$
$I_{D} = 0.5 m A$
Now, since all the transistors are connected in series, thus
$I_{D} = \frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{1} (V_{G S 1} - V_{T})^{2}$
$0.5 \times 10^{- 3} = \frac{36 \times 10^{- 6}}{2} (1.75) (V_{G S 1} - V_{T})^{2}$
$0.5 m A = \frac{36 \times 10^{- 6}}{2} (1.75) [10 - V_{x} - 1]^{2}$
$V_{x} = 5.01 V$
$I_{D} = \frac{μ_{n} C_{o x}}{2} {(\frac{W}{L})}_{2} (V_{G S 2} - V_{T})^{2}$
$0.5 \times 10^{- 3} = \frac{36 \times 10^{- 6}}{2} {(\frac{W}{L})}_{2} (V_{x} - 2 - 1)^{2}$
${(\frac{W}{L})}_{2} = 6.83$

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Consider the circuit shown in the figure below: If (WL)1=1.75 and (WL)3=27.8, then the value of (WL)2=____. (Assume μnCox=86 μA/V2 and VTN=1V for all the transistors) 6.83

Consider the circuit shown in the figure below:

If ${(\frac{W}{L})}_{1} = 1.75$ and ${(\frac{W}{L})}_{3} = 27.8$ , then the value of ${(\frac{W}{L})}_{2} =$ ____. (Assume $μ_{n} C_{o x} = 86 μ A / V^{2} a n d V_{T N} = 1 V$ for all the transistors)

6.83