Consider the circuit shown below Both the transistors have parameters as follows VTN-0-8 volts-k-n-30- A-V2- If width to length ratios of M1 and M2 areWL1-WL2-40- The output voltage V0 is Volts-

Since both the transistors are identical and work in saturation. Thus K^'_n_1(WL)_1(V_GS_1 V_Th)^2=K^'_n_2(WL)_2(V_GS_2 V_Th)^2 K^'_n_1(V_GS_1 V_Th)=K^'_n_2(V ...

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Consider the ...

Question

Consider the circuit shown below

Both the transistors have parameters as follows $V_{T N} = 0.8 v o l t s, k_{n}^{^{'}} = 30 μ A / V^{2}$ . If width to length ratios of $M_{1} a n d M_{2} a r e {(\frac{W}{L})}_{1} = {(\frac{W}{L})}_{2} = 40.$ The output voltage $V_{0}$ is ________ Volts.

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Solution

Since both the transistors are identical and work in saturation. Thus

$K_{n_{1}}^{^{'}} {(\frac{W}{L})}_{1} (V_{G S_{1}} - V_{T h})^{2} = K_{n_{2}}^{^{'}} {(\frac{W}{L})}_{2} (V_{G S_{2}} - V_{T h})^{2}$

$K_{n_{1}}^{^{'}} (V_{G S_{1}} - V_{T h}) = K_{n_{2}}^{^{'}} (V_{G S_{2}} - V_{T h})$

$∵ K_{n_{1}}^{^{'}} = K_{n_{2}}^{^{'}}$

$\Rightarrow V_{G S_{1}} - V_{T h} = V_{G S_{2}} - V_{T h}$

$V_{G S_{1}} = V_{G S_{2}} a s V_{T h_{1}} = V_{T h_{2}} = V_{T h}$

$\Rightarrow 5 - V_{0} = V_{0}$

$\Rightarrow V_{0} = 2.5 V$

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Consider the circuit shown below Both the transistors have parameters as follows VTN=0.8 volts,k′n=30μA/V2. If width to length ratios of M1 and M2 are(WL)1=(WL)2=40. The output voltage V0 is ________ Volts.

Since both the transistors are identical and work in saturation. Thus K′n1(WL)1(VGS1−VTh)2=K′n2(WL)2(VGS2−VTh)2 K′n1(VGS1−VTh)=K′n2(VGS2−VTh) ∵ K′n1=K′n2 ⇒ VGS1−VTh=VGS2−VTh VGS1=VGS2 as VTh1=VTh2=VTh ⇒ 5−V0=V0 ⇒ V0=2.5 V

Consider the circuit shown below

Both the transistors have parameters as follows $V_{T N} = 0.8 v o l t s, k_{n}^{^{'}} = 30 μ A / V^{2}$ . If width to length ratios of $M_{1} a n d M_{2} a r e {(\frac{W}{L})}_{1} = {(\frac{W}{L})}_{2} = 40.$ The output voltage $V_{0}$ is ________ Volts.