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<p>3.38 (∴I = 50 μA)</p>
<p>to find small signal reponse v<sub>0 </sub>, open the dc current source I and short the capacitive c<sub>1 </sub>and c<sub>2 </sub>also replace the diode with its small signla resistance:</p>
<p>r<sub>d </sub>= n V<sub>t </sub>/<sub> </sub>I (n=2)</p>
<p>v<sub>o</sub> = v<sub>s</sub> * r<sub>d</sub> / r<sub>d</sub>+R<sub>s</sub></p>
<p>= ( v<sub>s</sub> * n V<sub>t</sub> / I ) / ( nV<sub>t</sub> / I + R<sub>s </sub>) = v<sub>s</sub> n V<sub>t</sub> / ( n V<sub>t</sub> + I * R<sub>s</sub>)</p>
<p> </p>
<p>v<sub>o</sub> = 10 mV * 0.05 / ( 0.05 + 10<sup>3</sup> I ) = 0.476 mV ~ I = 1 mA</p>
<p> 3.333 mV ~ I = 0.1 mA</p>
<p> 9.804 mV ~ I = 1 μA</p>
<p> </p>
<p>for v<sub>o</sub> - 1/2 * v<sub>s</sub> = v<sub>s</sub> * 0.05 / (0.05 + 10<sup>3</sup>) I</p>
<p>∴ I = 50 μA</p>
<p> </p>
<p>3.57</p>
<p>(a) (∴ 아래에 있습니다) </p>
<p> </p>
<p>v<sub>o</sub> / v<sub>i </sub>= R / ( R + (2r<sub>d</sub> // 2r<sub>d</sub>) ) = R / ( R + r<sub>d )</sub></p>
<p> </p>
<p>where r<sub>d</sub> = V<sub>t</sub> / ( I / 2 ) = 2 V<sub>t</sub> / I = 0.05 V / I</p>
<p> </p>
<p>∴</p>
<p>I : v<sub>o</sub> / v<sub>i</sub></p>
<p>0 : 0</p>
<p>10<sup>-3</sup> : 0.167</p>
<p>10<sup>-2</sup> : 0.667</p>
<p>10<sup>-1 : </sup> 0.952</p>
<p>1.0 : 0.995</p>
<p>10 : 0.9995</p>
<p> </p>
<p>(b) </p>
<p>if the signal current is to be lemited to ± 10 I , the chanse in diode voltage △v<sub>d</sub> can be found from i<sub>d</sub> / I = e <sup>△vd / n Vt </sup>=> 0.9 to 1.1</p>
<p>thus, for n=1</p>
<p>△v<sub>d</sub> = -2.63 mV to + 2.38 mv or approximately ±29.5 mV</p>
<p> </p>
<p>for the diode current to remain within ±10% of their dc bias currents the signal voltage across each diode must be limited to 2.5 mV</p>
<p>now if v<sub>i peak</sub> = 10 mV</p>
<p>we can obtain the following situation</p>
<p> </p>
<p>figure</p>
<p> </p>
<p>we see that v<sub>o</sub> = 5 mV and i = 5 mV / 10kΩ = 0.5 μA</p>
<p> </p>
<p>thus, each iode is carrying a current signal of 0.25 mA for this to at most 10% of the dc current, the dc current in each diode must be al least 2.5 μA it follows that the minimum value of i must be 5 μA</p>
<p> </p>
<p>(c)</p>
<p>for i = 1 mA, i<sub>d</sub> = 0.5 mA, and for maximum signal of 10%, I<sub>d</sub> = 0.05 mA </p>
<p>thus i<sub>d</sub> = 2 i<sub>d</sub> = 0.1 mA and the corresponding maximum v<sub>d</sub> is 0.1 mA * 10 kΩ = 1 V</p>
<p>the corresponding peak input can be found by dividing v<sub>o</sub> by the transmisiion factor of 0.995, thus v<sub>i max</sub> = 1 V / 0.995 V = 1. 005 V</p>
<p> </p>
<p>3.43</p>
<p>opening the current cource we get the following samll-siganl circuit : (n=1)</p>
<p> </p>
<p>figure</p>
<p> </p>
<p>v<sub>o</sub> / v<sub>i</sub> = ( 1 / SC ) / ( 1 / SC + r<sub>d</sub>) = 1 / ( 1+ SC * r<sub>d</sub>)</p>
<p> </p>
<p>phase shift = -arctan ( ω c r<sub>d</sub> / 1)</p>
<p>= - arctan (2 π 10<sup>5</sup> * 10 * 10<sup>-9</sup> * 0.025 / I )</p>
<p> </p>
<p>for a phase shift of -45° we have </p>
<p>2 π 10<sup>5</sup> * 10 * 10<sup>-9</sup> * 0.025 / I = 1</p>
<p>I = 157 μA</p>
<p> </p>
<p>range of phase shift for I = 15.7 μA to 1570 μA </p>
<p>i<sub>o </sub>: -84.3° to - 5.71°</p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
<p> </p>
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