EE 321 Analog Electronics, Fall 2011Homework #5 sol ution3.37. Find the parameters of a piecewise-linear model of a diode for which vD=0.7 V at iD= 1 mA and n = 2. The model is to fit exactly at 1 mA and 10 mA.Calculate the error in millivolts in predicting vDusing a piecewise-linear modelat iD= 0.5, 5 and, 14 mAWe have iD= ISevDnVT, soIS= iDe−vDnVT= 1 × 10−3× e−0.72×25.2×10−3= 0.93 nANext,vD= nVTlniDIS= 2 × 25.2 × 10−3× ln10 × 10−30.93 × 10−9= 0.82 VThen we haverD=vD10− vD1iD10− iD1=0.82 − 0.710 − 1= 13 ΩandvD0= vD1− rDiD1= 0.7 − 13 × 1 × 10−3= 0.69 VWe can now calculate vDbased on the piecewise linear model asvD= vD0+ iDrDor based on the exponential modelvD= nVTlniDISand those are tabulated hereiDvD(exp) vD(lin) error0.5 0.67 0.70 0.035 0.78 0.76 0.0214 0.83 0.87 0.043.54. In the circuit shown in Fig. P 3.54, I is a DC current and vsis a sinusoidalsignal. Capacitors C1and C2are very large; their function is to couple the signalto and from the diode but block t he DC current from flowing into the signalsource or the load (not shown). Use the diode small-signal model to show thatthe signal component of the output voltage isvo= vsnVTnVT+ IRsIf vs= 10 mV, find vofor I = 1 ma, 0.1 mA, and 1 µA. Let Rs= 1 kΩ and n = 2.At what value of I does vobecome one-half of vs? Note that this circuit functions1as a signal attenuator whith the attenuation factor controlled by the value of theDC current I.The signal portion is t ransferred from input to output according to a voltage division betweenRSand rD,vo= vsrDRs+ rDwhererD=dvDdiDiD=I=diDdvD−1id=I=InVT−1=nVTIand thusvo= vsnVTIRS+nVTI=nVTIRs+ nVTFind vofor several values of I, and vs= 10 mV, Rs= 1 kΩ, and n = 2. It is tabulated belowI (mA) vs(mV)1 0.480.1 3.410−39.8Value of I f or which vo=vs2:nVTnVT+ IRs=12I =nVTRs=2 × 25.2 × 10−31 × 103= 5.0 × 10−5A = 50 µA3.59 Consider t he voltage-regulator crictuit show in Fig. P3.59. The value of Ris selected to obtain an output voltage Vo(across the diode) of 0.7 V .2(a) Use the diode small-signal model to show that the change in output voltagecorresponding to a change of 1 V in V+is∆Vo∆V+=nVTV++ nVT− 0.7This quantity is known as the line regulation and is usually expressed inmV/V.(b) Generalize the expression above to the case of m diodes connected in serisand the value of R adjusted so that the voltage across each diode is 0.7 V(and Vo= 0.7 m V).(c) Calculate the value of line regulation for the case V+= 10 V (nominally)and (i) m = 1, and (ii) m = 3. Use n = 2.(a) We can writeiD=V+− vDR= ISevDnVTand we are interested in findingdvDdV+. we can re-arrangeV+= RvDR+ ISevDnVT= vD+ RISevDnVTand then computedV+dvD= 1 +RISnVTevDnVT3Now, we are interested in evaluating this at a bias point vD= VD, for whichID= ISeVDnVTso we can insert that and getdV+dvD= 1 +RIDnVTat that bias point we have RID= V+− VD, so we can writedV+dvD= 1 +V+− VDnVTand we are asked to computedvDdV+=11 +V+−VDnVT=nVTnVT+ V+− VDWith the bias point VD= 0.7 V this is identical t o the expression we were asked tocompute.(b) In this case we just writeiD=V+− mvDR= ISevDnVTre-arrangeV+= RmvDR+ ISevDnVT= mvD+ RISevDnVT= vO+ RISevOmnVTand computedV+dvo= 1 +RISmnVTevOmnVT= 1 +RISmnVTevDnVTNow note that at the bias point, vD= VD=VOm, we can substituteID= ISevDnVTdV+dvo= 1 +RIDmnVT= 1 +V+− VOmnVT= 1 +V+− mVDmnVTFinally we can compute4dvodV+=11 +V+−mVDmnVT=mnVTmnVT+ V+− mVDAgain we use VD= 0.7 V.(c) (i)dvodV+=2 × 25.2 × 10−32 × 25.2 × 10−3+ 10 − 0.7= 0.0054(ii)dvodV+=3 × 2 × 25.2 × 10−33 × 2 × 25.2 × 10−3+ 10 − 3 × 0.7= 0.0193.61 Design a diode voltage regulator to supply 1.5 V to a 150 Ω load. Use twodiodes specified to have a 0.7 V drop at a current of 10 mA and n = 1. Thediodes are to be connected to a +5 V supply through a resistor R. Specify thevalue of R. What is the diode current with the load connected? What is theincrease resulting in the output voltage when the load is disconnected? Whatchange results if the load resistance is reduced to 100 Ω? To 75 Ω? To 50 Ω?First we compute ISfrom the 10 mA po int asIS=iDevDnVT=10 × 10−3e0.725.2×10−3= 8.64 × 10−15ANext compute the amount of current which is required to produce a 0.75 V drop across onediode.iD= ISevDVT= 8.64 × 10−15e0.7525.2×10−3= 72.8 mAWe have this current through the resistor, as well as the curent through the 150 Ω loadresistor. The current through the load isiL=vLRL=1.5150= 10 mAand the size of the resistor can then be found fromV = R (iD+ iL) + 2vDR =V − 2vDiD+ iL=5 − 1.572.8 + 10= 42.3 ΩIf the load is disconnected let’s assume that the additional small 10 mA goes through thediodes. Let’s compute t he diode resistance, rD,rD=diDdvD−1iD=72.8 mA=IDnVT=nVTID=25.2 × 10−372.8 × 10−3= 0.35 ΩThe change in voltage is then5∆vO= ∆iDrDIf the load is disconnected, ∆iD= 10 mA, and ∆vD= 10 × 10−3× 0.35 = 3.5 mVIf the load resistance is reduced to 100 Ω, ∆iD=1.5150−1.5100= −5 mA, and then ∆vD=−5 × 1 0−3× 0.35 = −1.8 mV.If the load resistance is reduced to 75 Ω, ∆iD=1.5150−1.575= −10 mA, and then ∆vD=−3.5 mV.If the load resistance is reduced to 50 Ω, ∆iD=1.5150−1.550= −20 mA, and then ∆vD=−7