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GT ECE 3050 - Current Sources

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c° Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Tec hnology, School of Electricaland Computer Engineering.Current SourcesCurren t mirrors are commonly used for current sources in integrated circuit design. This section covers othercurrent sources that are often seen.FET Current SourcesFigure 1(a) shows t wo FET current sources, one which uses an n-channel depletion mode MOSFET and theother which uses an n-channel JFET. The equivalent p-channel sources are shown in Fig. 1(b). Rememberthat the JFET is a depletion mode device. The analysis for the two sources is the same with the exceptionthat the transconductance parameter is denoted by K for the MOSFET and by β for the JFET.Figure 1: MOSFET (depletion mode) and JFET current sources. (a) n-channel. (b) p-channel.For the n-channel device, the MOSFET drain current and gate-source voltage are given byID= K (VGS− VTO)2VGS= −ISRS= −IDRSThe object is to solve for RSfor a desired drain current. When the equation for VGSis substituted into theequation for ID,weobtainID= K (−IDRS− VTO)2= K (IDRS+ VTO)2ThiscanbesolvedforRSto obtainRS=pID/K − VTOIDwhere VTO< 0.NotethatK = K0(1 + λVDS).IfVDSis not specified, an often used approximation isK ' K0.For the n-channel JFET, the drain current is give n byID= β (VGS− VTO)2It follo ws that the MOSFET solution for RScanbeusedwiththesubstitutionofβ for K to obtainRS=pID/β − VTOID1The output resistance is a figure of merit for a current source. Ideally, it should be infinite. The outputresistance is the resistance seen looking into the drain of each source. It is give n byrout= rid= r0µ1+RSrs¶+ RSrs=1gmr0=λ−1+ VDSIDwhere gm=2√KIDfor the MOSFET and gm=2√βIDfor the JFET.For the p-channel devices, the subscripts for the voltages are reversed, e.g. VGSbecome VSGand VDSbecomes VSD.Example 1 A depletion mode MOSFET has the parameters K0=5× 10−4A/ V2, λ =10−4V−1,andVTO= −2V. Calculate the value of RSand routif the transistor is to be used as a current source with acurrent ID=1.5mA. Assume VDS=8V.Solution.K = K0(1 + λVDS)=5.2 × 10−4A/ V2RS=pID/K − VTOID=2.47 kΩr0=λ−1+ VDSID=13.87 kΩgm=2pKID=1.766 × 10−3S rs=1gm=566.1 Ωrout= rid= r0µ1+RSrs¶+ RS= 745 kΩResistor RScauses routto be greater than r0by more than a factor of 5.One-BJT Current SourceFigure 2 sho ws npn and pnp BJT current sources. The object is to select the resistors in the circuit for adesired collector current IC.Figure 2: BJT current sources.2For the npn device, the steps can be summarized as follows:(a)ChooseavalueforRE. It should be small but large enough for good bias stability. It is common tochoose REso that the voltage across it is some multiple n of the base-emitter junction voltage VBE,wheretypically 1 ≤ n ≤ 4.Thelargern, the better the current stability. It follows thatIERE=ICαRE= nVBE=⇒ RE=αnVBEIC(b)ChooseavalueforR2. It is usually chosen so that the current I2is some multiple m of IB,wheretypically m ≥ 9. It follows thatI2R2= mIBR2= mICβR2=(n +1)VBE=⇒ R2=β (n +1)VBEmICIf m is too small, the uncertainty in IBcan cause errors if β is not known precisely or if β drifts withtemperature.(c) Solve for R1.I1R1=(m +1)IBR1=(m +1)ICβR1= V+−£V−+(n +1)VBE¤=⇒ R1= βV+− V−− (n +1)VBE(m +1)IC(d) Solve for rout.rout= ric=r0+ r0ekRE1 −αREr0e+ REr0=VA+ VCEICr0e=R1kR2+ rx1+β+ reFor the pnp device, the subscripts for the voltages are reversed, e.g. VBEbecome VEBand VCEbecomesVEC.Example 2 A BJT has the parameters β = 100, VA=75V,andrx=40Ω. The transistor is to be usedasacurrentsourcewithacurrentIC=1.5mA, V+=15V,andV−= −15 V. Calculate the values of R1,,R2,androutif I1=10IB(n = 10) and IERE=2VBE(m =2). Assume VBE=0.65 V and VCE=8V.Solution.RE=αnVBEIC=βnVBE(1 + β) IC= 858 ΩR1= βV+− V−− (n +1)VBE(m +1)IC= 170 kΩR2=β (n +1)VBEmIC=13kΩr0=VA+ VCEIC=55.33 kΩr0e=R1kR2+ rx1+β+ re=R1kR2+ rx1+β+αVTIC= 136.5 Ωrout= ric=r0+ r0ekRE1 −αREr0e+ RE= 380.4kΩResistor REcauses rout to be greater than r0b y almost a factor of 7.3Tw o-BJT Current SourceFigure 3 shows npn and pnp two-transistor BJT current sources that are simpler to design. The outputcurrent in each is the collector current IO. For the circuit of Fig. 3(a), the following equations can bewrittenV+− V−= I1R1+ VBE2+ VBE1+IC1α1RE1IC1= I1−IOβ2µIOα2−IC1β1¶RE2= VBE1+IC1α1RE1For the pnp circuit, the subscripts for the voltages are reversed, e.g. VBEbecome VEB.Figure 3: Two-BJT current sources.There is a positive feedback effect in these circuits which increases the output resistance seen looking intothe collector of Q1. To see this, consider the small-signal circuit with V+= V−=0and the collector of Q1driven by a small-signal test current source it.Ifitis positive, it causes a current to flow through r01fromits collector to its emitter. This forces the base voltage of Q2to increase. This is amplified by Q2to causeits collector voltage to decrease. This decrease is applied to the base of Q1to cause its collector voltage toincrease. This feedback effect causes the collector voltage of Q1to be larger than it w o uld be without thefeedback. Because resistance is voltage divided by current, it follows that the resistance seen looking intothe collector of Q1is increased.The circuit can be designed by specifying the currents I1and IC1and the resistor RE2. The current I1must be chosen so that it is much larger than the anticipated base current in Q1. RE2can be omitted, butit aids in preventing temperature drift of the currents. If it is too large, however, it reduces the gain aroundthe positive feedback loop, thus reducing the output resistance. T ypically, RE2mightbechosentohaveavalue such that VBE1≤ IE1RE1≤ 4VBE1.ResistorsR1and RE1are then given b yR1=V+− V−− (VBE1+ VBE2) −µI1−IOβ2¶RE1α1I1RE2=VBE1+IC1α1RE1IOα2−IC1β14Example 3 For V+=15Vand V−= −15 V, design the circuit in Fig. 3(a) for an output currentIC1=1.5mA.Solution. If we estimate β = 100, the base current in Q1is 0.015 mA.LetuschooseI1=20IB2=0.3mA.We estimate VBE1= VBE2=0.65 V. The current IC1is given by IC1= I1−IO/β =0.285 mA. The currentthrough RE1is IE1= IC1/α =0.288 mA.IfwechooseIE1RE1= VBE1, it follows that RE1, R1,andRE2are given b yRE1=0.650.288 mA=2.26 kΩR1=30 − 2 × 0.65 − 0.285 mA × 2.26 kΩ0.3mA=91.2kΩRE2=2 × 0.651.5mA= 867 ΩCurrent Sources Using an Op AmpFigure 4 shows two current sources that use an op amp as an error amplifier. The


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