Analog Integrated Circuit Design P2
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Analog Integrated Circuit Design P2
Before proceeding, it is worth discussing the terms weak, moderate, and strong inversion. As just discussed, a gatesource voltage greater than V,, results in an inverted channel, and drainsource current can flow. However, as the gatesource voltage is increased, the channel does not become inverted (i .e., nregion) suddenly, but rather gradually.
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 24 Chapter 1 IntegratedCircuit Devices and Modelling Triode region ,,: "DS Fig. 1.14 The I versus VDS curve for an ideal MOS tronsistor. For D VDs > VDs,, , I i s approximately constant. D Before proceeding, it is worth discussing the terms weak, moderate, and strong inversion. As just discussed, a gatesource voltage greater than V, results in an , inverted channel, and drainsource current can flow. However, as the gatesource voltage is increased, the channel does not become inverted (i .e., nregion) suddenly, but rather gradually. Thus, it is useful to define three regions of channel inversion with respect to the gatesource voltage. In most circuit applications, noncutoff MOS FET transistors are operated in strong inversion, with Veff> 100 m V (many prudent circuit designers use a minimum value of 200 mV). As the name suggests, strong inversion occurs when the channel is strongly inverted. It should be noted that all the equation models in this section assume strong inversion operation. Weak inversion occurs when VGS is approximately 100 m V or more below V,, and is discussed as subthreshold operation in Section 1.3. Finally, moderate inversion is the region between weak and strong inversion. LargeSignal Modelling The triode region equation for a MOS transistor relates the drain current to the gate source and drainsource voltages. It can be shown (see Appendix) that this relation ship is given by As ,V increases, I increases until the drain end of the channel becomes pinched off. , D and then bno longer increases. This pinchoff occurs for VDG = V,, , or approxi mately, VDS = VGS Vtn = Ve,, (1.66) Right at the edge of pinchoff, the drain current resulting from (1.65) and the drain current in the active region (which, to a firstorder approximation, is constant with
 1 .2 MOS Transistors 25 respect to V,, ) must have the same value. Therefore, the active region equation can be found by substituting (1.66) into (1.65), resulting in For V > Veff, the current stays constant at the value given by (1.67), ignoring ,, secondorder effects such as the finite output impedance of the transistor. This equation is perhaps the most important one that describes the largesignal operation of a MOS transistor. It should be noted here that (1.67) represents a squared currentvoltage relationship for a MOS transistor in the active region. In the case of a BJT transistor, an exponential currentvoltage reiationship exists in the active region. As just mentioned. ( 1.67) implies that the drain current, ID, is independent of the drainsource voltage. This independence is only true to a firstorder approximation. The major source of error is due to the channel length shrinking as V increases. To , see this effect, consider Fig. 1.15, which shows a cross section of a transistor in the active region. A pinchedoff region with very little charge exists between the drain and the channel. The voltage at the end of the channel closest to the drain is fixed at VGS Vtn = V e f f .The voltage difference between the drain and the near end of the channel lies across a short depletion region often called the pinchofS region. As ,V , becomes larger than V,,, this depletion region surrounding the drain junction increases its width in a squareroot relationship with respect to VDs This increase in the width of the depletion region surrounding the drain junction decreases the effective channel length. In turn, this decrease in effective channel length increases the drain current, resulting in what is commonly referred toas channellengrh modularion. To derive an equation to account for channellength modulation, we first make use of ( I . I 1) and denote the width of the depletion region by xd, resulting in where I Depletion region AL  &DS  v.ff + a. \ Pinchoff region Fig. 1.15 Channel length shortening For VDs> V,
 26 Chapter 1 IntegratedCircuit Devices and Modelling and has units of m/&. Note that NA is used here since the ntype drain region is more heavily doped than the ptype channel (i.e., N D >> N) By writing a Taylor ,. approximation for b around its operating value of VDs = VGs V,, = Veff, we find I to be given by D where I D the t current when VDs = V,f , or equivalently, the drain current is , drain , when the channellength modulation is ignored. Note that in deriving the final equa tion of (1.70). we have used the relationshp dL/aV,, = dx,/dV,,. Usually, (1.70) is written as where h is the output impedance constant (in units of V') given by Equation (1.71) is accurate until VDs is large enough to cause secondorder effects, often called shortchannel effects. For example, ( 1.71) assumes that current flow down the channel is not veloci9saturated (i.e., in~reasing electric field no longer the increases the camer speed). Velocity saturation commonly occurs in new technolo gies that have very short channel lengths and therefore large electric fields. If VDs becomes large enough so shortchannel effects occur, I increases more than is pre D dicted by (1.71). Of course, for quite large values of VDs, the transistor will eventu ally break down. A plot of I versus VDs for different values of VGSis shown in Fig. 1.16. Note D that in the active region, the small (but nonzero) slope indicates the small dependence of ID on V . ,, I Triode , Shortchannel : effects I I A Increasing v , , L ' 1VGS> Vtn Fig. 1.16 IDversus ,V , for different values of .,, V
 1.2 MOS Transistors 27 EXAMPLE 1.8 Find 1, for an nchannel transistor that has doping concentrations of ND= NA = 10", pnC,, = 92 p ~ / ~ W/L = 20 p m / 2 p m , VGs = 1.2 V, 2 . V = 0.8 V, and Vos = V e f f .Assuming h remains constant, estimate the new , value of 1, if V,, is increased by 0.5 V. Solution From (1.69), we have which is used in (1.72) to find h as 362 x lop9 h = = 95.3 x lo' VI 2 x 2 10~x&9 ~ Using (1.71), we find for VD, = Veff = 0.4 V, In the case where VDs = V, + 0.5 V = 0.9 V , we have ID, = 73.6 P A x (1 + h x 0.3) = 77.1 p A Note that this example shows almost a 5 percent increase in drain current for a 0.5 V increase in drainsource voltage. Body Effect The largesignal equations in the preceding section were based on the assumption that the source voltage was the same as the substrate (i.e., bulk) voltage. However, often the source and substrate can be at different voltage potentials. In these situa tions, a secondorder effect existsthat is modelled as an increase in the threshold voltage, V,, , as the sourcetosubstrate reversebias voltage increases. This effect, typically called the body eflect, is more important for transistors in a well of a CMOS process where the substrate doping is higher. It should be noted that the body effect is often important in analog circuit designs and should not be ignored without consid eration. To account for the body effect, it can be shown (see Appendix at the end of this chapter) that the threshold voltage of an nchannel transistor is now given by where V,,, is the threshold voltage with zero Vss (i.e., sourcetosubstrate voltage),
 28 Chapter 1 IntegratedCircuit Devices and Modelling and The factor y is often called the bodyef/ect constant and has units of f i .Notice that y is proportional to FA , l o so the body effect is larger for transistors in a well where typically the doping is higher than the substrate of the microcircuit. pChannel Transistors All of the preceding equations have been presented for nchannel enhancement tran sistors. In the case of pchannel transistors, these equations can also be used if a negative sign is placed in front of e v e q voltage variable. Thus, VGSbecomes VSG, VDs becomes VSD, Vtn becomes V,, , and so on. The condition required for con duction is now VSG> V where V,, is now a negative quantity for an enhancement , pchannel transistor." The requirement on the sourcedrain voltage for a pchannel transistor to be in the active region is VsD> VSG+ Vlp. The equations for I,, in both regions, remain unchanged, because all voltage variables are squared, resulting in positive hole current flow from the source to the drain in pchannel transistors. For nchannel depletion transistors, the only difference is that Vtd < 0 V. A typical value might be V = 2 V . , SmallSignal Modelling in the Active Region The most commonly used smallsignal model for a MOS transistor operating in the active region is shown in Fig. 1.17. We first consider the dc parameters in which all the capacitors are ignored (i.e., replaced by open circuits). This leads to the low frequency, smallsignal model shown in Fig. 118. The voltagecontrolled current source, g, ~ , is the most important component of the model, with the transistor transconduc tance g defined as , In the active region, we use (1.67), which is repeated here for convenience, 10. For an nchannel transistor. For a pchannel transistor, y is proportional to ND. 1 1 . It is possible to realize depletion pchannel transistors. but these are of little value and seldom worth the extra processing involved. Depletion nchannel transislors are also seldom encountered in CMOS microcir cuits, although they might be wonh the extra processing involved in some applications. especially if they were in a well.
 Fig. 1 .I7 The smallsignal model for a MOS transistor in the active region. Fig. 1.18 The lowfrequency, smallsignal model for an active MOS transistor.  and we apply the derivative shown in (1.75) to obtain or equivalently, where the effective gatesource voltage, V, , is defined as Veff  VGS Vtn. Thus, we see that the transconductance of a MOS transistor is directly proportional to Veff . Sometimes it is desirable to express g, in terms of I rather than VGS. From D ( 176), we have v , = vtn+ / ~ncox(w/L) The second term in (1.79) is the effective gatesource voltage, Veff, where
 30 Chapter 1 IntegrotedCircuit Devices and Modelling Substituting (1.80) in (1.78) results in an alternate expression for 9,. Thus, the transistor transconductance is proportional to D J for a MOS transistor, whereas it is proportional to Ic for a BJT. A third expression for g is found by rearranging (1.8 1) and then using (1.80) to , obtain Note that this expression is independent of pnC,, and W/L , and it relates the transconductance to the ratio of drain current to effective gatesource voltage. This simple relationship can be quite useful during an initial circuit design. The second voltagecontrolled currentsource in Fig. 1.18, shown as g , ~ , , models the body effect on the smallsignal drain current, id. When the source is connected to smallsignal ground, or when its voltage does not change appreciably, then this current source can be ignored. When the body effect cannot be ignored, we have a~, gs =     a~, av,, avss av,,av,, From (1.76) we have  Using (1.73), which gives V,, as we have The negative sign of (1.84) is eliminated by subtracting the current g,v, from the major component of the drain current, g,~,, , as shown in Fig. 1.18. Thus, using (1.84) and (1.86), we have Note that although g, is nonzero for V,, = 0, if the source is connected to the bulk, AVsB is zero, and so the effect of gs does not need to be taken into account. How ever, if the source happens to be biased at the same potential as the bulk but is not
 1.2 MOS Transistors 31 directly connected to it, then the effect of g, should be taken into account since AVsB is not necessarily zero. The resistor, rds, shown in Fig. 1.18, accounts for the finite output impedance (i.e., it models the channellength modulation and its effect on the drain current due to changes in V,d. Using (1.7 repeated here for convenience, l), we have where the approximation assumes h is small, such that we can approximate the drain bias current as being the same as IDsat. Thus, where and . k,, = 1 ' It should be noted here that (1.90) often empirically adjusted to take into account is secondorder effects. EXAMPLE 1.9 Derive the lowfrequency model parameters for an nchannel transistor that has doping concentrations of N = lo2', NA = , pnC,, = 92 p ~ / ~W/L,= 2 20 pm/2 p m , VGs = 1.2V, V,, = 0.8 V, and VDS = Veff. Assume y = 0.5 f i and VSg = 0.5 V. What is the new value of rdr if the drainsource volt age is increased by 0.5 V? Sdufion Since these parameters are the same as in Example 1.8, we have and from (1.87), have we
 32 Chapter 1 IntegratedCircuit Devices and Modelling Note that this sourcebulk transconductance value is about 116 that of the gate source transconductance. For rds,we use ( I 90)to find At this point, it is interesting to calculate the gain g,rds = 52.6, which is the largest voltage gain this single transistor can achieve for these operating bias conditions. As we will see, this gain of 52.6 is much smaller than the corre sponding singletransistor gain in a bipolar transistor. Recalling that V = 0.4 V, if VDs is increased to 0.9 V1the new value for , h is resulting in a new value of ,r given by An alternate lowfrequency model, known as a T model, is shown in Fig. 1.19. This T model can often result in simpler equations and is most often used by experi enced designers for a quick analysis. At first glance, it might appear that this model allows for nonzero gate current. but a quick check confirms that the drain current must always equal the source current, and, therefore, the gate current must always be zero. For this reason, when using the T model, one assumes from the beginning that the gate current is zero. Fig. 1.I9 The smallsignal, lowfrequency T model for an active MOS transistor (the body effect i s not mod elled).
 1.2 MOS Transistors 33 EXAMPLE 1.10 Find the T model parameter, r for the transistor in Example 1.9. . , Solution The value of r is simply the invclse of g , !, resulring'in The value of rd, remains the same, either 143 WZ or 170 kR,depending on the drainsource voltage. Most of the capacitors in the s~nallsignal rncx!el are related to the physical tran sistor. Shown in Fig. 1.20 is a cross section of a MOS rr,ansistor, where the parasilic capacitances are shown at the appropriate locations. The laryest capacitor in Fig. 1.20 i s Cg, . T h i s capacitance is primarily due ro the change in channel charge as a result of a chnnse in VGS.It can be shown [Tsividis, 19871 that Cg, is approx.irnately given by 2 C, , E ;WLC,, (1.93)  When accuracy is important, an additional term should be added to (1.93) to take into account the overlap between the p ~ t and solrl.ce junction, which sllould include c the,fiinging ctlpacitance (fringing capacitance is due to boundary effects). This addi tional componenl is given by vs, =0 V~~ ' "tn Polysilicon p substrate T Fig. 1.20 A cross section of an nchonnel MOS transistor showing !he smallsignalcopocitances.
 34 Chapter 1 IntegratedCircuit Devices and Modelling where Lo, is the overlap distance and is usually empirically derived. Thus, when higher accuracy is needed. The next largest capacitor in Fig. 1.20 is CJsb, capacitor between the source the and the substrate. This capacitor is due to the depletion capacitance of the reverse biased source junction, and it includes the channeltobulk capacitance (assuming the transistor is on). Its size is given by where A, is the area of the source junction, Ach is the area of the channel (i.e., WL) and Cj, is the depletion capacitance of the source junction, given by fi Note that the total area of the effective source includes the original area of the junc tion (when no channel is present) plus the effective area of the channel. The depletion capacitance of the drain is smaller because it does not include the channel area. Here, we have where fi and Ad is the area of the drain junction. The capacitance Cgd, sometimes cal led the Millercapacitor, is important when the transistor is being used in circuits with large voltage gain. Cgdis primarily due to the overlap between the gate and the drain and fringing capacitance. Its value is given by where, once again, Lo, is usually empirically derived. Two other capacitors are often important in integrated circuits. These are the source and drain sidewall capacitances, C, and Cdsw. , , These capacitances can be large because of some highly doped pi regions under the thick field oxide calledfield implants. The major reason these regions exist is to ensure there is no leakage current between transistors. Because they are highly doped and they lie beside the highly doped source and drain junctions, the sidewall capacitances can result in large addi tional capacitances that must be taken into account in determining Csband Cdb. The sidewall capacitances are especially important in modem technologies as dimensions
 1.2 MOS Transistors 35 shrink. For the source, the sidewall capacitance is given by ,, c = psc,, (1.101) where P, is the length of the perimeter of the source junction, excluding the side adjacent to the channel, and r\ It should be noted that C,, the sidewall capacitance per unit length at 0V bias volt age, can be quite large because the field implants are heavily doped. The situation is similar for the drain sidewall capacitance, Cd,, Cdsw = PdCisw where Pdis the drain perimeter excluding the portion adjacent to the gate. Finally, the sourcebulk capacitance, C,, , is given by Csb = C'sb + Cssw with the drainbulk capacitance, Cdb given by , Cdb = C'db + Cdsw  EXAMPLE 1.1 1 An nchannel transistor is modelled as having the following capacitance parameters: 2 Cj = 2.4 x pF/(prn) , Ci,, = 2.0 x 1 o4 p F / p m , Cox 1.9 x 1 0  ~ ~ ~ 1 ( ~ r n ) ~ , = 4 CgS, = C, , = 2.0 x 10 pF/pm. Find the capacitances Cg,, Cgd, Cdb,and Csb for a transistor having W = 100 m and L = 2 p m . Assume the source and drain junctions extend 4 prn beyond the gate, so that the source and drain areas are A, = 2 Ad = 400 ( p m ) and the perimeter of each is P, = Pd = 108 p m . ' Solution We calculate the various capacitances as follows: Csb= C,(As+ WL) + (Ci,, x P,) = 0.17 pF Note that the sourcebulk and drainbulk capacitances are significant compared to the gatesource capacitance. Thus, for highspeed circuits, it is important to
 keep the areas and perimeters of drain and source junctions as small as possible (possibly by sharing junctions between transistors, as seen in the next chapter). SmallSignal M d l i n g in the Triode and Cutoff Regions The lowfrequency, smallsignal model of a MOS transistor in the triode region (which is sometimes referred to as the linear region) is a resistor. Using (1.65), the largesignal equation for IDin the triode region, results in where rds is the smallsignal drainsource resistance (and g ds is the conductance). For the common case of VDSnear zero, we have which is similar to the IDversusVDs relationship given earlier in (1 60). EXAMPLE 1.12 For the transistor of Example 1.9,find the triode model parameters when V ,s is near zero. Solution From (1.108), we have Note that this conductance value is the same as the transconductance of the tran sistor, g, , in the active region. The resistance, rds, is simply l/gds, resulting in rds = 2.72 kQ. The accurate modelling of the highfrequency operation of'a transistor in the triode region is nontrivial (even with the use of a computer simulation). A moder ately accurate model is shown in Fig. 1.21, where the gatetochannel capacitance
 1.2 MOS Transistors 37 v, Gatetochannel capacitance Channeltosubstrate capacitance Fig. 1.21 A distributed RC model for a transistor in the active region. and the channeltosubs trate capacitance are modelled as distributed elements. How ever, the IV relationships of the distributed RC elements are highly nonlinear because the junction capacitances of the source and drain are nonlinear depletion capacitances, as is the channeltosubstrate capacitance. Also, if V,, is not small, then the channel resistance per unit length should increase as one moves closer to the drain. This model is much too complicated for use in hand analysis. A simplified model often used for small VDs is shown in Fig. 1.22, where the resistance, rds, is given by (1.108). Here, the gatetochannel capacitance has been evenly divided between the source and drain nodes, Note that this equation ignores the gatetojunction overlap capacitances, as given by (1.94), which should be taken into account when accuracy is very important. The channeltosubstrate capacitance has also been divided in half and shared between the source and drain junctions. Each of these capacitors should be added to the junction tosubstrate capacitance and the junctionsidewall capacitance at the appropriate Fig. 1.22 A simplified trioderegion model valid for small VDs.
 38 Chapter 1 IntegratedCircuit Devices and Modelling node. Thus, we have and Also, and It might be noted that CSbis often comparable in size to Cg, due to its larger area and the sidewall capacitance. When the transistor turns off, the model changes considerably. A reasonable model is shown in Fig. 1.23. Perhaps the biggest difference is that rds is now infinite. Another major difference is that Cg, and Cgdare now much smaller. Since the chan nel has disappeared, these capacitors are now due to only overlap and fringing capac itance. Thus, we have However, the reduction of C, and Cpd does not mean that the total gate capaci , tance is necessarily smaller. We now have a 'hew" capacitor, Cgb, which is the gate Fig. 1.23 A smollsignal model for a MOS FET that is turned off.
 tosubstrate capacitance. This capacitor is highly nonlinear and dependent on the gate voltage. If the gate voltage has been very negative for some time and the gate is accumulated, then we have C, = AchCO,= WLC,, , ( 1 . 1 15) If the gatetosource voltage is around 0 V, then Cgbis equal to C in series with the , channeltobulk depletion capacitance and is considerably smaller, especially when the substrate is lightly doped. Another case where Cgbis small is just after a transistor has been turned off, before the channel has had time to accumulate. Because of the complicated nature of correctly modelling C when the transistor is turned off, gb equation ( 1.115 ) is usually used for hand analysls as a worstcase estimate. The capacitors CSb Cdbare also smaller when the channel is not present. We and now have CS,O= AsCjo (1.1 16) and Cdb0 = Adcia 1.3 ADVANCED MOS MODEWNG In this section, we look at three advanced modelling concepts that a microcircuit designer is likely to encountershortchannel effects, subthreshold operation, and leakage cur rents. . ShortChannel Effects A number of shortchannel effects degrade the operation of MOS transistors as device dimensions are scaled down. These effects include mobility degradation, reduced out put impedance, and hotcarrier effects (such as oxide trapping and substrate currents). These shortchannel effects will be briefly described here. For more detailed model ling of shortchannel effects, see [Wolf, 19951. Transistors that have short channel lengths and large electric fields experience a degradation in the effective mobility of their carriers due to several factors. One of these factors is the large lateral electric field (which has a vector in a direction perpen dicular from the gate into the silicon) caused by large gate voltages and short channel lengths. This large lateral field causes the effective channel depth to change and also causes more electron collisions, thereby lowering the effective mobility. Another fac tor causing this degradation is that, due to large electric fields, carrier velocity begins to saturate. A firstorder approximation that models this carriervelocity saturation for electrons is given by where E is the electric field and E is the critical electrical field, which might be on , the order of 1.5 x lo6 V/m. Using this equation in the derivation of the IDV,,,
 40 Chopter 1 IntegratedCircuit Devices and Modelling chnracteriqtics of a MOS rrrinsistor. it can be shown [Gr:~y,19931 that the drain cur re.nt is now given by where e = I /(LIE,) and. for a 0.8pm tcchnalogy. might have a typical value of 0.6 VI. 11 can be shown that t h s mobility degradation is ecjui\~nlentt c ~ finitc scnes a source ~.csistancegiven by For p , , C = 90 p.A/'V2. this resistance might be on the order of 6 k!J per prn of width (again, for a O'.Rpmlong transistor). This equivalent series solrrce resist:lnce is typically larger than the physical vource resil;t;~nce.'I'his saturnlion c~tuscs square the law chan~ctcristicof the currentvoltage relntionship to be inaccurate, and the true relationsliip will be somewhere between lincar and square. In many voltagetocurrent conversion circuits that rely on the squarelaw characteristic, this inaccuracy can be a major source of error. Taking channel lengths larger than the rninin~um allowed helps to nlinimizc this degradation. Transistors wit11 short channel lengths also experience a reciuccd output impcd ance because depletion region variations at the drain end (~vhicb affect the effective cha.nnel length) have an increased proponiond effect on the drain current. In addition, a phenomenon known as draininduced barrier lowering (DIBL) effectjveIy lowers V, as V ,, is increnscd, thereby forther lowering the output impedance of a short channel device. This lower oiltput impedance is the main reason that cascode current mirrors are beconling increasingly popular. Another important shortchannel cffcct is due to hor catt.ir.r.v.These highvelocity carriers can cause harmful effects, such as the generation of electronhole pairs by impact ionization and avalanching. These extra eleclsonhole pairs can cause currents to flow from the drain to the subsrrnte, as sl~own Fig. 1.24. This effect can be mod in 0 0 Gate   current n" Punchthrough current p Draintosubstrate Q ccurrent   Fig. 1.24 Draintosubstrate current caused by electronhole poirs generoted by impact ionization ot drain end of channel.
 1.3 Advanced M O S Modelling 41 elled by a finite draintoground impedance. As a result, this effect is one of the major limitations on achieving very high output impedances of cascode current sources. In addition, this current flow can cause voltage drops across the substrate and possibly cause latchup, as the next section describes. Another hotcarrier effect occurs when electrons gain energies high enough so they can tunnel into and possibly through the thin gate oxide. Thus, this effect can cause dc gate currents. However, often more harmful is the fact that any charge trapped in the oxide will cause a shift in transistor threshold voltage. As a result, hot carriers are one of the major factors limiting the longterm reliability of MOS transis tors. A third hotcarrier effect occurs when electrons with enough energy punch through from the source to the drain. As a result, these highenergy electrons are no longer limited by the drift equations governing normal conduction along the channel. This mechanism is somewhat similar to punchthrough in a bipolar transistor, where the collector depletion region extends right through the base region to the emitter. In a MOS transistor, the channel length becomes effectively zero, resulting in unlimited current flow (except for the series source and drain impedances, as well as external circuitry). This effect is an additional cause of lower output impedance and possibly transistor breakdown. It should be noted that all of the hotcarrier effects just described are more pro nounced for nchannel transistors than for their pchannel counterparts because elec trons have larger velocities than holes. Finally, it should be noted that shortchannel transistors have much larger sub threshold currents than longchannel devices.  Subthreshold Operation The device equations presented for MOS transistors in the preceding sections are all based on the assumption that Vef, (i.e., VGS Vt ) is greater than about 100 m V and the device is in strong inversion. When this is not the case, the accuracy of the square law equations is poor. If V <  100 m V, the transistor is in weak inversion and is , said to be operating in the subthreshold region. In this region, the transistor is more accurately modelled by an exponential relationship between its control voltage and current, somewhat similar to a bipolar transistor. In the subthreshold region, the drain current is approximately given by the exponential relationship [Geiger, 19901 where and it has been assumed that Vs = 0 and VDs > 75 m V . The constant ID, might be around 20 nA.
 42 Chapter 1 IntegratedCircuit Devices and Modelling Although the transistors have an exponential relationship in this region, the trans conductances are still small because of the small bias currents, and the transistors are slow because of small currents for charging and discharging capacitors. In addition, matching between transistors suffers because it now strongly depends on transistor thresholdvoltage matching. Normally, transistors are not operated in the subthreshold region, except in very lowfrequency and lowpower applications. Leakage Currents An important secondorder device limitation in some applications is the leakage cur rent of the junctions. For example, this leakage can be important in estimating the maximum time a sampleandhold circuit or a dynamic memory cell can be left in hold mode. The leakage current of a reversebiased junction (not close to breakdown) is approximately given by where Aj is the junction area, ni is the intrinsic concentration of carriers in undoped silicon, T is the effective minority carrier lifetime, and x,., is the thickness of the , depletion region. .to is given by t where 5, and T,, are the electron and hole lifetimes. Also, xd is given by and ni is given by ni JN,NVe(Eg)'(kT) where Nc and Nv are the densities of states in the conduction and valence bands and Eg is the difference in energy between the two bands. Since the intrinsic concentration, ni, is a strong function of temperature (it approximately doubles for every temperature increase of I I "C for silicon), the leak age current is also a strong function of temperature. Roughly speaking, the leakage current also doubles for every 11 "C rise in temperature. Thus, the leakage current at higher temperatures is much larger than at room temperature. This leakage current imposes a maximum time on how long a dynamically charged signal can be main tained in a high impedance state. 14 BIPOLARJUNCTION TRANSISTORS In the early electronic years, the majority of microcircuits were realized using bipolar junction transistors (BJTs). However, in the late 1970s, microcircuits that used MOS
 1.4 BipolarJunctionTronsistors 43 transistors began to dominate the industry, with BJT mjcrucircuits remaining popular for highspeed applications. More recently, bipolar CMOS (I3iCMOS) ~echnologies, wherc both bipolar and MOS transistors are realized in the s ~ m e microcircuit^ have grown in populasity. BiCMOS ~echnologies particularly ntt1,nctiveformixed analog are digitill ~~pplications. thus it is irnpo~tantfrlr an ontrlog tlchig~~cr become h~uiliar and to with bipolar Jc\ iccs. Modern bipolar transistors can have unitygain I'i.tquencics as h g h as L5 to 45 GHz or more, compared to unitygain frequcncies of only I to 4 GHz for MOS transistors that use a technology with qirnilallithography resolution. I.in.fortunately, in bipolar ~r,ansistors, /)use control terrnini~lh:~s a nonzero input current whcn the the transistor is conducting current (from the coIlcctor to the emitter fi)r an npn transistor; from the emitter to the collector for a pnp transistor). Fortunately, at low frequencies, the base current is much smallcr than the collectortoemitter currentit may be only 11100 of the coltector current for an npn transistor. For lateral pnp transistors, the base current may be as larpe as 1/20 of thc' c~nittzrtocollectorcurrent. A typical cross aection of an npn bipolarjunction transisror is shown in Fig. I .25. Although this structure looks quite complicated. i t corresponds approximately to the equivalent structure shown in Fig. 1.26. In a good BJT transistor. the width of the base Base The base contact surrounds the emitter contact to minimize base resistanc 7 Substrate or bulk P Fig. 1.25 A cross section of on npn bipolarjunction transistor. Emitter co~.ctor Fig. 1.26 A simplified structure of an npn transisfor
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Ebook Digital integrated circuits prentice hall: Part 1
241 p  6  1

Ebook Digital integrated circuits prentice hall: Part 2
249 p  4  0