How to Make a BANDGAP VOLTAGE REFERENCE in ONE EASY Lesson V+ (Not to be confused with the Many Arduous Ones to become proficient) R=R + A - 8A VOUT A A. Paul Brokaw © Copyright 2011, A. Paul Brokaw and Integrated Device Technology, all rights reserved. All diagrams and figures are the sole property of the author and may not be used without permission. A voltage reference is not only a convenience, but is also a necessity for many electronic devices, for example a voltage DAC which converts a digital input to an output voltage. This result is the product of the digital word, which is scalar with no dimension, and some voltage to which the scaled output is referred. The DAC needs to “know” how big a certain voltage is, in order to set its full scale value. We need a component which “knows” how big a volt is. It turns out that the silicon all around us can be persuaded to give up its secret knowledge of the Volt if we will accommodate it. The base emitter voltage of a bipolar transistor versus temperature can, in theory, be extrapolated to equal a known physical constant which has the dimension Volt at a temperature of zero Kelvin. This constant is called the extrapolated bandgap of silicon. Unfortunately, this voltage is temperature sensitive, but predictably so. However, silicon has a second property which also relates temperature and voltage and this can be combined with the change in Vbe to almost cancel the temperature effects and make a voltage approximating the bandgap at all temperatures, or at least the temperatures most of us require. LET ME BEGIN with a sort of hand-waving description of PN junctions and bipolar transistors. This representation of a PN junction is shown forward biased. Positive voltage on the P-type anode (blue,) drives holes toward the junction with N-type (red) material. At the same time the negative voltage on the N-type material repels electrons toward the junction where many recombine with the holes. Some of the electrons may cross over into the P-type, where they are minority carriers that do not spontaneously arise in P-type silicon. Similarly, holes cross the junction and become minority carriers in the N-type. Minority carriers will recombine with carriers of opposite polarity, within the silicon. The carriers that enter the silicon and combine with opposite polarity carriers do not re-emerge at their injection terminal, and so constitute a net current flow. The quantities of the two minority carrier types aren’t necessarily equal, and the ratio is related to the amount of doping of neutral silicon with the respective dopants which make the regions P and N. The carrier injection will depend upon the voltage applied to the diode, and so the current flow will be voltage dependent, though not linearly so, as you should expect of a resistor. If we reverse the polarity of the applied voltage the current flow will be greatly reduced to near zero. This results from the carriers being pulled away from the junction, leaving the dopant atoms, which are fixed in the silicon lattice. These fixed dopants aren’t free to move and recombine. The small current that flows is due only to thermal ionization, and we will neglect it. 4 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Forward Biased P-N-Junction Anode + + + + + + - + - + + + - -- - - - + + +- - + + - - - + + - -++ - - - - Cathode + - + - I=Is(e Vq/kT -1) © APB junc1 Forward Biased P-N junction I~0 Reverse Biased P-N Junction + Reversed Biased P-N junction HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB junc2 5 Now, suppose we restore the forward bias to the junction, causing a current to flow. Next, let’s add another N-type region within the P-type, and drive it in until it is separated from the other N region by a thin layer of P-type material and apply to it a bias voltage greater than the voltage on the forward biased junction. Since the added N region is more positive than the P region, the new junction is reverse biased, and we might expect very little current to flow. However, some of the electrons pulled into the P region will be attracted by the added N region, and they will cross that junction, become majority carriers in the N region, and will constitute a current flow. If we take as our goal to maximize the electron flow from bottom to top, there are several things we can do. One is to reduce the size of the bottom N region so that all the minority carriers will be less likely to recombine in the P region. Another is to adjust the doping levels to favor more electrons and fewer holes crossing the lower junction. By now I’m sure you recognize the Emitter, Base and Collector of an NPN and we have maximized electron flow from emitter to collector by making the P-type base thinner and covering the emitter with the collector junction, and have reduced the base current by doping to reduce the number of holes injected into the emitter. This seems a good place to interject a personal preference and the reasons for it. First, please note that the story I just told shows the base voltage to be the cause of both base and collector currents. The geometry and doping is selected to minimize the base current as a fraction of emitter current, or to raise beta, the ratio of collector current to base current. Because the transistor construction will control beta, we are free to consider either base voltage or base current as the cause of collector current. 6 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON + + + + + - + - + + + - + - +- + + - + +- + + + - + + + - -- - - - - - - - - + Additional Diffusion Forms Additional diffusion forms Crude NPN Transistor Crude NPN Transistor © APB junc3 Ic=Is(e Vbe(q/kT) -1) Collector - + - - -- Base + - - - -+ - + + - - - - - - - + - Emitter Ib=Isb(e Vbe(q/kT) -1) Adjustments to size and doping makes improved NPN transistor HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 7 However, I will assert that most of the time (with exceptions granted) it will be easier to treat base voltage as the cause of collector current. Here’s an example of the principle involved. Suppose we measure the base and collector current of a transistor and record our measurements in Roman numerals. Then to determine beta we need only divide one by the other to discover that beta = CLXI+IV/X. Now, that result is perfectly valid and as accurate as our measurements, but wouldn’t it have been so much easier if we had used our modern number system? Well, that’s similar to doing an analysis using beta vs. Vbe. You can get the right answer either way, but as we’ll see one is much easier than the other. An additional issue is that in most applications you want to minimize sensitivity to beta or base current. Why would you want to minimize the effects of your major controlling transistor parameter? 8 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON VPLS POS + - NEG V1 V V Q1 V UA I1 ic =DCCCVII uA GND Beta Measurement beta = DCCCVI I ; V XX XX XX C I V ) DCCCVI I D C C C C C C V V II XX+XX +XX = LX & II / V = IV/ X = beta = CLXI + IV / X http://www.guernsey.net/~sgibbs/roman.html © APB romβ HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 9 With that observation in mind, let’s look at a simple current mirror. Q1 and Q2 have the same Vbe, so we can hope they will have the same collector current. The collector current of Q2 may be higher than Q1 since the collector voltage is higher. The collector current of Q2 may also be reduced from the current in R because the two base currents subtract from R current before it drives the collector of Q1. In this case we would treat the base currents as an error in current mirroring along with the error due to the mismatch of collector voltage. These errors have opposite signs, but let’s see if we can’t do better. Another simple circuit will reduce by 4X the input current mirrored by Q2, with the same approximations. Note that all the transistors are alike, with the same Vbe so that they all operate at the same current density or current per unit of active emitter area. 10 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON VPLS R POS NEG Ib1 + ib2 Q1 Ic=is(exp(Vbeq/kT)-1) Q2 Ic=is(exp(Vbeq/kT)-1) GND Simple Current Mirror Simple Current Mirror © APB cm1 +V R ic Q5 ic Q4 ic Q3 ic Q1 VBE Q2 ic Common A 4:1 mirror operates all transistors at the same current density HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 11 A sort of shorthand we might substitute for the explicit four transistors is to show a single transistor collector and base with four emitters. The significance of this symbol is shown in the next picture, which is a cross section of such a transistor. This cross section shows a P-type wafer with N-type epi and a device including an N-type well with buried layer and a P ring to isolate this region from the rest of the epi. A P-type base is diffused into the isolated well and heavily doped N+ dots are diffused into the base. At the left of the base the N+ dot, the P-type base, and the underlying n-epi form an NPN transistor. This transistor is confined to the region below the emitter and shares isolation and base with three others There are four emitter dots diffused into the base and so this device acts like four NPNs the base and collectors of which are joined by construction, and with four separate emitters. The emitters are separated, rather than combined into one large one, so they may be well matched to a device with a single identically sized emitter. Typically for bandgap circuit applications the four emitters will be joined by metallization, but in other applications they may be used individually. 12 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON +V R 4ic Q1 VBE Q2 4A ic A Common Pictorial shorthand Pictorial Shorthand Represents Construction represents construction © APB cm3 Junction Isolated NPN With Multiple Emitter Sites n epi P Isolation n epi n epi P Isolation P Isolation N+ n epi P N+ (Active) n epi (Active) N+ n epi P Isolation N+ Buried Layer P Wafer Junction Isolated NPN with multiple emitter sites HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB 4Xnpn 13 Even this simplification of the symbol becomes clumsy when more emitters are wanted and so a multi emitter device may be indicated by notation nA and to show what each emitter matches, one is designated A. In this diagram both bases are driven by a common voltage V1, so the circuit is no longer a current mirror, but is a source of two collector currents, instead. If we trust that the currents are a function of base voltage without evaluating the direct relation, we can note that some voltage V1 applied to Vbe 1 should result in a particular collector current from Q2. Since Q1 is effectively eight (I’ve chosen n=8, somewhat arbitrarily) transistors in parallel, all of which have the same Vbe as Q2, I expect the net collector current of Q1 to be eight times that of Q2. Since they share collector, base and emitter voltage, the systematic error in this ratio should be very small, and in particular, unaffected by base current, which is supplied to both transistors by V1. Now let’s introduce a resistor (1K) into the emitter connection of Q1. As the current is adjusted by changing V1 the corresponding voltage drop across R1 will disturb the ratio of collector currents. 14 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON +V 8ic 8A Q1 VBE Q2 ic A + V1 Variable - Common further condensation uses 8A and A to indicate an 8:1 ratio of AAtheFurther Condensation Uses 8A and A to two outputs of this voltage-driven current source Indicate an 8:1 Ratio of the Two Outputs of This Voltage Driven Current Source © APB 8:1 V ic 8A R1 1K Q1 VBE2 Q2 ic A + Common - V1 Variable An Added Resistor Changes the Emitter Voltage of Q1 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON An added resistor changes the emitter voltage of Q1 15 © APB 8:1R For small values of V1 where currents are relatively small, the voltage across R1 will be small and will hardly interfere with Vbe of Q1, so the collector currents will stay very nearly in the ratio eight to one. I hope you will take my word for it below 350mV, and note that at 450mV the eight to one ratio is only starting to be reduced by the difference in the two Vbes resulting from ~15 nA in R1. i(Q1,C) i(Q2,C) 16 14 12 xle-9 10 8 6 4 2 0 0 50 100 150 200 250 variable, xle-3 300 350 400 450 WhenWhen the the Resistor Voltageis small, is Small, resistor voltage the Colloector Current Ratiois close is Close the collector current ratio to 8:1to 8:1 V ic 8A R1 1K 16 Q1 VBE2 Q2 ic A + Common - V1 Variable HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB But, as we will see, while the current in Q2 can rise exponentially with V1, the rise of current in Q1 will be asymptotic to a linear slope defined by R1. This means that as V1 rises, the ratio of collector currents falls until it is below one. In the figure this happens as V1 crosses ~720mV in a room temperature simulation. In order to more accurately set and maintain this crossover voltage an opamp may be used to adjust Vbe2 to the value that makes the collector currents equal. 160 i(Q1,C) i(Q2,C) 140 120 Exponential Rise xle-6 100 80 60 Slope Limited by R1 40 20 0 440 480 520 560 600 640 variable, xle-3 680 720 760 HigherVbe1<Vb2 Currents At higherAtcurrents andVbe1<Vb2 the current ratio falls and the Current Ratio Falls V ic 8A R1 1K Q1 VBE2 Q2 ic A + Common - V1 Variable HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB 8:1R_2 17 An op-amp drives the common base line to maintain Equal collector currents V R4 40K R3 40K 2 + 1 8A R1 1K - Q1 VBE2 Q2 A 3 Common An Op-Amp Drives the Common Base Line In this to arrangement both collector are drivenCurrents up and down together, Maintain Equalcurrents Collector but their different slopes result in a differential signal at the opamp © APBinput. 8:1force There is a negative feedback path including Q2, node 2, and the opamp, as well as a positive feedback path including Q1 with R1, node 1, and the opamp. R1 reduces the effective Gm in the positive feedback loop which reduces the gain of that path below that of the negative feedback path. So, when the two currents match and nodes 1 and 2 are equal in voltage, the negative feedback dominates and Vbe2 is maintained at the voltage which keeps the collector currents balanced. It will be important to note that by making the collector currents equal we have made the current densities unequal by a factor n = 8. While the feedback makes the two collector currents equal it does not answer the question of what value the currents actually have. However, it should be apparent that the common value for the currents is closely approximated by the current in R1, which itself is determined by the difference between Q1 Vbe and Q2 Vbe. 18 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON As indicated before, the temperature of the devices affects the result. This figure shows that V3, the voltage across R1, changes in proportion to temperature. And the resulting current appears at the two collectors. The currents track so well that the Q1 current line on the graph is obscured everywhere by the Q2 current trace. 80 i(Q1,C) i(Q2,C) xle-6 70 60 50 40 V R4 40K R3 40K 2 1 8A R1 1K + - Q1 VBE2 Q2 A 3 Common V(3) 80 xle-3 70 60 50 40 -60 -40 -20 0 20 40 tdegc 60 80 100 120 120 The two collector currents track almost perfectly as temperature is swept – And their common value increases with rising temperature as indicated by the voltage across R1 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB 8:1force_2 19 To see what and why the voltage across R1 is what it is, let’s examine a little experiment. Take two identical NPN transistors and bias each into conduction with a voltage applied to the base, using a different voltage for each device. Arrange to measure each collector current and the difference of the base voltages. Now, using the junction equation for the two transistors, the form is the same, but allowance is made for the ratios of current and area. One of the factors in the collector current is an exponential function of Vbe and q, electron charge, over k, Boltzmann’s constant, and T, Absolute Temperature, all minus one. Since the exponential will be on the order of 1e14 to 1e18 for the transistors and environment we will use, the subtraction of one can safely be neglected. Take the ratio of current densities and with a bit of manipulation you will find the difference of the Vbes or ΔVbe as it is generally referred to, depends upon the current ratio which is controlled in the circuit, the emitter area ratio, Kelvin temperature and the constants. 20 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON V ic1 A1 Q1 VBE1 VBE2 + s Q2 + A2 ic2 s V2 V1 VBE1 Common VBE2 Current Density vs. Base Voltage Difference Current Density vs. Base Voltage Difference © APB ∆Vbe V ic1 Ic1 = A1Is(e Vbe1q/kT ic2 -1) A1 Ic1 ~ A1Is(eVbe1q/kT) s Q1 VBE1 V1 + VBE1 - Common VBE2 + V2 - VBE2 Q2 A2 Ic2 = A2Is(eVbe2q/kT-1) s Ic2 ~ A2Is(eVbe2q/kT) Then: (ic1/A1)/(ic2/A2)=(e Vbe1q/kT)/(e Vbe2q/kT)=e (Vbe1-Vbe2)q/KT So: (q/KT)(Vbe1-Vbe2)=In((ic1/A1)/(ic2/A2)) ∆Vbe=Vbe1-Vbe2=(kT/q) In((ic1/A1)/(ic2/A2))=(kT/q) In((ic1/ic2)(A2/A1)) Current Ratio HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Area Ratio 21 The Bottom Line: DVbe = (kT/q)ln((ic1/ic2)(A2/A1)) ...and if the Currents are Equal: DVbe = (kT/q)ln(A2/A1) Setting the current ratio equal to one and noting that the area ratio and the constants are unlikely to change, we find that DVbe is Proportional To Absolute Temperature … 22 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON ... or PTAT. (as it's often referred to) HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 23 Current Density vs. Base Voltage Difference V(Vbe2) V(VBE2, 3) 1.2385 1.25 1.2 1.15 1.1 1.05 1 .95 .9 .85 .8 .75 .7 .65 .6 .55 .5 .45 -60 -40 -20 0 20 40 60 tdegc 80 100 120 140 If we extrapolate temperature to zero Kelvins we should find that ΔVbe also goes to zero despite the fact that the transistors may be differently sized and operated at different currents. This figure shows a linear interpolation of Vbe back to zero. Ideally this extrapolation would go to VG0, the extrapolated bandgap of silicon. But the actual value of the two Vbes would meet at a slightly lower voltage than the extrapolation because the slope of Vbe vs. temperature is not quite linear. That is, the linear extrapolation reaches about 1.2385V at zero Kelvin, but tracing out the complete theoretical characteristics of actual transistors would find they meet at a slightly lower voltage. The curvature of Vbe is difficult to see in the figure, unless you look closely to see the departure from the linearity of the extrapolation in the region where the transistors were accurately simulated as indicated by the colored traces. In practice this curvature over the temperature range of common interest is small enough to often neglect. However, if better temperature behavior is needed several methods may be used to compensate it, one of which will be mentioned later. 24 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB Xtrap For now, let me neglect Vbe curvature and treat it as a linear function of temperature. As we have seen, the PTAT voltage across R1 can be scaled to be larger as it is by R3 (40K) and R4 (40K). So, we can generate a PTAT voltage scaled as we wish to add to the Vbe of Q2. Near our temperature range of interest we should be able to combine Vbe2 with a PTAT voltage scaled to give a wide range of results for the sum. However, as we approach zero Kelvin the PTAT voltage approaches zero, so that whatever Vbe it is combined with will approach VG0, the extrapolated bandgap voltage. The combined Vbe2 and PTAT voltage, being the sum of two functions linear in temperature, should itself be such a linear function. And since we know one value it will take on, we can make that function constant at all temperatures by adjusting the PTAT voltage to make the sum the same as VG0 at any other temperature, including room temperature. And since these voltages complement one another over temperature Vbe2 is often called Complementary To Absolute Temperature, or CTAT. 1.2385 1.2385- (tdegc+273) *2m (tdegc+273) *2m 1.3 1.2 Temperature Invariant Sum of Linear Functions of Temperature 1.1 1 .9 .8 .7 .6 A PTAT Voltage Scaled to be Complemented by Vbe A Vbe With Negative Slope Complementing A PTAT Voltage (CTAT) .5 .4 .3 .2 .1 0 -300 -200 -100 0 100 tdegc, from -273 to 345 200 300 © APB Invar Current Density vs. Base Voltage Difference HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 25 A simple circuit can be used to generate the sum of both the PTAT and CTAT voltages. The amplifier used to drive VBE2 until its inputs balance will try to do that even if there is some additional impediment, such as R2, in the way. That is, the amplifier will drive VBE2 positive and the current necessary to drive R3 and R4 will be supplied by a rising voltage on R2. When the two collector currents match, their sum will pass through R2 (actually a bit more when we include the base currents of Q1 and Q2, but they can often be neglected, or made negligible by additional circuitry) making it take on the PTAT voltage characteristic of current in R1. Since we are free to make R2 whatever value we choose, it can be made large enough to be complemented by the CTAT Vbe at the output of the amplifier. Since the CTAT voltage is not strictly linear, a slight additional PTAT voltage component will tilt up the downward curvature for a best fit to our desired temperature range. This linear addition acts like the extrapolation which, as we said before, passes through a voltage which is a small amount larger than the bandgap. This small amount will be process dependent and the total is often called the “Magic Voltage” for the process. The magic voltage is the output voltage setting at some selected temperature which gives the best temperature performance. Designs to hit the magic voltage can also be trimmed, for example by adjusting R2 at a single temperature. 26 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Disclaimer: This configuration is optimized for presentation, not necessarily for integration V Bandgap Voltage R4 40K R3 40K (well, almost) 2 + 1 8A PTAT ∆Vbe R1 1K R2 4.619K - Q1 VBE2 Q2 A 3 R5 20K CTAT Vbe 4 Adding R2 provides a fully adjustable PTAT Voltage Common HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Multiple Image of ∆Vbe © APB Invar_3 27 PTAT, CTAT, and Resulting ZTAT (~OTC) Sum V R4 40K R3 40K 2 1 8A R1 1K + - Q1 VBE2 3 4 R2 4.619K Common Q2 A 1.25 1.2 1.15 1.1 1 1.05 .95 .9 .85 .8 .75 .7 .65 .6 .55 .5 .45 .4 .35 .3 .25 .2 .15 .1 .05 0 -60 V(4) V(VBE2, 4) V(VBE2) -40 -20 0 20 40 60 tdegc 80 100 120 140 An example of the simulated result for the process used in the other examples shows the low Temperature Coefficient (TC) result at the node VBE2, of adding Vbe2 to the voltage across R2. The apparent TC, at this viewing scale, appears flat and so is often abbreviated ZTAT. The magic voltage for a given process may be less than some desired reference level such as 1.5V or 2V. And/or it may be desirable to make a design easily reusable in other processes which have different magic numbers. 28 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON A Small Additional Burden on the OP-Amp Makes Available Any Desired Output Voltage Greater Than the Bandgap V R4 40K R3 40K 2 + 1 8A R1 1K - Q1 VBE2 Q2 3 A OUT R5 20K Vout=VBE2*(R5/R6+1) R6 20K 4 R2 4.619K Common This requirement can often be accommodated with the addition of a simple voltage divider. As we saw before the amplifier will do whatever it can to balance the two collector currents. So, attenuating the output before driving the common base line will require the amplifier output to rise in proportion to the attenuation to restore the magic voltage to VBE2. As shown in the figure, a resistor divider of one half will roughly double the output voltage. The green trace shows twice the magic voltage at VBE2 while the red trace shows the simulated output. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 29 2.5 V(Out) 2*V(VBE2) 2.0 1.5 1.0 .5 0 -60 -40 -20 0 20 60 40 tdegc 80 140 120 100 V R4 40K R3 40K 2 + 1 8A R1 1K - Q1 VBE2 Q2 A OUT R5 20K R6 20K 3 2X 4 R2 4.619K Common 30 © APB multpl_2 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Our Little Circuit Can Reveal Two Fundamental Error Sources V R4 40K R3 40K 2 + 1 8A R1 1K Q1 - VBE2 Q2 A OUT R5 20K R6 20K 3 4 R2 4.619K Common One is Due to Base Currents in R5, with No Corresponding Current in R6 The Other is Due to Non-linearity of Vbe as a Function of Temperature, the So-called Bandgap Curvature Two errors contributing to non-zero TC can be visualized with this circuit. The effects of base current on the basic reference can often be neglected. However, depending on the current you can afford to spend making the voltage divider from small value resistors, the base current effects on the multiplied output may not be negligible. We can also use this example to investigate curvature compensation. Up until now we have been looking at results on a scale where the total error spanned a pixel or two and looked pretty small. But if we examine the results and the scale of the errors we’ll find they look worse and may require reduction to meet our performance specification. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 31 The green trace represents about as well as we can do with the simple cell and the process at hand. It’s multiplied by two for comparison with V(Out). A similar arrangement should help you determine both the magic voltage and the intrinsic curvature due to your process. The blue curve is twice the magic voltage, which is ideally tangent to the real or simulated result near the midrange temperature. The red trace is the simulated output with a 2X feedback attenuation. The reason it is higher than the others is that the base currents of both transistors flow in R5, slightly raising its voltage, without a corresponding component of current in R6. Neglecting a few things, like the possibility that the base currents differ because of collector current density, we can make an approximate correction for this error. 32 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 2.500 V(Out) 2*V(VBE2) 2*1.2385 2.495 2.490 2.485 2.480 2.475 2.470 2.465 -60 -40 -20 0 20 40 tdegc 60 80 100 120 140 V R4 40K R3 40K 2 + 1 8A R1 1K - Q1 VBE2 Q2 A OUT R5 20K R6 20K 3 2X 4 R2 4.619K Common HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON © APB multpl_433 R7 introduced between Q1 and Q2 carries the base current of only Q1 and introduces a slight offset into the loop, making ΔVbe, which changes the current in R1 and, hence in R2. The formula given in the figure gives a good first approximation which can be refined with a bit of experimentation. V R4 40K R3 40K 2 + 1 8A R1 1K - Q1 VBE2 Q2 R7 2.164K 3 A OUT R5 20K R6 20K 4 R2 4.619K Common Adding R7=(R1/R2)*R5*R6/(R5+R6) creates a base current proportional voltage within the ∆Vbe Loop, reducing the reference voltage, slightly, to compensate the error due to base currents in R5 34 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 2.478 2.477 2.476 2.475 2.474 2.473 2.472 2.471 2.470 2.469 2.468 2.467 2.466 OUT 2.465 2.464 R5 2.463 20K 2.462 2.461 2.460 R6 20K 2.459 2.458 2.457 2.456 2.455 2.454 2.453 2.452 2.451 2.450 2.449 2.448 2.447 2.446 2.445 0 -60 Ideal V(Out) 2*(-1:V(VBE2)) 2*V(VBE2) Previous 2X VBE2 V R4 40K R3 40K 2 1 8A R1 1K - Q1 VBE2 Q2 A R7 2.164K 3 + 4 R2 4.619K Common The Approximation Used For R7 Results in Slight Over-correction 2X Error Correcting Output -40 -20 0 20 40 60 tdegc 80 100 120 140 © APB βcomp_2 Using the formula with the schematic values gives a value for R7. The simulation in black shows the effect of R7 and base current on the nominal voltage at VBE2. The red trace shows the simulated output of the circuit to compare with twice the value of VBE2 from the previous simulation. The black trace shows twice the pre-compensated value to compare with the simulated output which results from base current in R5. The simulated output is a bit less than twice the nominal magic voltage, since the approximation of the formula slightly overcompensated the base current. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 35 R7 can be reduced slightly to take into account a well-modeled difference in the base currents of Q1 and Q2 as well as the approximations in the formula for R7. V R7 Reduced to Avoid Overcompensation R4 40K R3 40K 2 1 8A R1 1K + - Q1 VBE2 Q2 R7 2.009K 3 See JSSC, Vol SC-9, #6, Dec 1974 page 390 A OUT R5 20K R6 20K 4 R7 Value Empirically Optimized for Best Fit (Carefully Tinkered in Simulation) R2 4.619K Common © APB βcomp_3 The result of this adjustment can be seen here to almost exactly match the simulated output to an ideal doubling of the output of the single bandgap, indicating that the output error due to base currents has been greatly reduced. The pre-distorted voltage at VBE2 is also doubled and shown in black to compare with the ideal doubling. As caveat, I want to point out that while this error reduction technique works well in practice you may prefer to find an alternative if noise at the output is an issue. The addition of R7 can easily double the intrinsic minimum noise of this circuit and must be considered along with other substantial noise sources. All of the outputs we have seen show a non-linear voltage vs. temperature component. Unlike PTAT errors, which can be corrected with PTAT voltages that are easily generated, the PTAT output adjustment can only correct for the linear baseline of the nonlinearity resulting from Vbe. 36 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON V(Out) 2*(-1:V(VBE2) 2*1.2385 2*V(VBE2) 2.480 2.475 Ideal 2.470 Compensation Adjusted 2.465 2.460 2.455 2.450 2.445 -60 -40 -20 0 20 40 tdegc 60 80 100 120 140 V R4 40K R3 40K 2 + 1 8A R1 1K - Q1 VBE2 Q2 A R7 2.009K 3 OUT R5 20K R6 20K 4 R2 4.619K Common HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 37 © APB βcomp_4 So what about the Curvature? Where does THAT come from?? The origin of the curvature is with Vbe, as is almost everything in this bandgap circuit. An expression for Vbe derived from first principles allows us to compare Vbe of a transistor with a second transistor which matches it or with itself when operated under different conditions. For a given process and device design, then, the expression for Vbe lets us say “If you’ve seen one, you’ve seen ‘em all.” Vbe of a Transistor as a function of Current and Temperature may be inferred from a measurement of Vbeo, its nominal value at temperature T=To, and current i=io, using the relation: Vbe = VG0 +(T/To)(Vbeo-VG0)+(kT/q)ln(i/io)+m(kT/q)ln(To/T) Where VG0 is the Bandgap Voltage extrapolated to 0K and m is a property of the Transistor determined by its design and processing. m is approximated by XTI in the GP SPICE Model Note that Vbe Consists of: • a Constant • a Term Linear in Temperature • a Term that May be Zero or nonlinear • a Term which is a non-zero, non-linear, function of temperature This last is the origin of the “Bandgap Curvature.” 38 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Since Vbeo will be less than VG0, the dominant linear term will fall with increasing temperature ... And, the third term, (kT/q)ln(i/io), will be zero, if the current is held invariant at io. However — if i=io(T/To), as is the PTAT current in our bandgap cell, then this term may be re-written as: (kT/q)ln(io(T/To)/io)) = - (kT/q)ln(To/T) which may be combined with the fourth term, which is the curvature term, to yield: (m-1)(kT/q)ln(To/T) The Vbe of Q2 Operating with PTAT Current is combined with the voltage across R2 scaled up from ΔVbe to produce the nominal output voltage: Vout= VG0 +(T/To)(Vbeo-VG0)+(m-1)(kT/q)ln(To/T)+ R2((kT/q)ln8)/R1 The second and fourth terms are both temperature proportional and are of opposite sign, so their sum may be made zero by setting the R2/R1 ratio. This leaves Vout as: Vout= VG0+(m-1)(kT/q)ln(To/T) VG0 is the ideal bandgap voltage of silicon extrapolated to zero, and would be a fine reference value if the total was not corrupted by the second, curvature, term. This term is generally curving downward in the temperature range of interest to us so we add a PTAT component (by adjusting R2/R1 to leave a residual + PTAT) to remove the net negative slope of the curvature. As a result we see the peak of the curvature in our temperature range, slightly above the ideal VG0. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 39 A Diferent Perspective on the Basic Circuit Using a Different Scale V(VBE2) 1.2390 1.2385 1.2380 1.2375 1.2370 1.2365 1.2360 1.2355 1.2350 1.2345 -60 -40 -20 0 20 40 tdegc 60 80 100 120 140 V R4 40K R3 40K 2 1 8A R1 1K + - Q1 VBE2 Q2 A 3 4 R2 4.619K Common 40 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Going back to the basic bandgap circuit and looking at the output on a different scale shows the curvature as if superimposed on our desired temperature invariant circuit. Although the curvature term is a logarithmic function of temperature in a series expansion it has a large quadratic term, for which we can compensate and reduce the remainder to a third and higher order error. The way in which the current variable term, (kT/q)ln(i/io), reduces the final curvature term may be exploited by generating a larger multiple to combine with the fundamental curvature of the same form to reduce the sum to zero. A more simple and cost effective compensating curvature can be made by adding a resistor with a more positive TC to the PTAT voltage multiplier. Note that the residual curvature appears to be a parabola, indicating that it has a large quadratic component. A temperature proportional resistor driven by the PTAT current results in a T2 voltage which can be sized to cancel the quadratic component of the residual curvature. The small remainder will be seen to be dominated by a third order term. If we put a PTAT current in a temperature proportional resistor, the result is a quadratic function of temperature. We can do that by putting a diffused resistor with a, relative to R1 and R2, large positive TC in series with R2. Of course R2 must be reduced, not only to “make room” for the R8 voltage, but to also reduce the total voltage across R2 and R8. As we will see, curvature correction gives us better performance, and also brings the magic voltage closer to the actual extrapolated bandgap. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 41 This figure hides the zero TC component of the previous plotted output voltage to show the curvature due to Vbe of Q2. Over the same temperatures we can see the quadratic compensating voltage appearing across R8, an added +TC resistor. The summation of the quadratic correction with Vbe curvature substantially reduces the remaining temperature error. A Quadratic Function Approximates the Curvature Function V Output Curvature Component R4 40K R3 40K 2 1 8A R1 1K - Q1 VBE2 Q2 A 3 4 R2 4.085K R8 414 5 + / PBASE Common A Small + TC Resistor Produces a Squared Response to the PTAT Current 4.50 4.25 4.00 3.75 3.50 3.25 3.00 2.75 2.50 2.25 2.00 1.75 1.50 1.25 1.00 .75 .50 .25 0 -.25 -.50 -.75 -1.00 -1.25 -1.50 -1.75 -2.00 -2.25 -2.50 -2.75 -3.00 -3.25 -3.50 -3.75 -4.00 -4.25 -60 nonlin(V(VBE2, 4)) nonlin(V(5)) -40 -20 0 20 40 60 tdegc 80 100 120 140 © APB curvpm_2 42 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON V Uncorrected Result R4 40K R3 40K 2 1 8A R1 1K + - Q1 VBE2 Q2 A 3 4 R2 4.085K R8 414 5 / PBASE Common Net Corrected Result Approximate Compensation for Curvature Causes “Magic Voltage” to Approach the Bandgap 1.2390 1.2385 1.2380 1.2375 1.2370 1.2365 1.2360 1.2355 1.2350 1.2345 1.2340 1.2335 1.2330 1.2325 1.2320 1.2315 1.2310 1.2305 1.2300 1.2295 1.2290 1.2285 1.2280 1.2275 1.2270 1.2265 1.2260 1.2255 1.2250 1.2245 -60 (-1:V(VBE2)) V(VBE2) -40 -20 0 20 40 60 tdegc 80 100 120 140 © APB curvpm_3 Note that the magic voltage that results from curvature correction is substantially reduced, making it closer to VG0 where it would be without the effects of Vbe curvature. So far, the example circuits and simulations have been based on the SPICE models of an analog bipolar process. Although the Bandgap Reference principles were developed using similar transistors, the need for voltage references persists on circuits made on processes that lack isolated vertical transistors. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 43 ... and Now, for the Device Deprived … From time-to-time some of us may find ourselves Deprived of real (vertical, isolated, bipolar) transistors and forced to make do with crumbs from the CMOS table. Fortunately, even parasitic vertical substrate transistors will exhibit the ΔVbe = (kT/q)ln((ic1/ic2)(A2/A1)) relationship, and the Emitter Current ratio MAY be a satisfactory substitute for (ic1/ic2) Moreover, these ill suited thick-base substitutes may be more uniform in manufacture than their thin-base counterparts that serve us so well. This uniformity is welcome, though it hardly compensates for the lack of individually usable collector currents. At the heart of the bandgap reference is a junction, or two or three. On CMOS processes, diffusions may be arranged to form transistors that have a common collector in the substrate. The opportunity to directly control the collector current ratio by comparing them is not available. Additionally, since the collector currents are not driving a second stage, ΔVbe must be measured directly rather than inferred from a higher gain common base input stage. 44 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON This paper shows in some detail a way to build a temperature stable reference using parasitic Vertical PNP transistors. And to generate an output voltage which may be freely chosen within a useful range. A similar and instructive circuit can be used for illustration. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 45 Simple sub-Bandgap uses CMOS Parasitic PNPs VIN MP13 1/40 1s 4s MP1 MP2 MP3 80/4 80/4 80/4 4s 4s 2 1 PP 5P MP7 C1 MP5 MP6 2s 40/4 2s 4 5 3 40/4 40/4 MN2 MN6 10/1 80/2 1s ZTAT MN5 VPLS - VREF MP7 10/.6 R4 100K R3 123.06K / TF / TF R1 123.06K / TF Q1 Q2 Startup MN1 80/2 PTAT 7 16 CTAT GND R2 15.96K / TF 6 MV 4s Diff In ZTAT Driven to 0 ZTAT V3 + 2s MN4 CTAT 1 MV 4s 6/2 MV 80/2 4s 9 8 MN3 12/2 1s Core 2s Amp This figure shows a relatively simple scheme to derive a 1V reference using VPNP transistors Q1 and Q2. They are part of the core, the center section of the diagram, which is flanked by an Amplifier and a Startup circuit. To simplify the explanation, we can strip away the Startup and part of the core. But First, A Simplified Sub-Block VIN 2 4s MP2 MP3 80/4 80/4 4s 2 MP5 MP6 2s 40/4 2s 4 5 3 0.4750942 0.7882932 0.4750931 40/4 40/4 1.08813 MN2 80/2 Equal Currents In Unequal Sizes Result In V3 + ∆ Vbe Forced VPLS Across R2 2s 6 1.087784 MV 4s Diff In Driven to 0 / TF R2 18.25K 7 0.3622341 GND 24 MN5 MN1 80/2 4s 80/2 MV 9 PTAT Q1 Q2 Core 46 PP 5P MP7 C1 0.2357213 1 1s MN4 6/2 MV 4s 8 MN3 0.2363653 12/2 2s Amp HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON The Amplifier, at the right, drives the gates of two matched PMOS transistors MP2 and MP3, causing them to deliver equal currents to Q2 and Q1 which are in the ratio 24:1 of matched emitters. As long as the currents in the two transistors are equal and non-zero the difference in their Vbes should be given by: ΔVbe = (kT/q)ln24 depending only on the ratio of the currents, not their magnitude. That means that when the matched currents are small the emitter voltage of Q1 is larger than the emitter voltage of Q2 by almost ΔVbe. This will make the gate voltage of MN1 more positive than the gate of MN2 and the drain of MN1 will pull down the gate of MP3 as well as of MP2 connected to node 5. Increasing MP3 current will, of course, increase Vbe of Q1 as a positive feedback. However, as currents increase, so does the voltage across R2 until it eventually is equal to ΔVbe. At that point, the difference between the gate voltages of MN1 and MN2, the inputs to the amplifier, will be zero. Considering the gain of the path from MP3 gate (node 5) to node 1 at the emitter of Q1 it will be limited by the transconductance of MP3 and the impedance of Vbe1. The gain of the path from MP2 gate to node 2 at the top of R2 will be limited by MP2 Gm and the combined impedance of R2 with Q2. Since the impedance of Q1 and Q2 are the same at equal currents, the negative feedback path will dominate the positive feedback path and the circuit will rest at equilibrium with currents set by ΔVbe/R2. Simulation over temperature shows the PTAT voltage appearing across R2. The second trace is the input voltage of the amplifier which is within 1uV over temperature. The amplifier topology yields these results, but of course is corrupted by device matching errors. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 47 We can be certain that when the MP3 and MP2 currents are equal, the amplifier inputs will be balanced and bring their drains to the voltage required to set up the PTAT currents. Developing a PTAT voltage across R2 by forcing equal Node 2 and 3 voltages V(2,7) 115 110 105 100 95 90 85 80 75 70 65 60 550 xle-6 ∆Vbe = (kT/q)ln((i1/1)/(i2/24)) Q1 and Q2 operate at equal currents set by MP2 and MP3, but they are sized 1:24, resulting in a 24X current density ratio and a difference in Vbe given by: ∆Vbe = (kT/q)ln((i1/1)/(i2/24)) The amp forces ∆Vbe to appear across R2 V(2,3) Equal currents in MP2 and MP3 increase the R2 voltage to balance the amp xle-6 1.1 1.0 .90 .80 .70 .60 .50 .40 .30 .20 .10 0 -.10 -.20 -.30 -.40 -60 -40 -20 0 20 40 60 80 100 120 2 MP2 PP 5P MP7 C1 MP3 80/4 4s 80/4 4s 2 40/4 2s 3 0.4750942 4 0.4750931 MP5 MP6 2s 0.7882932 40/4 40/4 5 1.08813 MN5 V3 + / TF R2 18.25K PTAT 7 0.3622341 GND 24 Q1 Q2 Core 48 MN1 MV 80/2 4s 9 0.2357213 MN4 1 1s 6/2 80/2 2s 6 MN2 1.087784 80/2 MV Diff In Driven to 0 VPLS - 140 vpls VIN 4s MV 4s 8 MN3 0.2363653 12/2 2s Amp HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Loading Nodes 2 and 3 requires the Amp to make more MP2 and MP3 Current to preserve equilibrium VIN 4s MP1 MP2 MP3 80/4 80/4 80/4 4s 4s 2 PP 5P MP7 C1 2s 4 5 3 0.4750941 MP5 MP6 2s 40/4 0.4750931 0.8520052 40/4 40/4 1.042865 MN2 80/2 ZTAT MN5 VPLS - VREF 0.9983732 R4 100K GND R3 123K / TF / TF CTAT R2 18.25K 24 MN1 MV 80/2 / TF PTAT 7 / 0.3622341 TF 6 1.04257 MV 4s Diff In ZTAT Driven to 0 ZTAT V3 + 2s Q1 Q2 1 R1 123K CTAT 4s 80/2 9 0.2155659 MN4 1s 6/2 Core MV 4s 8 MN3 0.2159596 12/2 2s Amp Kids! Don't try this at home! (It's a non-starter) If we load, equally, nodes 2 and 3 the amplifier must drive MP2 and MP3 to supply this additional current in order to maintain the input voltages which force the PTAT current. Equal valued R1 and R3 added to the schematic will require more current from MP2 and MP3, and the amplifier will drive them to provide it. Notice that the currents in Q1 and Q2 should remain PTAT while the currents in R1 and R3 are proportional to Vbe and must, therefore, be CTAT. MP2 and MP3 are supplying both. Remembering that the sum of a CTAT and PTAT quantity can be made temperature invariant by using the correct proportions, we can adjust either or both of R2 with R1and R3 to make a temperature invariant current in MP2 and MP3. Such a quantity is sometimes referred to as ZTAT, in this context. The ZTAT current should also flow in MP1, a transistor matched to MP2. This current can drive the load resistor, R4, to a voltage resulting from the ZTAT current flowing in it. HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 49 Loading CTAT Voltages With R1 and R2 Requires Complementary Currents From MP2 and MP3 This figure illustrates the PTAT and CTAT currents and their almost constant sum at scale that reveals an error and a consequence. The error is due to the curvature of Vbe vs. temperature, and will remain in the final output. The consequence is a slight net positive TC to the current, due to a small TC in the thin film resistors used for simulation. This TC should not appear in the output, since R4 has the same TC. The output voltage results from approximately 10µA flowing in R4, so that it can be set over a wide range by changing the nominal value of R4. The small difference seen between i(R4) and i(R2) + i(R3) is mostly a result of the difference in the drain voltage of MP1 from that of MP2 and MP3. In many applications this may be neglected, or if Vin has a wide range of values, the three ZTAT transistors can be cascoded to reduce supply voltage sensitivity. This circuit shares a common “gotcha” with most self biasing circuits. In addition to the desired behavior described here, the circuit can take on another stable state when all currents and the reference voltage are at zero. Once in this state the circuit will remain there. 50 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON i(R2) i(R3) 7.0 R2 Current is Still PTAT 6.5 6.0 xle-6 5.5 5.0 4.5 4.0 3.5 New R3 Current (and R1, too) is CTAT 3.0 i(R2) + i(R3) i(R4) 10.10 10.08 Sum of PTAT and CTAT and as Mirrored by MP1 xle-6 10.06 10.04 10.02 10.00 9.98 9.96 V(VREF) 1.0020 1.0015 1.0010 1.0005 1.0000 .9995 Nominal 1V Ref over Temp .9990 (Your Results May Vary) .9985 .9980 -60 -40 -20 0 20 40 vpls 60 80 100 120 140 VIN MP1 4s MP2 80/4 80/4 4s PP 5P MP7 C1 MP3 80/4 4s 2 2s 40/4 2s 4 5 3 0.4750941 MP5 MP6 0.4750931 40/4 40/4 1.042865 MN2 0.8520052 80/2 ZTAT MN5 VREF 0.9983732 R4 100K GND R3 123K / TF / TF CTAT R2 18.25K 24 MN1 MV 80/2 / TF PTAT 7 / 0.3622341 TF 6 4s Diff In ZTAT Driven to 0 ZTAT V3 + VPLS - 2s 1.04257 MV Q1 Q2 Core 1 R1 123K CTAT 4s 9 0.2155659 1s 80/2 MV 4s 8 MN4 MN3 0.2159596 6/2 12/2 2s Amp HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON 51 So, most self-biased circuits require a startup which is sometimes easily arranged, other times not. My personal preference is to avoid dynamic starts, since once started and switched off, for example by noise, your circuit may then fail to restart. This circuit includes a starter circuit that operates when power is applied to VIN. If the circuit does dynamically start (depending on the dV/dt of VIN among other things), fine, but if not, VREF will be low ~ 0. As a result, MP13 will come on as power ramps up. It will turn on M6 which drives the common PMOS gate line near the top turning on MP2, 3, and-so-forth. Once there are some non-zero currents the circuit will come on regeneratively. The current in MN6 is likely to be very much larger than the nominal current provided by MN1 to node 5, and it is important to prevent it from corrupting the reference. As the circuit starts and VREF rises, it will drive MN7 on. Note that while MN7 has a relatively large gate aspect ratio, MP11 has a small one and is easily overcome by MN7. When node one is pulled low, MN6 will be turned off and the small current from MP13 will continue to flow in MN7. This arrangement works well and can be used in other applications. But, a word of warning: The startup circuit works by triggering the positive feedback in the core and amplifier loop. However, the startup circuit includes another negative feedback path in addition to the one that stabilizes the bandgap output. Normally this path is unstable and gives control to the desired loop. However, it may be stabilized by the addition of a low frequency dominant pole, such as one formed by a large capacitor from VREF to GND that might be added to reduce noise. That can cause the startup to stabilize VREF at a voltage lower than the main loop. But, assuming that problem is avoided, the startup can be observed in a sweep of VIN. 52 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Origin of the expression “Awakened with a Start” VIN MP13 1/40 MP1 1s 4s MP2 80/4 80/4 4s 80/4 4s 2 1 PP 5P MP7 C1 MP3 MP5 MP6 2s 40/4 2s 4 5 3 40/4 40/4 MN2 MN6 10/1 80/2 1s ZTAT MN5 VPLS - MN7 10/.6 GND Startup VREF R4 100K R3 123K / TF CTAT / TF / TF R2 18.25K 24 MN1 MV 80/2 / TF PTAT 7 6 MV 4s Diff In ZTAT Driven to 0 ZTAT V3 + 2s Q1 Q2 1 R1 123K CTAT 4s 80/2 9 1s Core HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON MV 4s MN4 MN3 6/2 12/2 8 2s Amp 53 Initially R4 holds VREF low for any small VIN voltage. When VIN is below the threshold of any MOS device, they must all be off or operating sub-threshold at low currents. The green trace shows the simulator’s guess at the node 1 voltage for this condition. As VIN reaches the threshold voltage of MP13 it comes on and pulls node 1 to VIN. Initially this voltage is insufficient to turn on MN6, but as VIN continues to rise MN6 connects node 5 (the common PMOS gate line) to node 1 at Q1. VIN continues to rise so that MP2 and MP3 start to turn on the core, and when VIN reaches about 1.6V the positive feedback of the main loop comes on and drives VREF to the stable operating point. As VREF rises it turns on MN7 which pulls node 1 low and cuts off MN6. This circuit using VPNPs does not explicitly express the bandgap voltage, although other arrangements can be made to do so. A simple thought experiment may give a different perspective on how the bandgap still remains the reference. At zero Kelvin we should not expect the circuit to work correctly, but we can extrapolate room temperature behavior to that temperature. 54 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON Start-up and Regulation vs. Supply Voltage V(VREF) V(V1) V(VIN) 5 4 3 2 1 0 0 .5 1 1.5 2 2.5 vpls 3 3.5 4 4.5 5 MN6 energizes the PMOS Gate Line until VIN provides headroom to start. Then VREF turns on MN7 to sink small drain current of Long Channel MP13 VIN MP13 1/40 MP1 1s 4s MP2 80/4 80/4 4s 80/4 4s 2 1 PP 5P MP7 C1 MP3 MP5 MP6 2s 40/4 2s 4 5 3 40/4 40/4 MN2 MN6 10/1 80/2 1s ZTAT MN5 VPLS - MP7 10/.6 GND Startup VREF R4 100K R3 123K / TF CTAT / TF / TF R2 18.25K 24 80/2 / TF PTAT 7 6 MV 4s Diff In ZTAT Driven to 0 ZTAT V3 + 2s Q1 Q2 1 R1 123K CTAT MN4 6/2 MN1 MV 4s 80/2 9 1s Core HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON MV 4s MN3 12/2 8 2s Amp 55 In a Thought Experiment Extrapolating to 0K ZTAT Currents Remain ~ 10µA VIN MP13 1/40 MP1 1s 4s MP2 80/4 80/4 4s 80/4 4s 2 1 PP 5P MP7 C1 MP3 MP5 MP6 2s 40/4 2s 4 5 3 40/4 40/4 MN2 MN6 10/1 80/2 1s ZTAT MN5 VPLS - MP7 10/.6 GND VRef Remains ~ 1V Startup VREF R4 100K R3 123K / TF CTAT / TF / TF R2 18.25K 24 MN1 MV 80/2 / TF PTAT 7 6 MV 4s Diff In ZTAT Driven to 0 ZTAT V3 + 2s Q1 Q2 1 R1 123K CTAT 4s 80/2 9 1s Core MV 4s MN4 MN3 6/2 12/2 8 2s Amp PTAT Currents ~ 0 CTAT Voltage ~ 1.23V ~ “Magic Voltage” Suppose we take this circuit and imagine the linear approximation of the PTAT and CTAT voltages across R2 and R1, R3, respectively, will continue to zero Kelvin. As we approach zero the PTAT component will approach zero volts, while Vbe should approach the linear extrapolation to VG0. If we also imagine that the ZTAT current will remain invariant at 10µA, all of this current must now be in each of R1 and R3. Using the values which gave the “correct” ZTAT current at room temperature we see that the 10µA in 123K indicates the Magic Voltage should be about 1.23V, for these transistors. 56 HOW TO MAKE A BANDGAP VOLTAGE REFERENCE IN ONE EASY LESSON R=R VOUT + A - ll Folks ! A s ’ t ha T h-th A th-t 8A I hope this presentation has been useful and answered a few questions – and I'd like to think it may have evoked even more questions. VIN We have some of the answers at IDT MP13 – perhaps you could help us find more. 1/40 1s MP1 MP2 4s 80 /4 4s 80 /4 1 2 MN6 10 /1 80 /4 2s 3 4 ZTAT VPLS - VREF R4 100K R3 123K ZTAT / TF / TF D ZTAT Driveiff In n to 0 R2 18.25K MN5 7 / PTAT TF HOW TO MAKE IN ONE EASY Q2 LESSON GNAD BANDGAP VOLTAGE REFERENCE CTA Startup 4s 1s V3 + MP7 10 /.6 MP3 T Q1 24 Core / TF 80 /2 R1 123K 57 1 CTAT Amp Trick 9 1s MV 4s MN4 6// 2 V+ V+ R=R VO + A - VIN MP13 1/40 A 8A 1s 1 MN6 10 /1 1 V3 + VPLS - MP7 10 /.6 1 GND Startup VOUT 1s MP1 4s MP2 80 /4 4s 4s 2 6 1 MP3 80 /4 PP 5P MP7 C1 80 /4 2s 3 ZTAT R4 100K 2 4 1s VREF 40 /4 R3 123K / TF CTAT ZTAT / TF / TF 24 D ZTAT Driviff In en to 0 R2 18.25K 7 MN5 / TF PTAT Q2 Q1 Core 5 1 R1 123K CTAT 80 /2 Amp Trick 9 1s MV MN1 4s 80 /2 MN4 6// 2 MV 4s MN3 12/2 Amp 2 V+ + R=R + A - 8A A Integrated DeviceTechnology Learn more: www.idt.com VOUT