IV. Diodes. 4.1 Energy Bands in Solids

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1 I. Dodes We start our study of nonlnear crcut elements. These elements (dodes and transstors) are made of semconductors. A bref descrpton of how semconductor devces work s frst gven to understand ther v characterstcs. You wll see a rgorous analyss of semconductors n the breadth courses. 4.1 Energy Bands n Solds In every atom, the nucleus (postvely charged) s surrounded by a cloud of electrons. Intally scentsts envsoned the atomc structure to be smlar to the solar system: the nucleus n the center (smlar to the sun), electrons revolvng on orbts around the nucleus (smlar to planets), and the electrc attracton force between the nucleus and electrons beng smlar to the gravtatonal force between the sun and planets. However, accordng to Maxwell s equatons, electrons that revolve around a nucleus should emt electromagnetc radaton. As such, the electrons should lose energy gradually and ther orbt should decay (n a fracton of a second!). Ths means that there s a mechansm that would not allow the electrons to lose energy gradually and s one of man reasons that the quantum theory was developed. Accordng to quantum theory, electrons revolvng around an atom can have only dscrete levels of energy as s shown below. Electrons are not allowed to gradually lose or gan energy. They can only jump from one level to another (by absorbng or emttng a quanta of energy, usually a photon). Furthermore, there are only a fnte number of electrons that are allowed n each state (Paul s Prncple). Therefore, both the energy levels for electrons and the number of electrons at each energy level are specfed. Because any system would tend to be n a mnmum energy state, electrons n an atom start by flng the lowest energy level. Once all allowable slots s flled, electrons start fllng the next energy level and so on. Therefore, the electronc structure nclude flled energy levels (all slots are taken by electrons), empty energy levels (postons are avalable but no electron s present), and partally flled levels (there are some electrons but there s also room for more electrons). Isolated Atoms A Sold ECE65 Lecture Notes (F. Najmabad), Wnter

2 When atoms are arranged n a sold, the nter-spacng between atoms can become comparable to the sze of electron orbtals of each atom (electrons at each energy level are confned to a regon n space called the orbtal and the hgher energy levels have larger orbtal szes). In ths case, the outer orbtals merged nto energy bands. Electrons n these bands are not ted to an atom, rather they are free to move around the sold (f space s avalable per Paul s Prncple). In addton, nstead of dscrete energy levels, these shared electrons can have contnuous values of energy wthn a band of energy. As before, there are range of energes that no electron can occupy. These range of energes are called Forbdden Gaps. (see the fgure above). The electrc propertes of metals, semconductors, and nsulators are can be understood wth ths pcture. As these propertes are ted to energy bands and forbdden gaps, we only focus on these regons as are shown n the fgure below. In metals, one of the energy bands s only partally flled. Obvously, all energy bands below (.e., wth lower energes) are completely flled and all energy bands above are completely empty. Electrons n the partally flled band can easly move around the sold wth smallest amount of energy as there are lots of spaces avalable. As such, metals can conduct electrcty (and heat) very well. In a semconductor, no partally flled band exsts. In the flled energy band, there are a lot of electrons, but there are no avalable slot to move nto. In the empty band, there are a lot of slots but no electrons. One would expect that semconductors to be perfect nsulators (for both electrcty and heat). However, the sze of the forbdden gap n a semconductor s small. At room temperature, the thermal energy n the materal s suffcent to provde suffcent energy to a select number of electrons to move from the flled energy band nto the empty energy band. Empty Energy Band Forbdden Gap Partally Flled Band Forbdden Gap Flled Energy Band Metals Empty Energy Band Forbdden Gap Flled Energy Band Semconductors These electrons that are promoted to the empty energy band now can carry electrcty and heat because there are a lot of slots avalable n ths band. In addton, these electrons leave spaces (or holes ) n the orgnally flled energy band. As such, electrons n the orgnally flled energy band can also move around the sold by movng from one hole to another hole and partcpate n the conducton of electrcty. Obvously t s easer to keep track of the small number of holes n the flled energy band as opposed to the large number of electrons n that band (t s much easer to fnd the attendance n a almost full class room by countng the number of empty seats rather the number of students!). In ths pcture, when electrons move for example to the left to fll a hole, t would look lke the hole s movng to the rght drecton. So, n descrbng semconductors we usually keep track of negatvely charged ECE65 Lecture Notes (F. Najmabad), Wnter

3 electrons (those that are promoted to the empty energy band) and postvely charged holes (spaces left behnd). nce at room temperature the number of electrons that are promoted nto the energy band ( to ) s much smaller than number of electrons n the partally flled band of a metal ( to ), semconductor conduct electrcty but have a much lower conductvty (thus, the name semconductor). Overall, the propertes of semconductors are very senstve to temperature as the number of electrons than can jump over the Forbdden gap ncreases exponentally. Ths s an mportant effect and should be taken nto account n the desgn of electronc crcuts. Lastly, nsulator have a smlar energy band structure as semconductors (no partally flled band) wth one excepton. The sze of the forbdden gap n an nsulator s several tme larger than that of a semconductor. As such, the number of electrons that can be promoted to the energy band s very small ( 10 7 at room temperature). So, these materal are very poor conductors. 4.2 Semconductor Devces Semconductor materal are manly made of elements from group IB of the perodc table lke C (damond),, Ge, C. These materal have 4 electrons n ther outer most electronc shell. Each atom can form a covalent bond wth four of ts neghbors sharng one electron wth that atom. In ths manner, each atom sees eght electrons n ts outer most electronc shell (4 of ts own, and one from each neghbor), completely fllng that shell. It s also possble to form ths type of covalent bond by combnng elements from group IIIB (sharng three electrons) wth element from group B (sharng fve electrons). Examples of these semconductors are GaAs or AlGaAs and are usually called 3-5 semconductors. We focus mostly on semconductors n ths class. Fgure below shows ths covalent bond structure for. A par of electrons and holes are slow shown. Note that form a tetrahedron structure and an atom n the center of the tetrahedron share electrons wth atoms on the each vertex. Fgure below s a two-dmensonal representaton of such a structure. The left fgure s for a pure semconductor and an electron-hole par s depcted. Both electrons and holes are called moble carrers as they are responsble for carryng electrc current. If we add a small amount of an element from group B, such as P, to the semconductor, we create a n-type semconductor and the mpurty dopant s called a n-type dopant. Each of these new atoms also form a covalent bond wth four of ts neghbors. However, as a n-type dopant has 5 valance electron, the extra electron wll be located n the empty energy band. As can be seen, there s no hole assocated wth ths electron. In addton to electrons from the n-type dopant, there are electron-har par n the sold from the base semconductor ( n ECE65 Lecture Notes (F. Najmabad), Wnter

4 the above fgure) whch are generated due to temperature effects. In a n-type semconductor, the number of free electrons from the dopant s much larger than the number of electrons from electron-hole pars. As such, a n-type semconductor s consderably more conductve than the base semconductor (n ths respect, a n-type semconductor s more lke a resstve metal than a semconductor). Hole Electron Electron Hole P B Pure Semconductor n-type semconductor p-type semconductor In summary, n a n-type semconductor there are two charge carrers: holes from the base semconductor (called the mnorty carrers) and electrons from both the n-type dopant and electron-hole pars (called the majorty carrer). mlarly, we can create a p-type semconductor by addng an element from group IIIB, such as B, to the semconductor. In ths case, the p-type dopant generate holes. We wll have two charge carrers: majorty carrers are holes from the p-type dopant and electron-hole pars and mnorty carrers are electrons from the base semconductor (from electron-hole pars). The charge carrers (electrons and holes) move n a semconductor through two mechansms: Frst, charge carrers would move from regons of hgher concentraton to lower concentraton n order to acheve a unform dstrbuton throughout the semconductor. Ths process s called dffuson and s characterzed by the dffuson coeffcent, D. Second, charge carrers move under the nfluence of an electrc feld. Ths moton s called the drft and s characterzed by the moblty, µ. The rato of D and µ s ndependent of the dopant or the base semconductor materal and s gven by Ensten s Equaton D µ = kt q T 11, 600 T where k s the Boltzmann s constant, T s temperature n Kelvn, and q s the electron charge. Parameter T s called the volt-equvalent of temperature and appears n most semconductor formulas. At room temperature, T ECE65 Lecture Notes (F. Najmabad), Wnter

5 v 4.3 The Juncton Dode D The smplest semconductor devce s a pnjuncton dode as s shown n the fgure wth p- D type materal on the left and n-type materal on the rght and metal contacts are attached to both sdes. Crcut symbol for the devce s also shown wth the nward arrow depctng the p-type materal sde. The conventon for the drecton of the dode current and the voltage across the dode are also shown. We see that n the n-type materal sde, there s a large concentraton of free electrons. As such, these electrons would lke to dffuse to the p-sde where there are few electrons. mlarly there s a large concentraton of holes n the p-type materal and they would lke to dffuse to the n-type materal sde. When these electrons and holes meet n the vcnty of the juncton, they combne and are neutralzed. nce ths regon s depleted of moble carrers, t s called the depleton regon (also referred to as the transton or spacecharge regon). The wdth of ths regon s small, typcally 0.5µm. nce the n-type materal has lost electrons, t becomes postvely charged. mlarly the p- type materal becomes negatvely charged. As a result, an electrc potental and an electrc feld appears across the depleton regon. Ths electrc feld (or the voltage barrer) mpedes the flow of electrons from the n-type materal and flow of holes from the p-type materal. The heght of ths voltage barrers s related to the forbdden gap of the base semconductor materal (and not the amount of dopant). v Characterstcs Let us attach a voltage source to ths juncton dode such that the postve termnal of the voltage source s attached to the n-type materal. Ths confguraton s called reverse bas. In ths case, some of the free electrons from the n-type materal wll travel to the voltage sources makng the n-type materal sde more postvely charged. mlarly, some of the holes from the p-type materal would go the voltage source, makng the p-type materal more negatvely charged. As a result, the depleton regon becomes wder and the heght of the potental barrer ncreases and no majorty carrers can cross the depleton regon. ECE65 Lecture Notes (F. Najmabad), Wnter

6 A small current however, flows n the crcut because of the mnorty carrers. On the n-type materal sde, electron-hole par are created due to thermal energy. Electrons move to the voltage source and holes move to the depleton regon. mlarly, on the p-type materal sde, electron-hole pars are created, holes are neutralzed by electrons from the voltage source and electrons move to the depleton regon. These electrons from the p-type materal and holes from the n-type materal (mnorty carrers) combne n the depleton regon leadng to the flow of a net current whch s called the reverse saturaton current, I S. In the crcut above, D = I S. I S s very small: at room temperature t s about several na (10 9 A) for semconductors. alue of I S strongly depends on the temperature (as t s related to the electron-hole producton rate) I S doubles for every ncrease of 7 C n temperature. Now, let us attach the voltage source to the dode such that the postve termnal s attached to the p-type materal. Ths confguraton s called forward bas. v D D everse Bas v D D Forward Bas Electrons are njected nto the p-type materal makng that regon less postvely charged whle electrons are taken from the n-type materal makng t less negatvely charged. As a result, the depleton regon become thner and the heght of the potental barrer decreases. In ths case the dffuson process can gradually overcome the force of the electrc feld and some of majorty carrers can cross the juncton and a postve current flows through the dode. As the voltage source voltage s ncreased, the potental barrer heght become smaller leadng to larger and larger currents (dode current ncreases exponentally wth the appled voltage). The v characterstcs of a juncton dode s gven by: D = I s ( e v D /η T 1 ) where the constant η s called the emsson coeffcent (η = 2 for and η T = ). Fgure below (left) shows the plot of D versus v D for a typcal dode. Note that because I S s very small (na range), v D should become large enough such that large value of exp(v D /0.052) compensate for the small value of I S. As can be seen, the dode current sharply rses when v D s n the range of 0.6 to 0.7. If we zoom near the orgn and expand the current scale to na range (fgure to the rght), we can see the behavor of the dode n the reverse-bas regon. ECE65 Lecture Notes (F. Najmabad), Wnter

7 8 7 6 I D (ma) I D (na) D () D () Dode Lmtatons: In the forward bas regon, the power P = D v D s dsspated n the dode n the form of heat. The dode packagng provdes for the conducton of ths heat to the outsde and dode s cooled by the ar. If we ncrease v D (or D ) dode s heated more. At some pont, the generated heat s more than capablty of the dode package to conduct t away and dode temperature rses dramatcally and dode burns out. As v D changes slowly, ths pont s usually characterzed by the current D,max, maxmum forward current, and s specfed by the manufacturer. Heat generaton s not a problem n the revers bas as D 0 (and, thus, P 0). However, f we ncrease the reverse bas voltage, at some voltage, a large current can flow through the dode. Ths voltage s called the reverse breakdown voltage or the Zener voltage. Ths large reverse current s produced through two processes. Frst, n the reverse bas, mnorty carrers enter the depleton regon. These mnorty carrers are accelerated by the voltage across the depleton regon. If the reverse bas voltage s hgh enough, these mnorty carrer can accelerate to a suffcently hgh energy, mpact an atom, and dsrupt a covalent band, thereby generatng new electron-hole pars. The new electron-hole pars can accelerate, mpact other atoms and generate new electron-har pars. In ths manner, the number of mnorty carrers ncreases exponentally (an avalanches process), leadng to a large reverse current. Ths s called the avalanche breakdown. Second, when the strength of the electrc feld across the juncton becomes too large, electrons can be pulled out of the covalent bonds drectly, generatng a large number of electron-hole pars and a large reverse current. Ths s known as the Zener effect or Zener breakdown. egular dodes are usually destroyed when operated n the Zener or reverse breakdown regon. These dode should be operated such that v D > v Z. A specal type of dodes, Zener dodes, are manufactured specfcally to operate n the Zener regon. We wll dscuss these dodes later. In Zener dodes, heat generaton sets a maxmum for the allowable dode current n the Zener regon. ECE65 Lecture Notes (F. Najmabad), Wnter

8 4.4 Solvng Dode Crcuts Wth dode v characterstc n hand, we now attempt to solve dode crcuts. Consder any lnear crcut wth a dode. The box labeled the rest of the crcut n the fgure can be replaced by ts Thevenn equvalent, gvng the smple dode crcut below. est of Crcut D v D Because the three elements are n seres, current D flows through all elements. Wrtng KL around the loop we have: T T = D T v D Ths s an equaton wth two unknowns ( D and v D ) as T and T are known. The second needed equaton (to get two equatons n the two unknown) s the dode characterstc equaton: T D v D D = I s ( e v D /η T 1 ) The above two equatons n two unknown cannot be solved analytcally as the dode v equaton s non-lnear. PSpce solves these equatons numercally. As analytcal solutons can provde nsght n the crcut behavor and may be also needed for crcut desgn, we develop two methods to solve dode crcuts wthout numercal analyss. Load Lne The v characterstc of the dode s plotted n the fgure. The D and v D values of any pont on the curve satsfes the dode equaton above. Also plotted s the lne representng T = D T v D. Ths lne s called the load lne. The D and v D values of any pont on the load lne satsfes the frst equaton above. nce the soluton to the above two equatons n two unknowns s a par of D and v D that satsfes both equatons, such a pont should be both on the v plot of the dode and on the load lne,.e., at the ntersecton of the two. Ths pont s called the Q-pont or the Quescent pont (or the operatng pont) of the dode. The load lne technque s not accurate n fndng numercal values of D and v D. However, t s a powerful tool to get qualtatve measures of the crcut behavor e.g., ensurng that the dode operatng pont s safety away from the maxmum forward current T / T 5 4 I Q I D Q Load Lne Q T D ECE65 Lecture Notes (F. Najmabad), Wnter

9 Pecewse Lnear Model The problem n arrvng at an analytcal soluton to the dode crcut s that the dode v equaton s nonlnear. An approach would be to try to approxmate the dode v equaton by a lnear equaton (.e., the dode v characterstcs curve wth a lne) I D D As can be seen from the fgure, t s NOT possble to approxmate the v characterstcs curve wth ONE lne. However, t s possble to do so wth TWO (or more) lnes. Ths type of approxmaton s called a pecewse lnear model,.e., peces of the curve are replaced by lnes. One such model for dodes s shown n the fgure wth equatons: Dode ON: v D = v γ, and D > 0 Dode OFF: D = 0 and v D < v γ, Parameter v γ s called the cut-n voltage and t s related to the forbdden gap of the base semconductor (not what dopant s used and how much). For, v γ For GaAs, v γ Note that the above equatons are vald only for a certan range of parameters, (or range of valdty) e.g., Dode ON equaton s vald only f D > 0. An ssue that arses n solvng crcuts wth dodes s that we do not know a pror the state of the dode (ON or OFF) and so we do not whch equaton to use. The followng recpe can be used to solve dode equatons: ecpe for solvng dode crcuts: 1. Wrte down all crcut equaton wth D and v D as parameters and smplfy as much as possble. 2. Assume dode s n one state (ether ON or OFF). Use the dode equaton for that state to solve the crcut equatons and fnd D and v D. 3. Check the range of valdty nequalty wth the values of D and v D that was found. If D and v D values satsfy the range of valdty, the assumpton was correct. Otherwse, the assumed dode state s ncorrect. Go to step 2 above and assume that the dode s n the other state. ECE65 Lecture Notes (F. Najmabad), Wnter

10 Let s use ths method for the dode crcut shown wth v s = 5 and s = 1 kω. S Step 1: KCL tells that current D flow n all elements and by KL: S D v D v s = D s v D 5 = 10 3 D v D Step 2: Assume dode s OFF: Dode OFF: D = 0 and v D < v γ Substtutng for D = 0 n the crcut equaton, we get: 5 = v D v D = 5 Step 3: nce v D = 5 > 0.7 = v γ, dode v D does NOT satsfy range of valdty and the assumed dode state s ncorrect. Step 2: Assume dode s ON: Dode ON: v D = v γ and D > 0 Substtutng for v D = v γ = 0.7 n the crcut equaton, we get: 5 = 10 3 D 0.7 D = 4.3 ma Step 3: Although we know that snce dode was not OFF, t should be ON, t s a good practce to check the range of valdty agan n case we mght have made a math mstake. For ths case, D > 0 and satsfes range of valdty. So dode s ON and v D = 0.7 and D = 4.3 ma. Parametrc soluton of dode crcuts It s very useful f we can derve the crcut soluton parametrcally,.e., wth values of varous crcut elements as parameters. Ths approach would allows us to solve the crcut ONCE. For example, for nvertng amplfer, we solved the crcut parametrcally and found the gan to be A = 2 / 1. So, f we encounter a crcut wth 1 = 1 kω and 2 = 10 kω, we can calculate the gan, A = 2 / 1 = 10 wthout solvng any crcut equatons. An example would the dode crcut above n whch we would lke to fnd D and v D n terms of v s and s. The problem s that dode can be ether of ts two states whch n prncple would depend on the values of crcut parameters (v s and s n the example above). As a result, snce the dode has two states, we wll fnd TWO solutons to the crcut, each beng vald over a range of parameters as descrbed n the recpe below: ECE65 Lecture Notes (F. Najmabad), Wnter

11 ecpe for solvng dode crcuts parametrcally: 1. Wrte down all crcut equaton wth D and v D and smplfy as much as possble. 2. Assume dode s n one state. Use the dode equaton for that state to solve the crcut equatons and fnd D and v D. Use the range of valdty of that state to fnd a range of valdty for the crcut parameters. 3. epeat step 2 for all possble states. For example, consder the crcut above wth v s and s as parameters. Followng our recpe we get: KCL tells that current D flow n all elements and by KL: v s = D s v D Case 1: Dode s OFF: D = 0 v D = v s v D < v γ v s < v γ Case 2: Dode s ON: v D = v γ D = v s v γ s D > 0 v s v γ s > 0 v s > v γ Therefore, the parametrc soluton of the crcut can be summarzed as: v s > v γ Dode n ON v D = v γ and D = v s v γ s v s < v γ Dode n OFF v D = v s and D = 0 Note that f we had solved the crcut parametrcally frst, we can get the answer for the case v s = 5 and s = 1 kω mmedately: v s = 5 > v γ = 0.7, therefore, dode s ON, v D = v γ = 0.7 and D = (5 0.7)/10 3 = 4.3 ma. We wll use the above two methods to explore some dode crcuts. ECE65 Lecture Notes (F. Najmabad), Wnter

12 4.5 Dode Logc Gates You have seen bnary mathematcs and logc gates n ECE25. We wll explore some electronc logc gates n ths course. Bnary mathematcs s bult upon two states: 0, and 1. We need to relate the bnary states to currents or voltages as these are the parameters that we can manpulate n electronc crcuts. mlar to our dscusson of analog crcuts, t s advantageous (from power pont of vew) to relate these the bnary states to voltages. As such, we choose two voltages to represent the bnary states: L for state 0 or Low state and H for state 1 or Hgh state (for example, 0 to represent state 0 and 5 to represent state 1). These voltages are qute arbtrary and can be chosen to have any value. We wll add subscrpt O and I to these voltages to denote nput and output voltages to the gate, e.g., OL s the output voltage of the gate at the low state. mlar to analog crcuts, we can plot the voltage transfer characterstcs of a gate. For example, the voltage transfer characterstcs of an deal nverter s shown n the fgure, when the nput s low, the output s hgh and the when the nput s hgh, the output s low. We can see a dffculty rght away. In a practcal crcut, there would be an output voltage for any nput voltage, so the output voltage makes a smooth transton for the hgh voltage to the low voltage as the nput voltage s vared. We have to be careful as t s extremely dffcult, f not mpossble, to desgn an electronc crcut to gve exactly a voltage lke 5 (what f the nput voltage was 4.99?). So, we need to defne a range of voltages (nstead of one value) to represent hgh and low states. We wll try very hard to make sure that the output of our gates to be as close as possble to OH and OL. But, we wll desgn our gates such that they respond to a range of voltages,.e., the gate would thnk that the nput s low f the nput voltage s smaller than IL and would thnk that the nput s hgh f the nput voltage s larger than IH. Wth ths defntons, the voltage transfer characterstcs of a practcal nverter s shown. The range of voltages, OL to IL and IH to OH are called the nose margns. The range of voltages between IL to IH s the forbdden regon as n ths range, the output of the gate does not correspond to any bnary state. The maxmum speed that a logc gate can operate s set by the tme t takes to traverse ths regon as the nput voltage s vared from one state to another state. H L OH IL O O L OL IL IH H OH I I ECE65 Lecture Notes (F. Najmabad), Wnter

13 Logc crcuts are typcally constructed from basc logc gates lke NO or NAND. You have seen n ECE25 that all hgher level logc gates, e.g., flp-flops, can be made by a combnaton of NO gates or NAND gates. So, for each logc gate that we wll work on, we have to remember that the output of the logc gate s attached to the nput of another logc gate. Dode AND Gate: Let us now to proceed to a smple logc gate, a dode AND gate as s shown. To study the behavor of the gate we wll consder the state of the crcut for dfferent values of v 1 and v 2 (ether 0 or 5 correspondng to low and hgh states). To ad the analyss, let s assume CC = 5 and A = 1 kω. We note that by KCL, A = 1 2 (assumng that there s no current drawn from the crcut). Case 1, v 1 = v 2 = 0: nce the 5- supply wll tend to forward bas both D 1 and D 2, let s assume that both dodes are forward based. Thus, v D1 = v D2 = v γ = 0.7 and 1 > 0, 2 > 0. In ths case: v v 1 2 D D 1 2 A 1 2 CC A v o v o = v 1 v D1 = v 2 v D2 = 0.7 A = CC v o = = 4.3 ma A 1, 000 Current A wll be dvded between two dodes by KCL, each carryng one half of A (because of symmetry). Thus, 1 = 2 = 2.1 ma. nce dode currents are postve, our assumpton of both dode beng forward based s justfed and, therefore, v o = 0.7. So, when v 1 and v 2 are low, D 1 and D 2 are ON and v o s low. Case 2, v 1 = 0, v 2 = 5 : Agan, we note that the 5- supply wll tend to forward bas D 1. Assume D 1 s ON: v D1 = v γ = 0.7 and 1 > 0. Then: v o = v 1 v D1 = 0.7 v o = v 2 v D2 v D2 = 4.3 < v γ and D 2 wll be OFF ( 2 = 0). Then: A = CC v o = = 4.3 ma A 1, = A 2 = = 4.3 ma nce 1 > 0, our assumpton of D 1 beng forward based s justfed and, therefore, v o = 0.7. ECE65 Lecture Notes (F. Najmabad), Wnter

14 So, when v 1 s low and v 2 s hgh, D 1 s ON and D 2 s OFF and v o s low. Case 3, v 1 = 5, v 2 = 0 : Because of the symmetry n the crcut, ths s exactly the same as case 2 wth roles of D 1 and D 2 reversed. So, when v 1 s hgh and v 2 s low, D 1 s OFF and D 2 s ON and v o s low. Case 4, v 1 = v 2 = 5 : Examnng the crcut, t appears that the 5- supply wll NOT be able to forward bas D 1 and D 2. Assume D 1 and D 2 are OFF: 1 = 2 = 0, v D1 < v γ and v D2 < v γ. Then: A = 1 2 = 0 v o = CC 1 A = 5 0 = 5 v D1 = v o v 1 = 5 5 = 0 < v γ and v D2 = v o v 2 = 5 5 = 0 < v γ Thus, our assumpton of both dodes beng OFF are justfed. So, when v 1 and v 2 are hgh, D 1 and D 2 are OFF and v o s hgh. Overall, the output of ths crcut s hgh only f both nputs are hgh (Case 4) and the output s low n all other cases (Cases 1 to 3). Thus, ths s an AND gate. Ths analyss can be easly extended to cases wth three or more dode nputs. Ths s actually not a good gate as for nput we used low states of 0 and the output low state was 0.7. We need to make sure that the nput low state voltage s smlar to the output low state voltage so that we can put these gates back to back. (You can easly show that f we had assumed low states of 0.7 for nput, the output low state would have been 1.4.) Ths gate s not usually used by tself but as part of dode-transstor logc gates that we wll dscuss n the BJT secton. Dode O Gate: Show that ths crcut s an O gate. v 1 D 1 v 2 D 2 v o A ECE65 Lecture Notes (F. Najmabad), Wnter

15 4.6 ectfer Crcuts Ths crcut s smlar to the crcut we have solved n page 86. There, we found: v s < v γ Dode n OFF: D = L = 0 and v L = 0 S v D D L L v L v s > v γ Dode n ON: D = L = v s v γ L and v L = v s v γ Consder a case n whch v s s a snusodal waveform, v s = s cos(ωt). In the postve half of the nput perod wth v s > v γ, dode s ON and v L = v s v γ and n the rest of the nput perod, v s < v γ, dode f OFF and v L = 0 as s shown. Exercse: Explan why the output voltage s NOT exactly v L = v s v γ at every pont. The above crcut s a smple method to convert AC nput voltages to a DC output voltage and s used n AC to DC converter part of power supples. Such crcuts are called rectfer crcuts. Because the output of the crcut s only one half of the nput waveform, the above crcut s called a half-wave rectfer crcut. A better crcut would be a full-wave rectfer n whch both porton of the AC nput waveform s turned nto a DC output (so that we do not throw away half of the nput). An example of such a full-wave rectfer s the brdge rectfer shown below wth four dodes. We can see that n the porton of the nput perod n whch v s > 2v γ, dodes D1 and D3 are ON and a postve voltage appear across L. Dodes D2 and D4 are OFF. In the porton of the waveform n whch v s < 2v γ, Dode D2 and D4 are ON and dodes D1 and D3 are OFF and agan a postve voltage appears on L. The resultng voltage across L s shown below. ECE65 Lecture Notes (F. Najmabad), Wnter

16 4.7 Peak Detector Another popular dode crcut s the peak detector whch s smlar to the half-wave rectfer crcut wth the addton of a capactor. Followng the same procedure, we fnd: v D = v s v C > v γ v s > v C v γ Dode n ON v D = v s v C < v γ v s < v C v γ Dode n OFF S v D C C L L L S γ C C L L L L C S C L L Dode s ON Dode s OFF So, when v s > v C v γ and dode s ON, we have a parallel C crcut and capactor wll be charged up (mddle crcut). When v s < v C v γ, we have an C crcut (crcut to the rght) and the capactor dscharges n the resstor wth a tme constant τ = C. The output voltage wave form s shown f the value of the capactor s chosen such that τ = C T (T s the perod of the nput AC voltage). To understand the shape of output waveform, let s assume that voltage across the capactor was zero at t = 0. As the nput voltage ncreases n the frst quarter wave perod (v γ < v s < s ), the capactor charges up as v s > v C v γ and dode s ON. In the second quarter of the wave perod, the ECE65 Lecture Notes (F. Najmabad), Wnter

17 nput voltage v s s decreasng. In ths case, dode s OFF and the capactor slowly dscharges n L. Ths contnues untl the next perod when v s becomes agan larger than v C v γ and capactor charges for a short perod of tme. Exercse: What would have f τ = C T where T s the perod of the nput AC voltage? Ths crcut has two applcatons. Frst, by addng a large capactor (10s to 100s of µf) to our rectfer crcut, we get a relatvely smooth DC output. mlarly such a capactor can be added to a full-wave rectfer crcut. (All AC to DC converters have such a large capactor). Second, the above crcut can be used as a peak-detector crcut. For example, consder the sgnal transmtted from an AM staton. Such a sgnal ncludes a carrer wave (rado staton frequency). The ampltude of the carrer s modulated accordng the sound sgnal. An example of such a sgnal s shown below (assumng that the sound sgnal s a trangular wave). If we apply ths modulated voltage to our peak-detector crcut above and choose the value of capactor such that τ = C T rf where T rf s the perod of rado-frequency carrer wave but τ = C T so where T so s the perod of the sound wave, the output of the crcut would be an approxmaton of the the ntal sound wave as s shown below. The crcut s called the peak-detector because the output voltage s the envelope of the peak ampltudes of the nput sgnal. 4.8 Zener Dodes Zener dodes are specally manufactured to operate n the Zener regon. These dode are made by means of heavly doped regons near the metal contacts to the semconductor. The hgh densty of charge carrers provdes the means for a substantal reverse breakdown current to be sustaned. These dodes are useful n applcatons where one would lke to hold some load voltage constant, for example, n voltage regulators. The crcut symbol, the v characterstcs of Zener dodes and a pece-wse lnear model are shown below: ECE65 Lecture Notes (F. Najmabad), Wnter

18 D v Z v γ v D The pece-wse lnear characterstcs equatons are: Dode ON: v D = v γ, and D > 0 Dode OFF: v D < v γ, and D = 0 Dode n Zener egon: v D = v Z, and D < 0 D _ v D Crcut soluton technques developed n ths sectons for dodes can be appled drectly to Zener dodes. The only change s that Zener dodes can have three states (nstead of two for regular dodes). A smple Power Supply An ndependent voltage source s a crcut n whch the output voltage s constant regardless of the current drawn from the crcut. Ths crcut (wth v s > 0) s such an ndependent voltage source,.e., the output voltage, v L s constant for a range of L s. It s a smple power supply crcut f we replace the voltage source v s wth the rectfer (and capactor) crcut that we developed before. _ v S _ v D v L We wll use the parametrc method to analyze ths crcut. Frst, we wrte down the crcut equatons: D L L _ KCL: KL: KL: = D L v s = v D v L = v D ECE65 Lecture Notes (F. Najmabad), Wnter

19 Case 1: Dode s n the Zener regon. In ths case, v D = v Z and D < 0. Substtutng for v D = v Z n the thrd equaton above, we fnd v L = v Z and s ndependent of L. So, f the dode s n the Zener regon, ths crcut would behave lke an ndependent voltage source. To fnd the range of L for whch the dode s n the Zener regon, we calculate D from equatons above: = v s v D = v s v Z D = L < 0 L < = v s v Z Therefore, as long as L s smaller than the value L,max = (v s v Z )/, the dode would reman n the Zener regon and the crcut would act as an ndependent voltage source. Case 2: Dode s n the reverse bas regon. In ths case, D = 0 and v Z < v D < v γ. Substtutng for D = 0 n the above crcut equatons, we get: KCL: = D L = L KL: v s = v D v D = v s L KL: v L = v D = v s L Thus, n ths regon the output voltage does not reman constant, rather t decreases wth L. The dode s n the the reverse bas regon as long as v D > v Z or L > L,max = (v s v Z )/. Case 3: Dode s n the forward bas regon. nce v s > 0, the dode CANNOT be n the forward bas regon. The v characterstcs of ths crcut s shown n the fgure v L v Z Dode n everse Bas Dode n Zener egon L v / L,max S ECE65 Lecture Notes (F. Najmabad), Wnter

20 4.9 Wave-form shapng Crcuts A wde-varety of waveform shapng crcuts are used n electroncs. These crcuts are used to transform one waveform nto another. Two examples of such crcuts made wth dodes are below. Clpper Crcut Dodes can be used to clp off a porton of the nput sgnal. Consder the crcut below: v = D v D v DC v o = v D v DC = v D Dode ON: v D = v γ and D > 0 v o = v γ v DC and D = v v γ v DC > 0 v > v γ v DC Dode OFF: D = 0 and v D < v γ v o = v 0 = v and v D = v D v DC < v γ v < v γ v DC So, when v < v γ v DC, the dode s OFF and v o = v : the output sgnal s exactly the same as the nput sgnal. However, when v exceeds ths value of v γ v DC, the dode s ON and v o = v γ v DC. As a result, the nput sgnal s clpped when t exceeds v γ v DC. The shape of the output sgnal for a snusodal nput sgnal s shown below. v v o v v DC DC o mlarly, the crcut below wll clp the nput at v γ v DC - v - v DC v o S DC o v ECE65 Lecture Notes (F. Najmabad), Wnter

21 And the two crcuts can be combned to clp both the postve and negatve values. v v o v v DC - v - v DC S o DC1 DC2 It s usually more convenent to use zener dodes nstead of v DC voltage sources. The three crcuts below clp the nput, respectvely from left to rght, at v γ v Z1, at v γ v Z2, and at both v Z1 and v Z2. S D2 o S D2 o S D1 D2 o Exercse: Prove that crcuts above clp the nput voltage at the values gven. Clamp Crcut Another popular dode crcut s the dode clamp whch s smlar to the peak-detector crcut wth the locaton of dode and capactor swtched as s shown below. C C C S L L S γ C S C L L Dode s ON Dode s OFF From the crcut we have, v D = v s v C. Dode s ON f v s > v C v γ. In ths case, the crcut smplfes to the crcut n the mddle and capactor charges up (because the dode s ON and t has a zero resstance, the current n L would be zero and L does not appear n the crcut). Dode s OFF when v s < v C v γ. In ths case, the crcut smplfes to the crcut to the left wth dode beng open crcut. ECE65 Lecture Notes (F. Najmabad), Wnter

22 When nput s a snusodal wave, durng the postve half cycle when v s > v C v γ dode s ON and the capactor charges up (see crcuts above). Durng the negatve half cycle when dode s OFF, the capactor dscharges lttle f τ = C T (smlar to the peak detector crcut). Eventually capactor wll become fully charged such that v C = s v γ where S s the ampltude of the nput sgnal, v s = s cos(ωt). From then on, dode remans n the OFF poston durng both negatve and postve half cycle. nce v L = v s v C, the effect of the constant DC voltage across the capactor (v C = s v γ ) s to produce an output sgnal whch s smlar to the nput sgnal wth an addton of a DC voltage (negatve n ths case) as s shown above. The clamp crcut above produces an output sgnal whch s shfted downward by the value of s v γ. Dfferent values of downward shft can be produced by the addton of a DC voltage crcut n seres wth the dode. Alternatvely, an upward shft can be produced by swtchng the polarty of the dode as s shown n crcuts below (Note that DC can be negatve). C C C S L L S L DC L S DC L L down-shft by s v γ DC up-shft by s v γ up-shft by s v γ DC mlar to the clpper crcut, the voltage sources n the clamp crcuts can be replaced by zener dodes as s shown below. C C S L L S L L down-shft by s v γ v z up-shft by s v γ v z ECE65 Lecture Notes (F. Najmabad), Wnter

23 4.10 Exercse Problems In crcut desgn, use commercal resstor values (1, 1.1, 1.2, 1.3, 1.5, 1.6, 1.8, 2, 2.2, 2.4, 2.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5.1, 5.6, 6.2, 6.8, 7.5, 8.2, 9.1) and commercal capactor values (1, 1.5, 1.8, 2. or 2.2, 3.3, 4.7, and 6.8) values. You can also use zener dodes wth any Zener voltage. Problems 1 to 9. In crcuts below wth dodes and v Z = 5 for Zener dodes, A) Fnd v and/or for v s = 10, B) Fnd v and/or for v s rangng from 20 to 20. S _ S S 5 Problem 1 Problem 2 Problem 3 S 10k _ S 2k 1k S 2k 1k 10k Problem 4 Problem 5 Problem 6 S 10k _ 10k 1k vs v 1k v S Problem 7 Problem 8 Problem 9 1k 1k v ECE65 Lecture Notes (F. Najmabad), Wnter

24 Problems 10 to 12. Fnd v o for v s = s sn(ωt). ( dodes and v Z = 5 for Zener dodes) S o S o S o Problem 10 Problem 11 Problem 12 Problem 13. Desgn a clpper crcut that clps voltages above 5 usng a) DC power supples only, b) Zener dodes only. Problem 14. Desgn a clpper crcut that clps voltages above 5 and below 3 usng a) DC power supples only, b) Zener dodes only. Problem 15. Desgn a clpper crcut that clps voltages above 5 and below 3 usng a) DC power supples only, b) Zener dodes only. Problem 16. Consder a snusodal source wth v = 15 sn(ωt). Desgn a clamp crcut that adds a DC offset of 5 to the nput voltage usng a) DC power supples only, b) Zener dodes only. Problem 17. Consder a snusodal source wth v = 15 sn(ωt). Desgn a clamp crcut that adds a DC offset of 5 to the nput voltage usng a) DC power supples only, b) Zener dodes only. Problem 18. Crcut below s a voltage doubler. Show that f v s = s sn(ωt), v o = 2 (Assume that capactor C s large such that t dscharge very lttle per cycle and s v γ.) Problem 19. Crcut below s a smplfed verson of an electronc flash for cameras. Swtch S2 s controlled by the shutter and s closed to operate the flash bulb. Normally swtch S2 s open. The crcut to the left of swtch S2 charges the capactor to about 100 usng a 1.5 battery. Swtch S1 s an electronc swtch that s opened and closed at about 10 khz. As a result, the capactor s charged to about 100 usng a 1.5 battery. Explan how the charge-up crcut works. C 1 A D 2 S D 1 C 2 L L 1.5 S 1 C S 2 Problem 18 Problem 19 ECE65 Lecture Notes (F. Najmabad), Wnter

25 4.11 Soluton to Selected Exercse Problems Problem 2. In crcut below wth dodes, A) Fnd v and/or for v s = 10, B) Fnd v and/or for v s rangng from 20 to 20. Crcut equatons: KCL: KL: KL: s = D v s = s v D v = v D = S S Part A: v s = 10. Assume dode s ON: v D = v γ = 0.7, D > 0. Substtutng n the above equatons we get: v s = s v D 10 = s 0.7 s = 1.86 ma v = v D = 0.7 v = v D = = = 0.14 ma s = D = D D = 1.72 ma nce D = 1.72 ma > 0, our assumpton of dode beng ON s correct and = 0.14 ma and v = 0.7. Part B: Parametrc Soluton: For dode beng ON: v D = v γ = 0.7, D > 0. Substtutng n the crcut equatons we get: v s = s v D v s = s 0.7 s = (v s 0.7) A v = v D = 0.7 v = v D = = = 0.14 ma s = D D = s = (v s 0.7) D > (v s 0.7) > 0 v s > 1.4 For dode beng OFF: D = 0 and v D < v γ = 0.7. From crcut equatons, we get: s = D s = ECE65 Lecture Notes (F. Najmabad), Wnter

26 v s = s v D = s = = 10 4 v s v = v D = v = v D = 0.5v s v D < v γ = v s < 0.7 v s < 1.4 Therefore, for v s < 1.4, dode s OFF v = 0.5v s and = 10 4 v s. For v s > 1.4, dode s ON, v = 0.7, and = 0.14 ma. Problem 4. In crcut below wth dodes, A) Fnd v and/or for v s = 10, B) Fnd v and/or for v s rangng from 20 to 20. Crcut equatons: KL: v s = v D D1 KL: v s = 10 3 D2 v D2 S D 1 2k 1k D 2 v = 10 3 D1 and = D1 Part A: v s = 10. Assume both dodes are ON: v D1 = v D2 = v γ = 0.7, D1 > 0, and D2 > 0. Substtutng n the above equatons we get: KL: v s = v D D1 10 = D1 D1 = = 4.65 ma KL: v s = 10 3 D2 v D2 10 = D2 D2 = = 9.3 ma v = 10 3 D1 v = 4.65 nce both D1 and D1 are postve, our assumpton of both dodes beng ON s correct and v = 4.65 and = D1 = 4.65 ma. Part B: Parametrc Soluton: Here, we have to consder four cases. Case 1: D1 and D2 are both ON: v D1 = v D2 = v γ = 0.7, D1 > 0, and D2 > 0. KL: v s = v D D1 v s = D1 D1 = (v s 0.7) KL: v s = 10 3 D2 v D2 v s = D2 D2 = 10 3 (v s 0.7) v = 10 3 D1 v = 0.5(v s 0.7) D1 > (v s 0.7) > 0 v s > 0.7 D2 > (v s 0.7) > 0 v s > 0.7 ECE65 Lecture Notes (F. Najmabad), Wnter

27 So, for v s > 0.7, both dodes wll be ON wth v = 0.5(v s 0.7) and = D1 = (v s 0.7) = 0.5(v s 0.7) ma. Case 2: D1 and D2 are both OFF: D1 = D2 = 0, v D1 < v γ = 0.7, and v D2 < v γ = 0.7. KL: v s = v D D1 v D1 = v s KL: v s = 10 3 D2 v D2 v D2 = v s v = 10 3 D1 v = 0 v D1 < v γ = 0.7 v s < 0.7 v D2 < v γ = 0.7 v s < 0.7 So, for v s < 0.7, both dodes wll be OFF wth v = 0 and = D1 = 0 Note that snce for v s > 0.7, both dodes wll be ON and for v s < 0.7, both dodes wll be OFF, one would expect that t would not be possble for one dode to be ON and one to be OFF. We wll demonstrate ths below. Case 3: D1 s ON: v D1 = v γ = 0.7, and D1 > 0 whle D2 s OFF: D2 = 0 and v D2 < v γ = 0.7. KL: v s = v D D1 v s = D1 D1 = (v s 0.7) KL: v s = 10 3 D2 v D2 v D2 = v s v = 10 3 D1 v = 0.5(v s 0.7) D1 > (v s 0.7) > 0 v s > 0.7 v D2 < v γ = 0.7 v s < 0.7 We see that for D1 to be ON ( D1 > 0), we need v s > 0.7 and for D2 to be OFF (v D2 < v γ ) we need v s < 0.7. These cases are mutually exclusve so we cannot have smultaneously D1 ON and D2 OFF. mlarly, we can fnd D1 s OFF and D2 ON case s not possble. ECE65 Lecture Notes (F. Najmabad), Wnter

28 Problem 6. In crcut below wth dodes, A) Fnd v and/or for v s = 10, B) Fnd v and/or for v s rangng from 20 to 20. Part A: v s = 10. Wth the excepton of smple crcuts, t s usually more useful to redraw the crcut based on the state of the dodes as ths can smplfy the crcut consderably. Note that when a dode s ON, v D = v γ for D > 0 and the dode resembles an deal voltage source wth the strength of v γ. When the dode s OFF, D = 0 and the dode resembles an open crcut. Ths can be seen n crcuts below: S 10k 1 _ D2 2 S 10k _ 3 D2 2 S 10k 1 _ D2 2 S S 10k 1 10k D 1 10k 3 D2 2 D 2 D1 10k D1 10k D1 10k D1 10k D1 ON D1 ON D1 OFF D1 OFF D2 ON D2 OFF D2 ON D2 OFF Case 1: Assume D1 s ON and D2 s ON. The above crcut can be solved by node voltage method. Ths crcut has four nodes, v 1, v 2, v 3, and v s = 10. All nodes are supernodes as they are attached to a voltage source. So we have only we KCL contanng nodes v 1, v 2, and v 3 : Supernode: v 1 v 2 = 0.7 Supernode: v 3 v 2 = 0.7 KCL at v 1, v 2, v 3 : v 1 v s v 3 v s v v = 0 3 v 1 v s 2(v 3 v s ) 2v 1 v 2 = 0 3v 1 v 2 2v 3 = 3v s = 30 We substtute for v 1 and v 3 from the frst two equatons n the last equaton to get: 3(v 2 0.7) v 2 2(v 2 0.7) = 30 v 2 = 4.42 and v 1 = v 3 = 5.12 ECE65 Lecture Notes (F. Najmabad), Wnter

29 To check the valdty of our assumpton of D1 and D2 ON, we need to compute D1 and D2. D2 s the same as the current n the 5 k resstor on the top of the crcut and D2 can be found by KCL at node v 1 : D1 = v s v v = = 0.54 ma < D2 = v s v = = 0.98 ma > nce D1 < 0, our assumpton s ncorrect and we should consder other cases. Case 2: Assume D1 s ON and D2 s OFF. The above crcut can be solved by node voltage method. Ths crcut has three nodes, v 1, v 2, and v s = 10 (v 3 = v s = 10 as D2 = 0). All nodes are supernodes as they are attached to a voltage source. So: Supernode: v 1 v 2 = 0.7 KCL at v 1, v 2 : v 1 v s v v = 0 3 v 1 v s 2v 1 v 2 = 0 3v 1 v 2 = v s = 10 We substtute for v 1 from the frst equaton n the second equaton to get: 3(v 2 0.7) v 2 = 10 v 2 = 1.98 and v 1 = 2.68 To check the valdty of our assumpton of D1 ON and D2 OFF, we need to compute D1 (KCL at node v 1 ) and v D2. D1 = v s v v = = 4.06 ma > v D2 = v 3 v 2 = = 8.02 > 0.7 nce v D2 > 0.7, our assumpton s ncorrect and we should consder other cases. Case 3: Assume D1 s OFF and D2 s ON. The above crcut can be solved by node voltage method. Ths crcut has four nodes, v 1, v 2, v 3, and v s = 10. All nodes except v 1 are supernodes as they are attached to a voltage source. So: Supernode: v 3 v 2 = 0.7 KCL at v 2, v 3 : v 3 v s 5 10 v = 0 3 2v 3 20 v 2 = 0 KCL at v 1 : v 1 v s v = 0 v v 1 = 0 v 1 = 3.33 ECE65 Lecture Notes (F. Najmabad), Wnter

30 The frst two equatons can be solved to fnd v 2 and v 3 : 2(v 2 0.7) v 2 = 20 v 2 = 6.2 v 3 = 6.9 To check the valdty of our assumpton of D1 OFF and D2 ON, we need to compute v D1 and D2. v D1 = v 1 v 2 = = 2.87 < 0.7 D2 = v s v = = 0.62 ma > nce v D1 < 0.7 and D2 > 0, our assumpton s correct. So D1 s OFF, D2 s ON and = D2 = 0.62mA and v = v s v 3 = = 3.1. Ths case could have been solved more smply by notng that when D1 s OFF, the two branches of the crcut become ndependent (we have two separate crcuts) and v 1 and v 2 can be drectly calculated: v 1 = 10 = v s 0.7 = D2 = = 0.62 ma and proceed to compute v D1 < 0.7 to show that our assumpton was vald. Case 4: Assume D1 s OFF and D2 s OFF. Whle we proved that D1 s OFF and D2 s ON, let s proceed to solve ths case. Agan, we see that two branches of the crcut are ndependent. In addton, snce = 0, v 2 = 0 and v 3 = v s = 10. From voltage dvder, we have v 1 = 3.33 (smlar to case 3). To check our assumpton, we note v D1 = v 1 v 2 = = 3.33 > 0.7 so D1 cannot be OFF and assumpton was ncorrect. As can be seen, the smplest crcut s when dodes are OFF as the crcut usually dvdes nto several smpler crcuts. As such, f one cannot make an educated guess regardng the state of the dodes, t s usually best to start the analyss wth dodes beng OFF. ECE65 Lecture Notes (F. Najmabad), Wnter

31 Part B: Parametrc Soluton: Here, we have to consder four cases. As dscussed above, we wll start wth smplest cases (dodes OFF). Case 1: D1 and D2 are both OFF: we see that two branches of the crcut are ndependent. In addton, snce = 0, v 2 = 0 and v 3 = v s. From voltage dvder, we have v 1 = v s /3. To fnd the range of valdty of ths soluton, v D1 = v s /3 < v γ v s < 2.1 v D2 = v 3 v 2 < v γ v s < 0.7 For both dodes to be OFF, we need v s < 0.7 (the most restrctve of both condtons). So, for v s < 0.7, both dodes wll be OFF, = 0 and v = 0 Case 2: D1 OFF and D2 ON: we see that two branches of the crcut are ndependent. From voltage dvder, we have v 1 = v s /3. = D2 = v s = (v s 0.7) v = = 0.33(v s 0.7) To fnd the range of valdty of ths soluton, D2 = (v s 0.7) > 0 v s > 0.7 v D1 = v 1 v 2 < v γ v s < v s 0.67(v s 0.7) < 0.7 v s > 0.7 For D1 OFF and D2 ON, we need v s > 0.7 (the most restrctve of both condtons). So, for v s > 0.7, D1 OFF and D2 ON, = (v s 0.7) and v = 0.033(v s 0.7). nce for v s < 0.7, both dodes wll be OFF and for v s > 0.7, D1 s OFF and D2 s ON, there s no range of values for v s when other cases (D1 ON and D2 OFF or both ON) are possble. Exercse: Solve crcut equatons drectly to show that D1 ON and D2 OFF Case or both dodes ON Cases are not physcally possble. ECE65 Lecture Notes (F. Najmabad), Wnter