nanoHUB-U Principles of Nanobiosensors/Tutorial 3.2: Amperometric and Cantilever Sensors Homework Solutions
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[Slide 1] Ok, so we are this week we discussed two special type of sensors amperometric sensors and cantilever sensors you remember that there are basically three that we planned to discuss one was in the previous model it was all about potentiometric sensors is the various limits associated with the potentiometric sensors meaning screening pH based the interference at the sensor surface this week primarily we are thinking about amperometric sensors which have a slightly better selectivity and cantilever sensors which is mass based so I'll give you a few examples and hence because the solution manuals all the available Also, the solutions are all available to you and so you should-- take a look but let me just workout a few of the homework so that you know what to look for. So I'll have maybe a two examples from the amperometric sensors and a few from the cantilever sensors and then--
[Slide 2] we'll be done. Now amperometric sensors, of course, depends very sensitively on the various electrodes we have in it's very important, you see to get comfortable with the electrode-- electrochemical potential associated with the with various types of sensors so remember there are two deviations two classes one is this spontaneous reaction where the electrons essentially want to go from higher electrochemical potential to lower electrochemical potential oxidation potential and in this case for example that's the case for the battery and the first example in 4.1, we essentially wanted to explore that particular problem spontaneous reaction where the-- chemical energy is being converted into an electrical energy of course you remember, the Butler-Volmer and the other equations there you had to apply a voltage before the reaction got initiated so this example is not that it is a spontaneous reaction and so this particular example says that we have a-- two electrodes one is Cadmium and then another was Antimony so this could be side by side like this a nearest the respective solution and in the respective solution Cadmium two plus can immediately get into-- Cadmium and Antimony need two plus can get to and and Antimony so these ionic changes are possible, so the question that was which metal-- will be cathode which metal will be anode? Write the half-reaction going on in the separate chamber, remember the cadmium electrode is in one chamber the antimony electrode is in the other chamber and we are asking that write down the reaction for this chamber, this one half reaction write down the direction for this chamber which is another half reaction and that's the half reaction and what is that electromotive force Net force for this-- particular cell so that's the question being asked
[Slide 3] and the solutions are relatively easy You can look at the standard reduction potential Reduction potential means that somebody is accepting electron that means its getting reduced. So therefore in general you see the Cadmium would accept, it will donate its electron, right? It donates its electron and antimony accepts it and so this is the this is the whole reaction and and therefore the half reaction would be that in here the cathode-- in the cathode for example it will accept the electrons, so if you look at the lecture slides, you will see that the electron goes from the anode after getting oxidized it goes to the-- cathode and there it deposits its electrons, so this would be the half reaction associated with Antimoney and the Cadmium since it gives away electrons so therefore-- and in the process gets oxidized and so this would be the-- anode, it will be called an anode and the corresponding EMF would be the difference of these two potentials so you have in one case .136 for the cathode in this particular case it is less negative and then correspondingly minus .403 but we are looking at the difference, so once you take the difference, we'll have about .67-- .267 volts for the electromotive force. The bottom line is that you always look at this reaction look at the big table which has-- a list of all the reduction potentials in the chemistry book you will find--you'll always have to be very careful whether its the oxidation potential or reduction potential and once you have sort of looked up from the handbooks or from Google these respective numbers-- these respective numbers then you can decide that which one is anode and which one is cathode, which one wants to accept electron and which one wants to donate electron right? So this is something there are various sources from which you can get more insight into this problem
[Slide 4] Alright the other problem now which is 4.3 was associated with current passing through an electrode of complex geometry and you remember that when the redux reactions were going on between one electrode and and the next the important point was that between two electrodes for example the important point was that we got CDSS in this expression for arbitrary geometry if you have a cylinder then you have to write the corresponding diffusion equivalent capacitance associated with the cylinders if you have two spheres, then you have to write the corresponding diffusion equivalent capacitance of the two spheres and what is very important to check that I did all the deviations for planar electrode and then simply generalized to in here but in this expression if you put back the planar electrode expression you should get this back and make sure that that happens
[Slide 5] and that's what it is done you see in case of CDSS for the planar electrode you assume the two electrodes are the same size so AB's equal to AA that they are equal and also and so therefore you have to put and also the CDSS is given by epsilon naught A over D or in this particular case epsilon is replaced by the diffusion coefficient D so you put this and put it in here and you will see that immediately you get back the expression that we had before AD over W here and another 00:07:50,930 --> 00:07:53,979 D over W because there's the A here the A will cancel the corresponding A in the CDSS and so therefore you have a D over W in this particular case so in general this-- these equations therefore are completely equivalent and that gives us the confidence, of course, the saw many examples that we did in the in the lecture itself where this equivalence was shown to be correct. Okay so on amperometric sensor there are one or two other examples that you should be you should be checking out
[Slide 6] So one example for the cantilever sensor was that where you can weigh a virus. Now that does something that interesting because of course as things have gotten small the cantilevers have gotten very very small many people have reported that they can detect virus what I want to emphasize today by through this example is that you have to be a little careful as you-- as you look at those papers, read those papers So, let's say a typical virus weighs about tens of atto-gram. Atto-gram is 10 to the power minus 18 gram-- atto-gram and if you have a silicon cantilever beam with the Young's modulus of lets say 150 density of 2300 couple of microns long and wide and relatively thin let's 25-13 nanometer. Then let's see whether we can detect this type of a typical virus because that would be good because most of the viruses have this type of range but there are other viruses which are considerably bigger which are rare but considerably bigger and these viruses are for example are about thousand times bigger So, 10 femto-gram, whereas typical viruses are 10 atto-gram and let's see whether we can calculate the corresponding frequency shift now you remember depending on the support of that cantilever we'll have different types of alpha-- alpha 1 and alpha 2 value when you look at the formula. So these are already given so we don't have to we don't have to worry about them so let's see how much-- how much change would you expect in both cases
[Slide 7] So the formula that we work with-- we'll be working with is delta f is delta m over 2m naught multiplied by f naught you have omega naught is equal to square root of km 00:10:45,630 --> 00:10:49,190 take log on both sides and take a differential that will give you this formula, of course this formula completely neglects the delta k variation, right? There's no delta k over k naught in here, why not? Because you see delta k has to change the spring constant like a starch, right? Like a starch it has to stiffen the beam itself before delta k and then you have to account for delta k and that stiffening requires that you have many biomolecule, sort of cross-linked with each other making sure that there is this bending is more difficult let's say but when you have a single virus biomolecule, a single one in that case there's no delta k because it's not sort of, cannot change on its own change the stiffness So there's no delta k here. Of course you can calculate the mass is the density WLH is, you know, the volume. Spring constants you already know and this has been said to be alpha 2 is equal to 1 and that is this corresponding value for I. Now if you calculate the whole thing you put it in you know put this value of m naught in here, put this value of-- Yeah, put this values and correspondingly for f naught, you will have to take his value of k naught from here and put it in there and then you will get some formula which is good, delta f is 195 00:12:20,329 --> 00:12:23,700 proportional to delta m, so should be able to directly see the mass, the initial mass and of course the the longer-- So you want it to be relatively small length and relatively small width you remember it's a micrometer size if you want to see a larger delta f the interesting thing is that for normal virus if you put all those numbers together you'll only have 120 hertz shift you remember what is the typical resonance frequency, couple mega hertz and so if you have a hundred hertz shift for typical viruses that would be essentially be undetectable especially when you have noise-- limit so I think to this day still even with nanoscale sensors A normal virus, a typical virus, it would be difficult to detect by cantilever sensors. So what is it when people say that they can detect viruses, well they are talking about these larger viruses, this is a thousand times heavier and so therefore this is thousand times big-- there time is thousand times big-- and so therefore is about 120 kilo-hertz Or .12 mega hertz shift that is something you can easily detect and so therefore larger viruses are possible but day-to-day viruses are difficult to detect the only way you can do the day-to-day viruses is to use another-- like a tag, a heavier tag with it so that when this little molecule that along with the tag they come together and bind to the sensors surface well in that case you can detect them together but that requires the label or a tag for the biomolecule we are not measuring the biomolecule itself but that tag associated with it okay its something important to
[Slide 8] to remember and if you see experimentally this is indeed what is seen I mean in this paper for example like, I took this data from this paper you have two biomolecules, one is in here shown here and move on and then other one is here you can see how big-- how big this biomolecule, it's length is about a micron so this size is this some something that is on most comparable It will be a fraction of a micron so this can be very very substantial and of course the most sensitivity will happen when the biomolecule is in here towards the end compared to closer to the base and you can indeed see that there is a shift in frequency the shift in frequency it is originally as 1.27 mega hertz and then eventually it is that 1.21 mega hertz so .06 mega hertz, about 60 kilo hertz shift in frequency. You can see still it can be detected but if you try to go farther down, right? one kilo hertz or so then it's not, not so easy now if you could somehow improve the noise up such a system then of course you could in principle, you can even detect a single DNA base pairs then let me let me show you-- what is what might be possible
[Slide 9] Ok, so lets assume that you want to measure even something smaller how small can you go. So here is an example so for example here instead of using the the classical cantilever which is about you know one 1.5 micron width Let's use a DNA-- a carbon nanotube, the carbon nanotube only has 2 nanometer diameter, w has sort of effectively gotten very very small, thickness previous case 25 nanometer, here let's say we make it 2 nanometer, again this is a this is a cylinder right so in that case it is considerably smaller cantilever and in that case we may be able to measure something that we couldn't measure, even with this silicon-based cantilever. So, this diameter is 2 nanometer lets say the length is about 2 micron very similar to the silicon cantilever, carbon nanotube has very large Young's modulus one tera-pascal and then now one important thing is to remember that when you look at the mass of a carbon nanotube, people generally specify it as a linear density, so you don't have it rho, rho is weight per volume kilogram per centimeter--cube let's say but here often as specified in terms of weight per unit length, integrated over the cross-section So here, for example, it is 5 times 10 to the power minus 15 kg per meter it's very lightweight, alright and alpha 2 and alpha 1, all these values are given let's say somehow you have very high signal processor-- signal processing capability that can detect 500 hertz shift in the frequency, let's say now of course how would you do it we'll discuss it later in the selectivity-- selectivity lectures that you can at actually average your signal average over signal and noise for a certain duration take many samples and in that case one can in principle improve the signal to noise ratio so that even a smaller shift becomes visible
[Slide 10] So in the solution this is fully discussed remember single biomolecule so we don't have to delta k term the decay time is gone this is a cylindrical type of sensor not a rectangular cross section so therefore your I is slightly different and-- and then correspondingly-- mass simply the linear mass density multiplied by L So you put them together calculate the change in the delta f the bottom line is you get about a kilohertz-- kilohertz, now this is simply for a DNA of having Let's see, there is four base pairs, assume AGCT So, there are four base pairs, so each one of them has about-- the total is about 1230 Daltons Daltons, you multiply with mass of a proton and in that case it looks like that it is detectable but of course you have to do signal averaging you have to have a DNA biosensor-- carbon nanotube based biosensors but you can see the ultimate limit that even a small tiny segment of DNA may become visible if you have sort of, if you can try if you can fabricate a DNA based cantilever sensor okay.
[Slide 11] Now often, you know we have worked with analytical-- analytical results I changed everything so that you can we can analyze everything in terms of Spin Mass System but in that case we need to know the alpha 1 and alpha 2 you know that doesn't change the results fundamentally the values of alpha 1 and alpha 2 order magnitude is always correct but if you needed a full simulation you can quickly do it by this additional simulator called MEMSLab the MEMSLab you can solve the the same problem
[Slide 12] and what you see that you can add the mass-- before adding the mass You apply your pulse and then there is this resonance Let's say the resonance is 7.41 megahertz for this particular problem and then you add the mass where ever you are and then the mass after the mass has been added the frequency has gone down remember the delta m in fact is always negative 7.16-- 7.16 megahertz about 300 kilohertz or so chain so you can do the simulators if you want more sort of quantitive answer but other than that these qualitative answers is that we discussed previously most of the time we'll get you into right ballpark
[Slide 13] Now one final thing they wanted to mention about to this cantilever based biosensor is that there are this new version, right? is called a fletcher fit and fletcher fit essentially combines the-- biosensing are associated with the simple cantilever and puts and electrostatics interactions with the cantilever bends to a point where the sensitivity is much higher and the key to enhance sensitivity is the reduced K are the reduced spring constant and so this particular homework looks into it looks into that one thing is that this y over y naught you should not go beyond that one third of a distance, so this is something I want to I want to explain to you-- y over y naught there's a critical value I want to show you where that comes from
[Slide 14] So you may remember that this is something we did in the class if you have this green being being pulled by this pink bottom electrode then what happens that for every voltage you apply it generally goes down oscillates then stops at the point and then if you keep doing it beyond beyond about-- around about 20 volts for this particular example then it will snap and snap shut so for all biosensing applications we have to sort of stay below this voltage in this particular case 20 volts but you can calculate for every every cantilever it will be slightly different voltage called the pull-in voltage you have to stay below it and you can see at the transition point this top side is one third and this bottom side is about too third so we cannot pull it below one-third because there are more than one-third because then we'll be pulled in and then we don't have a electrostatic-- we don't have a sensor So in order to get it to get this analysis right what has to be done is that at every point we have to balance the spring force associated with this spring, spring is trying to pull it in the opposite direction and electrostatic forces trying to, FE trying to pull it downward and so we have to solve these two forces and the solution will give us the steady-state position But the only point I want to make and it's something we have discussed the only point I want to make is that this critical point where beyond which you cannot-- it is snap shut reqires two conditions and this is something often in the beginning people miss the first condition is the force has to be equal that's fine that's this curve and the rate curve that it has to be a solution but that the point of transition the slope the tangent of the curve, the slope of this curve also has to be equal to each other and if you do that only then with this additional condition you will get the critical point so let me show you how it works
[Slide 15] So for example for example this is the balancing between the spring force and the electrostatic force so you can see that this a cubic equation that's fine that's from the force balance but that tangent requirement that a tangent is equal requires that we take a derivative of this which gives you k and a derivative this which gives you minus y cubed and so these two are equal thats why the minus is gone so you have k is equal to epsilon over y cubed V squared now these two equations have to be solved simotaniously, how do you do that? well what is done is that you can take one-- this one y from here put it up there so you have epsilon a a over y square v square and that you can replace it in here so did this to become a k y for this whole thing and once you put it in there, k will cancel, k go away and you can have a relationship like this otherwise if you try to put cross multiply and then solve that makes things a little complicated and by solving this you can see that the pulling point is two-third y naught and once you have that two-third y naught then now you can essentially put this value in put this value in in both cases and correspondingly find out the pull-in voltage. Okay so this is something you should work out but I just wanted to emphasize that there is this little trick that makes your life a little easy.
[Slide 16] Now, one thing about this curve that I wanted to say that this spring remember this weakening of the spring is the key thing that we are trying to explore in this homework and the final formula is the effective spring constant goes as 3 minus 2 y naught over y the whole thing as multiplied by the k now as you're pulling things down the y is less than y naught and so therefore the whole thing is gradually becoming larger and larger until it becomes the equal to three so for example let's say this original gap is y naught without applying a voltage and the new gap is y you know now two-third y is still here but let's say a certain value is y and so for example if the y is .9 y naught so it's just pulled down a little bit .1 1 y naught in that case is still seventy-seven percent of the original spring constant so that would be somewhere here seventy-seven percent of this original spring constant but if you change little bit let's say to .8 this .2 and this .8 in that case this spring constant value is a half the original value, increase it a little bit lets say .7 it will immediately strongly and very quickly drop towards 0 so what we really want is that we want to sort of operate somewhat close to here if you're operating with a little bit of bias in here then you will not get significantly enhancement in the sensitivity so therefore working out this example tells you how close to the pull-in point you have to bias Of course if you're too close then you can be pulled in so you don't want to be exactly a two- third y naught little bit in the back that that should be good so this is the purpose of this design but example will help you decide at what point to bias this particular flectcher fit so that's about it that's the homework that I wanted-- or the homework that I wish to discuss this week and we will end here. Okay, thank you.