Now, we finally reached the most [inaudible] part of our lecture which is the learning of MOSFET devices. MOSFET is the metal-oxide semiconductor field-effect transistors. Here are some MOSFET, here's the gate and gate dielectric oxide, and p-type semiconductor. There's a source and drain. Normally source is grounded zero voltage and you apply V_D and gate is V_G. In MOSFET device there is two important IV curve, which is I_d-V_d curve and the other is I_d-V_g curve which is the transfer curve. I_d-V_d curve, if your increasing on drain voltage then current increasing between the source to drain. Also, if your increasing gate voltage, drain voltage is increasing. But if you're applying the less voltage than threshold voltage current is now flowing even if you're applying high drain voltage. Transfer curve, where the I_d-V_g curve is measure the drain voltage which is the drain voltage at constant in I_d. It can be at one volt, it can be at 0.1 volt. Then you're increasing a small gate voltage, you expel the majority hole and underneath the oxide layer makes the depletion. Then adding on more positive voltage slightly minority weak inversion is so caries but they are totally regrettable compared to the background of p-type semiconductor which is called the weak inversion. At threshold the voltage where surface potential is two Pi F, which means the minority electron [inaudible] concentration is comparable to the background of the p-type silicon, that is the starting point of current flowing of digital one or transistor onset which is the threshold voltage here and of all threshold voltage, electron channel is formed between the source to drain, electron flowing from the source to drain. As you know, through the electron channel, minority channel is separated by p-type silicon by depletion region that's why minority carrier can survive in oceans of the majority hole because the depletion region, they cannot be combined. Another interesting point of the I_d-V_d curve is that I_d curve at certain V_D voltage, current is saturating. Even if you are applying more drain voltage, current is saturating. Why? This is so important we're going to learn in later section. The other, current saturation point is quadratically or scale proportional. So means that, current increase amount 1 volt, 2 volts, 3 volts, 4 volts is quadratically increasing. Why? We are going to learn later. Another interesting point of correlating the I_d-V_d curve and transfer curve which is the I_d-V_g curve is there. What is the transfer curve in I_d-V_d curve? Transfer curve in I_d-V_g curve is this. At constant V_d of the one volt or 0.1 volt, you increasing the gate voltage and below the threshold voltage current is not flowing. If you apply the gate voltage above the threshold voltage current is flowing, which is corresponding to here, more gate voltage than increasing I_d current. This transfer curve actually came from my research paper during my graduate school and this is the transfer curve. Transfer curve is important that you can find out the most important characteristic of the MOSFET devices. This is the transfer curve and the black is the linear value. Here's the threshold voltage. So threshold voltage at 0.7 then curve transistor is on, current is flowing. So threshold voltage is 0.7, this is measured at constant V_d 0.1 voltage. Later on, we're going to learn that the slope of the transfer curve is related with the mobility. As I said, mobility is the most important characteristic of semiconductor. If a mobility is one, 1,000 less current flowing compared to the silicon transistor. So organic semiconductor may be mobility is one, silicon transistor it's mobility around 500 or 1,000. Then thousands more current is flowing if your mobility is higher. This means that some of your smartphone require to turn transistor on at one million pair, maybe in silicon transistor you need lengths of one micrometer is required to turning on the one million pair. Maybe some less mobility transistor maybe requires 1,000 more tunnel width to accomplishing one million pair. So high mobility is one of the most important thing. Second, this is the linear transfer curve, but if you're changing this to the log scale, it becomes like this and equals the log scale transfer curve. From this log scale when transistor on and transistor off, on/off ratio can be determined here, maybe 10_6. From this transfer curve, we can get the mobility from the slope of the transfer curve. Slope means how much current can be flowing. More higher slope then more current is flowing, so that is the mobility. On/off ratio can be found in a log scale and threshold voltage can be found on turn-on voltage. Let's learn more detail about the MOSFET device, especially later on once I conclude that this is the scale relationship, second order relationship, and saturation region is really achieved or not. First, this is the source and this is the drain. You're applying the gate voltage above the zero voltage and below the threshold voltage. Then if you're applying V_g above positive voltage in p-type silicon, you expel majority carrier holes makes the depletion region underneath silicon oxide region. Then what is this depletion source drain region because the gate voltage potential is not influencing, but they still have a depletion region. Actually, this depletion region came from the p-n plus junction, p-n junction depletion region. P-n junction originally have depletion region. Now, you applying gate voltage above the threshold voltage. But source to drain voltage is zero, what happen? Then in addition to the depletion region, you formed minority electron inversion layer underneath silicon dioxide layer. So it makes the electron channel, bunch of electron in a source region electron channel, and drain region. But those minority electron channel is separate by the p-type majority hole, preventing the combination separated by depletion region of a fixed charge. This depletion charge is also negative of the boron. But since you didn't apply the positive drain voltage, current is not flowing at this condition, even if you have a inversion charge. Now, you're applying small V_D voltage when gate voltage above the threshold voltage. Now, you apply drain voltage, electron move source to drain, and current flowing. How much current flowing? Current, I_DS, source drain current is proportional to the channel width, how wide of the channel width, a Q inversion charge times velocity of both carriers. How much inversion charge is formed in here? Q is normally CV, so Q inversion is insulator capacitance C_i times V_G, gate voltage minus threshold voltage. Because if you're applying the gate voltage up to the threshold voltage, you're only forming that depletion charge, no inversion charge. Inversion charge is forming above the threshold voltage. Only above the threshold voltage, so voltage should be compensate and then to become V_G minus the threshold voltage. Velocity of [inaudible] is mobility times electric field. So final is the ZC_i V_G minus V_th mobility V_DS per L. Now look at this current and then let's think about why mobility is important. You need to have a certain current to be on current, maybe let's say that this is one milliampere. To achieve one milliampere, to increase I_D current, you may increase the Z channel width or reducing the channel length L. But if you're increasing the width or increasing the size of transistor, you cannot integrate many transistor. That's not a good option. You want to make it smaller and smaller. That's the idea of the integrated circuit. You don't want to make a very wide or very big sides of MOSFET transistors. So increasing width is not a strategy. Or you can increase the C_i capacitance, maybe deducing the oxide thickness, increasing capacitance, but right now, modern electronics already reached the oxide thickness one nanometer. You cannot further decrease the thickness of oxide, so there is also limit there. Gate voltage and drain voltage. You don't want to use huge power to operate your transistor. That will waste your battery of your smartphone. So gate voltage and drain voltage also limited number, let's say the threshold voltage 0.1, then on drain voltage and gate voltage, 1.5. So changing the gate voltage, drain voltage is not a good option either. Only thing is matter is using very high mobility semiconductor material. Mobility is one of the most important thing. Other things also important, but the mobility is, I guess, most important factor of semiconductor devices. So in here, I_D is related with drain voltage. So it can be expressed by the I_D is the g V_D. G is the conductance of the channel core. So g is other than the I_D. G is 1 over resistance here, I_D over V_D is the resistance. G is 1 over R, and then also, derivative of I_D over the V_D, the slope. If you derivative I_D over the V_D, then it becomes like this. Conductance, slope of the I_D V_D curve is related with the mobility, and capacitance, and etc. I said I_D V_D curve, current will be saturated. The saturation point is called pinch-off. So pinch-off is pinching with your nail or whatever it is, pinch-off. Because some modern MOSFET device, there's a punch through, whatever it is. Later on, you are confused to either name, but this is the pinch-off, very important phenomenon of the pinch-off. Then after this pinch-off, current is saturating. Why this pinch-off is occur? Pinch-off is occur when, obviously, gate voltage is above the threshold voltage, when the current is on, transistor is on, and pinch-off is occur when V_D drain voltage is equal to the gate voltage minus threshold voltage. Let's see there why the drain voltage equal to the V_D minus V_th, then pinch-off is occur. Let's look at this red line. Then magnify this red line to here. Then here's the source is grounded, gate voltage is five volt, drain voltage is three volt, and threshold voltage is two volt. Then if you look at this red line, this drain region, silicon region, this silicon region potential is the same with the drain potential. Silicon region potential, potential is the relative value. So potential in this region is equal to the drain voltage because you are applying drain voltage, this will go to here, influencing here, so potential is equal to the drain, which is the three volt. What about the gate voltage? Although you apply the gate voltage, five voltage, which is above the threshold voltage, but the gate voltage up to the threshold voltage, which is two volt, doesn't generating inversion charge. But gate voltage above the threshold voltage only influencing or forming the inversion charge, so the effective gate voltage forming the inversion is 5 minus 2. So you effective gate voltage is the gate voltage, 5, minus threshold voltage, in this case, 2 volt. This is the exactly cases of that V_D is V_G minus V_th. Now you're applying the three volt to the gate and the potential of the silicon at the drain region is the three. Three volt to three volt, then zero volt, then no inversion on is formed at the end of that drain region. This go back to the onset of the inversion. You probably know that this is the source and then this is the grounded, so this is the zero volt, and then you're applying the drain voltage, three volt. Then here is also three volt, and then those three volt is applied in the very resistive area of the channel. This area is very low resistive. This is very low conductivity, very conductivity area, channel is very resistive area compared to the source and drain. So voltage applied to the drain region is consumed in channel region. So you can think that source region potential is zero and drain region is the three volt. So that's why I said here is the three volt. So conclusion is that when V_D equal to V_G minus V_th, we say this is the pinch-off region, and then after this pinch-off region, no inversion is formed in the drain region, and saturation drain current will be flowing after this saturation voltage. Then let's look at further of the saturation region. Saturation region is the gain voltage above the threshold voltage, and then drain voltage is much higher than V_G minus V_th. Then example of V_G is the five threshold voltage two volt, and V_G minus V_th three volt, you're applying the V_g with four volt which is higher than three volt. Then pinch-off region or saturation region go much deeper to the channel region and those rejoin you start depleting. So you can think this as a lot of electron located in source region and the channel region, huge minority mobile electron charge located here, and this area is depleted region. Then drain region is highly conductive region with a lot of electron. So this is the N plus, a lot of electron N, and the depleted region like a P, and N plus. This region is also because here's the depleted, P-N junction. Now also important thing is that, if this is the P-N junction, you're applying positive voltage to N. Is this forward bias or reverse bias? Applying positive voltage to N is reverse bias. You learn that P-N junction in reverse bias current is now flowing. But the part here, huge current flowing, why that is possible? What's the difference between the P-N junction, reverse bias, and most path register with a reverse bias in this region? Difference is like this; P-N junction, this is the under equilibrium, and then this is the P-N junction reverse bias, huge electric field. But although there is a huge electric field but carry limited number of the electron minority carrier located in a p-type region, then limited the number, therefore, almost no current flowing. Here is the same thing, huge electric field. If you're applying high drain voltage, huge electric field. But the difference is that there is huge number of minority electron in most the transistor. Therefore, those huge number of electron carrier concentration can sweep down to the drain region by the reverse bias of a P-N junction. However, there is no inversion at those region, current is not increasing, but they are saturating. Let's look at another different graph. Here, K voltage above the threshold voltage, inversion charge is form, and small V_DS voltage, those small V_DS does not influence the channel inversion charge, therefore, current is linearly increasing. But you further increase the drain voltage, those drain voltage influence the channel's potential as shown in the V_D previous graph. Then those V_ D influencing the drain region, reducing the inversion charge at the drain region. Then current is starting to reduce. At pinch-off where the V_D equal V_G minus V_th, there is no inversion at the drain region. This is the pinch-off. At the pinch-off, current is saturating and more higher voltage pinch-off region go farther inside the channel region, and then huge minor routine inversion electron will be swept down to the drain region by the huge diverse electric field. But there are no inversion, therefore current is saturating.
