Lecture 17 Field-Effect Transistors chroder: Chapters, 4, 6 1/57 Announcements Homework 4/6: Is online now. ue Monday May 1st at 10:00am. I will return it the following Monday (8 th May). Homework 5/6: It will be online later today. This one will be a little different, and will basically be a data analysis exercise. It will still carry the same weight (7.5% of overall grade). ue Monday May 8th at 10:00am. I will return it the following Monday (4 th June). /57 1
Other ources of Information It turns out that the information on FETs in chroder is not well organized. It is spread across multiple chapters. There are some other good sources of information on this subject if you are interested: Review Articles: https://pubs.acs.org/doi/abs/10.101/cr0501543 https://pubs.acs.org/doi/abs/10.101/cm049391x Books: https://www.springer.com/us/book/9783319000015 Course ECE599 Thin Film Electronics, John Labram, Fall 018. 3/57 Lecture 17 The Gradual Channel Approximation. Evaluating Field-Effect Mobility. 4/57
The Gradual Channel Approximation 5/57 Modelling FETs There is only one main model for describing FETs: The Gradual Channel Approximation. It is used throughout semiconductor research and industry. It does not depend on device architecture, geometry or semiconductor material type (generally). It is used to extract parameters of interest from current-voltage characteristics. We are going to derive it from first principles. We will derive the model for an n-type FET, but as usual the theory is general. 6/57 3
Valence Band Conduction Band 5/18/018 Charge Injection Let s take a look at our FET again. We ground our source electrode. This acts as our source of charge carriers. First let us consider what happens when we apply a positive gate voltage ( ). emiconductor ielectric Metal (Gate) + 7/57 Charge Injection Last time we saw that carriers will be injected. emiconductor ource If the energy offsets are correct ource e - E emiconductor ielectric + Metal e 8/57 4
Charge Injection Initially charges will accumulate under the injecting electrode (source). But these carriers will be repelled, making space for more carriers. Eventually uniform charge will accumulate across the interface. 9/57 emiconductor ielectric Metal (Gate) + As a first approximation, the electrode overlap area doesn t matter. Capacitance We can identify this situation as a capacitor. If we know the capacitance of the dielectric, C (we can measure or calculate this), then we can determine the charge accumulated (Q ) for a certain : emiconductor ielectric Metal (Gate) + Q = C The next slide explains why we use a prime (`) 10/57 5
Capacitance / Area Because it makes more sense when comparing devices of different size, we instead will talk about charge per unit area (Q = Q ΤA), and capacitance per unit area (C ox = CΤA): Q = C ox Q has units of C/cm. C ox has units of F/cm. emiconductor ielectric Metal (Gate) + 11/57 Trapped Charge Recall that for capacitors, a lot of charge is not mobile. ome injected charge will fill traps and be immobile. ome already-trapped charge will be released and become mobile. Also, the semiconductor, dielectric, and the interface may be not be electrically neutral at. ± ± ± ± emiconductor ± ± ± ± ielectric Metal (Gate) + ± ± ± 1/57 6
Threshold Voltage We are interested in the density of mobile charges at the interface. I.e. those that are capable of transport and representative of device operation. We define the mobile charge per unit area as Q mob. As with the flatband voltage in MO capacitors, trapped charge will shift voltage characteristics. We can quantify this voltage offset by a threshold voltage: V T And then say: Q = C ox Q mob = C ox V T 13/57 Applied rain Voltage Q mob = C ox V T This equation is valid for an FET with no applied drain bias (since it is basically a capacitor). However when a bias is applied, there is a clearly a variation in charge density as a function of position. ielectric Metal (Gate) V 14/57 7
Applied rain Voltage V V x x ielectric Metal(Gate) We can account for the application of a drain voltage by adding an extra voltage term (V x ) to our equation for mobile charge density: Q mob x = C ox V T V x 15/57 Applied rain Voltage Now lets take a topdown look at our FET: efine our semiconducting channel to have a length L and a width W (sometimes called Z) W We assume that W L and hence can neglect edge effects. At x = 0, V 0 = 0 (it is at the source terminal). At L = 0, V L = V (it is at the drain terminal). L x 16/57 8
Channel Conductivity Q mob x = C ox V T V x This tells us the charge density changes as a function of position in the channel (x). Therefore the conductivity of the channel must also depend on position: σ = σ x. For a sample with a homogenous distribution of carriers, we would evaluate conductivity as a function of charge density using the expression: conductivity Magnitude of fundamental unit of charge σ = e μ e n + μ h p mobility Carrier density 17/57 Channel Conductivity σ = e μ e n + μ h p For an n-type sample this would simplify to σ = eμn We drop the subscript and say μ e = μ Let s take another look at our channel. For now, we say the thickness is : L L W W 18/57 9
plit up the Channel in x But we know the carrier density is not uniform. L L W W Instead of considering L, we can instead consider a very small slice of the channel, of length δx. We will make the approximation that in this slice the charge distribution is uniform. As δx 0, this becomes a valid assumption δx 19/57 Channel Conductivity Consider carrier density in this strip of the channel: W δx ay this is now valid: σ x = eμn x If n is the carrier density, let N be the number of carriers. Therefore in a volume V: N x N x n x = = V Wδx Put back into conductivity: eμn x σ x = Wδx 0/57 10
Channel Conductivity We identify en x as total charge: Q mob x = en x x σ x = μq mob Wδx We also have an equation for -dimensional charge density as a function of position: Q mob x = C ox V T V x We can write -dimensional charge density (Q mob ) in terms of total charge (Q mob ) x Q mob x = Q mob Wδx Q mob x = Q mob x Wδx 1/57 Channel Conductivity ubstitute Q mob x into our equation for σ x σ x = μq mob x Wδx Wδx σ x = μq mob x Use our equation for resistivity / conductance from Lecture : resistivity resistor length ρ = RA L resistance resistor crosssectional area L /57 A 11
Channel Resistance In our system we identify the crosssectional area as: W. Identify length as δx. σ = 1 ρ = δx RW δr = W δx σ x W Note: We use δr because we are talking about a small part of the channel, rather than the resistance (R) of the whole channel. ubstitute in σ x : δx δx δr = δr = μq mob x W μq mob x W 3/57 δx Channel Resistance δr = δx μq mob x W Now consider what happens to the change in resistance as δx 0: dx dr = μq mob x W We can use the differential form of Ohm s law to describe the current arising from changes in resistance with respect changes in voltage: I = dv dr 4/57 1
ource rain Current We call the current flowing in our device the sourcedrain current. Often this is shortened to drain current, labelled: I. I = dv dr dr = dx μq mob x W Combine these two differential equations: I = dv dx μq mob x W ubstitute in our equation for Q mob x : I = dv dx μc ox V T V x W 5/57 ource rain Current I = dv dx μc ox V T V x eparate the differentials: W I dx = μc ox V T V x WdV Which we can evaluate by integration:? න? I dx = න?? μc ox V T V x We just need to chose the limits. WdV 6/57 13
Channel Integral Look at our whole channel again: L V V x x V 0 = 0 V L = V We need to integrate over the whole channel. o we need to integrate from x = 0 (where V x =0) to x = L (where V x = V ). 7/57 ource rain Current Put in these limits: L න 0 I dx = න 0 V μcox V T V x Current does not vary with position, so: WdV L න I dx = I x L 0 = I L 0 = I L 0 Hence we can say: I = W V L μc ox න VG V T V x dv 0 8/57 14
ource rain Current I = W V L μc ox න VG V T V x dv 0 and V T are just constants. o: I = W L μc ox V V T V V V 0 I = W L μc ox V V T V V I = W L μc ox V T V V 9/57 Evaluating Field-Effect Mobility 30/57 15
ource rain Current The master equation of the gradual channel approximation is: I = W L μc ox V T V V This can be used to approximate charge carrier mobility from current-voltage characteristics. In Lecture 10 we talked about why mobility it is important: witching speed. Amplification. 31/57 Mobility in Research A large motivation for a lot of work in emerging electronics is to improve charge carrier mobility. chlom et. al. Nature Materials 9, 881 883 (010) Labram et. al. mall 11, 547 (015) 3/57 16
Warning Recently there has been some controversy over reported mobilities in the literature. A number of groups are reporting inaccurate (too high) mobilities. Either using gradual channel approximation (GCA) incorrectly. Or applying it to situations when the assumptions we made are not valid. https://www.nature.com/articles/ncomms10908 http://science.sciencemag.org/content/35/693/151 https://www.nature.com/articles/nmat5035 33/57 Output Characteristics There are two main ways in which mobility is extracted from IV characteristics. In the linear regime linear mobility (μ lin ). In saturation regime saturation mobility (μ sat ). Recall what output curves look like: I aturation regime I ( A) 800 700 600 500 400 300 00 100 aturation 0 V -50-40 -30-0 -10 0 V (V) P-type transistor Linear 34/57 17
ource-rain Current (I ) 5/18/018 Output Characteristics Recall, this is due to the channel being pinched-off. E ielectric Metal (Gate) E ielectric Metal (Gate) E ielectric Metal (Gate) I I I V V V 35/57 Linear Regime Recall, for the device to have ~linear characteristics we require the field pointing ~down very small V. E ielectric Metal (Gate) V rain Voltage (V ) 36/57 18
Linear Regime We consider the FET in the linear regime when: V V T We can use this to say: V T V V Apply this approximation to the GCA Equation: I = W L μ linc ox V T V μ lin denotes this is only valid in the linear regime. 37/57 Linear Regime We actually tend to evaluate μ from transfer characteristics, rather than output characteristics. This is because it turns out we don t need to know V T. I = W L μ linc ox V T V I (A) 10-4 10-5 10-6 10-7 10-8 10-9 Transfer 10-3 V = 100V 10-10 -0 0 0 40 60 80 (V) V = 10V I ( A) 300 00 100 Output = 5V to 80V 0 0 0 40 60 80 100 V (V) 38/57 19
Linear Regime The way this is usual done is by numerically differentiating the transfer curve. Recall that the transfer curve is I plotted against : I = W L μ linc ox V T V I (A) 10-3 V = 100V 10-4 10-5 10-6 10-7 10-8 10-9 10-10 -0 0 0 40 60 80 (V) V = 10V di = W d L μ L di linc ox V μ lin = WC ox V d 39/57 Linear Mobility μ lin = L WC ox V di d Notice, we don t need to know V T now. We just need to know the channel dimensions (length and width), and the dielectric capacitance. These are normally very easy to determine. It is important to emphasize that this is only valid in the linear regime. Hence only if: V V T 40/57 0
Numerical ifferentiation You are all familiar with differentiating mathematical functions: y x = x Numerical differentiation is important to FET mobility analysis. And also many other techniques. If you have experimental data of I vs, how do you determine di d? dy dx = x But the concept of differentiating real data may be new to some of you. 41/57 Numerical ifferentiation When going from analytical functions to real data, we have to consider derivatives as difference equations: dy dx y x Where y and x are now finite steps. To find the derivative at a data point i we need to consider adjacent points i 1 and i + 1. y y The point we are interested in. i 1 x i + 1 y x = y i+1 y i 1 x i+1 x i 1 i erivatives for the end points are not defined 4/57 x 1
Linear Example ata Here is what some real data looks like differentiated: 10-4 10-5 10-6 I (A) 10-7 10-8 10-9 10-10 -0 0 0 40 60 80 (V) V = 10V di /d (A/V) 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-1 -0 0 0 40 60 80 (V) V = 10V 43/57 Linear Mobility di /d (A/V) 10-5 10-6 10-7 10-8 10-9 10-10 10-11 10-1 -0 0 0 40 60 80 (V) V = 10V The rest of the parameters are known: V, L, W, C ox. In the linear regime we can only consider data where V V T. Normally some average is taken in the correct part of the curve. Here: di d 1 10 6 A/V μ lin = L WC ox V di d 44/57
ource-rain Current (I ) 5/18/018 aturation Regime Recall, if V is very large we get saturation behavior. E ielectric Metal (Gate) V rain Voltage (V ) 45/57 aturation Regime V V x V 0 = 0 V L = V After pinch-off (saturation), a depletion region forms by the drain electrode. Here there are very few carriers, and this region is very resistive. Pinch-off occurs when V = V T. 46/57 3
aturation Regime V V x = V T Constant As the source-drain voltage is increased further the pinch-off point moves towards the source electrode and the depletion region will increase in size. The voltage drop between the source and the pinchoff point will remain constant at V x = V T. 47/57 aturation Regime V V x = V T V ( V T ) The additional voltage applied is dropped across the high-resistance depletion zone. ince carriers are only present in the left part of the channel, this is the effective device. The device behaves as if V = ( V T ) = constant. 48/57 4
aturation Regime We consider the FET in the saturation regime when: V V T The device behaves as if the voltage is constant at: V = V T Apply this approximation to the GCA Equation: I = W L μ satc ox V T μ sat denotes this is only valid in the saturation regime. 49/57 aturation Regime Notice we now have I. I = W L μ satc ox V T o we can t just differentiate the function as we did for the linear mobility. 10-3 V = 100V 10-4 10-5 10-6 10-7 10-8 10-9 10-10 -0 0 0 40 60 80 It turns out we have options to get mobility. I (A) (V) V = 10V Approach 1 Take square root of I and differentiate that. Approach ifferentiate I twice. 50/57 5
Approach 1 Take square root: I = W L μ satc ox V T I = V T W L μ satc ox Now take 1 st derivative: d I = W d L μ satc ox 51/57 Approach 1 quare both sides: d I = W d L μ satc ox d d I = W L μ satc ox μ sat = L WC ox d I d 5/57 6
Approach Multiply out: I = W L μ satc ox V T I = W L μ satc ox V T V T Take 1 st derivative: I = W L μ satc ox V T + V T di d = W L μ satc ox V T 53/57 Approach di d = W L μ satc ox V T Take nd derivative: d I dv = W G L μ satc ox Re-arrange: d I d = W L μ satc ox μ sat = L WC ox d I d 54/57 7
aturation Example ata Here is what some real data looks like differentiated: I (A) 10-3 10-4 10-5 10-6 10-7 10-8 10-9 -0 0 0 40 60 80 (V) V = 100V d d I d I d (di 1//d ) (A/V ) (A /V ) d I /d 10-6 10-7 10-8 10-9 10-10 10-11 10-1 10-13 -0 0 0 40 60 80 (V) 10 1 10 0 10-1 10 - V = 100V V = 100V 10-3 -0 0 0 40 60 80 (V) 55/57 aturation Mobility (di 1//d ) (A/V ) (A /V ) d I /d 10-6 10-7 10-8 10-9 10-10 10-11 10-1 10-13 -0 0 0 40 60 80 (V) 10 1 10 0 10-1 10 - V = 100V V = 100V 10-3 -0 0 0 40 60 80 (V) Here we can only consider data where V V T. Extracting mobility from saturation can be a little bit more subjective than linear mobility. Normally we look at the curve where the device is on ( V T ), but also in saturation. 56/57 8
Next Time Final Lecture on FETs eries Resistance V G G VG R m (Ω) R B I R R ΔL L (μm) Threshold voltage I A few other parameters V T 57/57 9