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Metal Semiconductor FET - MESFET� SMA5111 - Compound Semiconductors Lecture 9 - Metal-Semiconductor FETs - Outline • Device structure and operation Concept and structure:� General structure� Basis of operation; device types� Terminal characteristics Gradual channel approximation w. o. velocity saturation Velocity saturation issues Characteristics with velocity saturation Small signal equivalent circuits High frequency performance • Fabrication technology Process challenges: (areas where heterostructures can make life easier and better) 1. Semi-insulating substrate; 2. M-S barrier gate; 3. Threshold control; 4. Gate resistance; 5. Source and drain resistances Representative sequences: 1. Mesa-on-Epi; 2. Proton isolation; 3. n+/n epi w. recess; 4. Direct implant into SI-GaAs C. G. Fonstad, 3/03 Lecture 9 - Slide 1� BJT/FET Comparison - cont.� Base/ Emitter/ Collector/ Channel Source Drain 0 wB or L Controlled by vEB in a BJT; by vGS in an FET vCE or vDS C. G. Fonstad, 3/03 0 wB or L Lecture 9 - Slide 2 BJT/FET Comparison - cont.� OK, that's nice, but there is more to the� difference than how the barrier is controlled� BJT FET Charge minority majority carriers in base in channel Flow diffusion drift mechanism in base in channel Barrier direct contact change induced control made to base by gate electrode The nature of the current flow, minority diffusion vs� majority drift, is perhaps the most important difference.� C. G. Fonstad, 3/03 Lecture 9 - Slide 3� FET Mechanisms - MOSFET and JFET/MESFET� MOSFET� Induced n-type channel JFET Doped n-type channel C. G. Fonstad, 3/03� Lecture 9 - Slide 4 Junction Field Effect Transistor (JFET)� Reverse biasing the gate-source junction increases depletion width under gate and constricts the n-type conduction path between the source and drain. Ref: Fonstad, Mircoelectronic Devices and C. G. Fonstad, 3/03 Circuits, Chap. 10 - posted on SMA5111 Website Lecture 9 - Slide 5 JFET, cont.� Operating regions:� (a) linear (or� triode) well below� pinch-off� (a) (b) Vp < vGS < 0, vDS small (b) linear nearing pinch-off� Vp < vGS < 0, vDS� (d) (c) nearing (vGS - Vp) (c) (c) pinch-off� Vp �< vGS < 0,� vDS > (vGS - V�p) (d) cut-off� (b) (d) vGS < Vp < 0� (a) Ref: Fonstad, Chap. 10 C. G. Fonstad, 3/03 Lecture 9 - Slide 6 MEtal Semiconductor FET - MESFET� The operation is very similar to that of a JFET. The p-n junction gate is replaced by a Schottky barrier, and the lower contact and p-n junction are eliminated because the lightly doped p- type substrate is replaced by a semi-insulating substrate. C. G. Fonstad, 3/03 Lecture 9 - Slide 7� MESFET current-voltage (i-v) characteristic� We have two currents, iG and iD, that we want to determine as functions of the two voltages, vGS and vDS: ig (vgs,vds ) and id (vgs,vds ) If we restrict our model to drain-to-source voltages greater than zero, and gate-to-source voltages less than the turn-on voltage of the gate Schottky diode, then, say that the gate current is negligible and iG ≈0: ª £ ≥ ig (vgs,vds ) 0 if vgs Von , and vds 0. Our real problem then is to find an expression for the drain current, iD. The path for the drain current is through the channel to the source, and we model it using the gradual channel approximation (next slide). C. G. Fonstad, 3/03 Lecture 9 - Slide 8� MESFET current-voltage (i-v) characteristic� The model used to describe the drain current-voltage expression for an FET is the Gradual Channel Approximation. In this model we assume the current flow in the channel is entirely in the y-direction, and that the field lines terminating on the gate are entirely vertical. Gradual Channel Approximation Field lines above channel are strictly vertical Current flow is strictly horizontal C. G. Fonstad, 3/03 Lecture 9 - Slide 9� MESFET i-v: gradual channel approximation� If the Gradual Channel Approximation is valid we can solve two independent 1-d problems in sequence: (1) an electrostatics problem in the vertical (x-) direction to find the mobile charge sheet density at any point, y, in the� channel;� (2) a drift problem in the horizontal (y-) direction to relate the voltage drop along the channel to the current. We integrate this relation from y = 0 to y = L to get the final expression relating the drain current to the gate and drain voltage. Electrostatics problem in the x-direction Carrier drift problem in the y-direction C. G. Fonstad, 3/03 Lecture 9 - Slide 10� MESFET i-v: vertical electrostatics problem� vGS The sheet carrier density 0 at the position y along� Depletion the channel is: region � N (y) = N [a - x ()y ] xd(y) ch Dn d � NDn And the depletion width there is: x ()y = 2e {f - [v - v ()y ]} qN d S b GS CS Dn a v (y) CS Combining these yields: È ˘ x N (y) = N a - 2e {f - [v - v ()y ]} qN ch Dn ÎÍ S b GS CS Dn ˚˙ This is the information we wanted to C. G. Fonstad, 3/03 get from the vertical problem. Lecture 9 - Slide 11 MESFET i-v: horizontal drift current problem iD •�Define vCS(y) as the voltage in the channel relative to the source at y. vCS(y) varies from vDS at y = L to 0 at y = 0 the horizontal E-field in the channel, Ey, is - dvCS(y)/dy •�The drain current, iD, flows right to left in the channel: - iD is a constant and is not a function of y -�at any y, iD is the sheet charge density at that y, times its net drift velocity, times the channel width C. G. Fonstad, 3/03� Lecture 9 - Slide 12� MESFET i-v: horizontal drift current problem, cont� We write thus the current as: i =--[ qN (y)⋅W ⋅ s (y)] D ch y where -q is the charge per carrier; W is the channel� width; and Nch(y) is the sheet carrier density at point y along the channel, which we have already determined� by solving the vertical electrostatics problem:� È ˘ N (y) = N a - 2e {f - [v - v ()y ]} qN ch Dn ÎÍ S b GS CS Dn ˚˙ In the low- to moderate-field region, the average net carrier velocity in the y-direction is the drift velocity: dv (y) =-m = m CS sy (y) eFy (y) e dy C. G. Fonstad, 3/03 Lecture 9 - Slide 13� MESFET i-v: horizontal drift current problem, cont� Putting this all together we get:� È ˘ dv (y) =- - e {f - [ - ()]} ⋅ ⋅ m CS � iD qNDn Ía 2 S b vGS vCS�y� �qNDn ˙ W� e Î ˚ dy� We rearrange terms, multiply by dy, and integrate each side from y = 0, vCS = 0, to v = L, vCS = vDS: È 1/2˘ L vDS 2e � = m Í - S f - - ˙ Ú iD dy� eqNDnW�Ú a ( b [vGS vCS ()y ]) dvCS �0 0 � Î qNDn� ˚ Doing the integrals, and dividing both sides by L yields:� Ï ¸ W 2� 2e 3/2 3/2 i = a qm N Ìv -� S [(f - v + v ) - (f - v ) ]˝ D e Dn DS 2 � b GS DS b GS L Ó� 3�qNDn a ˛ C. G. Fonstad, 3/03 Lecture 9 - Slide 14� MESFET i-v: linear or triode region� Our result thus far: Ï ¸ W 2 2e 3/2 3/2 � = m Ì - � S (f - + ) - (f - ) ˝ iD a� q e NDn vDS 2 [ b vGS vDS b vGS ] L Ó� 3�qNDn a ˛ is only valid until xd(y) = a, which occurs at y = L when v = v - (f - qN a2 2e ) DS GS b Dn S Plotting this: Region of validity C. G. Fonstad, 3/03 Lecture 9 - Slide 15 MESFET i-v: saturation region� For larger v , i.e. when: v > v - (f - qN a2�2e ) DS DS GS ��b � Dn S � The current, iD, stays constant at the peak value, which is:� Ï È 3/2 3/2 ˘¸ Ô 2 �(f -V ) - (f - v ) Ô = Ì( - ) - �Í b �P�� ��b �GS�� ˙˝ iD Go ��vGS� �VP �� 1/2 , Ô 3ÎÍ (f -V ) ˚˙Ô Ó� � �b �P� ˛�� 2 W where V ≡ (f - qN a 2e ) and G ≡ a� qm N p b Dn S o� L� e Dn Plotting this: Saturated current C. G. Fonstad, 3/03 Lecture 9 - Slide 16� MESFET i-v: summary of characteristics We identify the pinch-off voltage, VP, and undepleted channel conductance, Go: W V ≡ (f - qN a 2�2e ) G ≡ a qm�N p b Dn S � o� L e Dn We can then write the drain current, iD, for each of the three regions: When vGS <VP the channel is Cutoff: pinched off for all vDS, and the ( - ) £ £ = device is cut-off,iI.e., iD = 0. When v�GS VP� 0 vDS , iD 0 Saturation:� Ï È 3/2� 3/2 ˘¸ Ô 2 �(f -V ) - (f - v ) Ô £ ( - ) £ = Ì( - ) - �Í b P � b GS ˙˝ When 0 v�GS �VP�� vDS , iD Go� vGS VP� 1/2 � Ô 3�ÎÍ (f -V ) ˚˙Ô Ó � b P � ˛�� Linear: Ï 3/2 3/2 ¸ Ô È (f - + ) �- (f - ) ˘Ô 2� b vGS vDS�� b vGS When 0 £ v £ (v -V ), i = G�Ì v - Í ˙˝� DS� GS� P D o�Ô DS� 3�ÎÍ (f -V )1/2 ˚˙Ô Ó b �P � ˛�� C. G. Fonstad, 3/03 Lecture 9 - Slide 17� MESFET characteristics - channel-length modulation� In practice we find that iD increases slightly with vDS in saturation. This is because the effective channel length decreases with increasing vDS above pinch-off. We model this by saying Æ - l L L(1 vDS )� Æ + l which means 1/ L (1 vDS )/L � and consequently G Æ (1 + lv )G o DS o iD Slope exagerated for emphasis Approx.
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