《Microelectronics Process》lecture6

3155/6.,152] Lecture6: IC Lab Testing Prof martin a schmidt Massachusetts Institute of Technology 9/24/2003

3.155J/6.152J Lecture 6: IC Lab Testing Prof. Martin A. Schmidt Massachusetts Institute of Technology 9/24/2003

Outline Review of process Structures to be tested Sheet resistance MOS Capacitor Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 2

Outline „ Review of Process „ Structures to be Tested „ Sheet Resistance „ MOS Capacitor Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 2

Our process Polysilicon gate(n-type )MOS Capacitor n-type substrate 250nm n-type polysilicon gate 50nm gate oxide Various size capacitors Polysilicon sheet resistivity monitor Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 3

Our Process „ Polysilicon Gate (n-type) MOS Capacitor „ n-type substrate „ 250nm n-type polysilicon gate „ 50nm gate oxide „ Various size capacitors „ Polysilicon sheet resistivity monitor Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 3

Resistance 工 R=pL/A=(p/t(L/W Resistivity Process Mask Q-cm Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 4

Resistance W L t R = ρ L/A = (ρ/t) (L/W) Resistivity Process Mask Ω-cm Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 4

Concept of sheet resistivity R=pl/A=(p/t(L/W Sheet Resistivity(rs) of Squares Q2/sq R Fall 2003-MA schmidt 3. 155J 6. 152]-Lecture 6-Slide 5

Concept of Sheet Resistivity R = ρ L/A = ( ρ/t) (L/W) Sheet Resistivity (R S) # of Squares Ω/sq L = W R = R S Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 5

Number of squares R=2R S R=Rs/2 R=Rs/3 R=8R S R=6.5R Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 6

Number of Squares R = 2RS R = R S/2 R = R S/3 R = 8RS R = 6.5RS Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 6

Measurement of sheet resistance DO D D DA A A CB R′=V DC /AB DA /CB Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 7

Measurement of Sheet Resistance + VDC ­ A B D C IAB A B D C + - VDA ICB R’ = VDC / IAB R’’ = VDA / ICB Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 7

Van der pauw R5=(π/n2)(R+R)fR/R) f(R/R") Correction factor Assumptions: Uniform thickness Continuous(no holes) 0.2 101001000 R/R Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 8

Van der Pauw RS = ( π/ln2) ½ (R’+R’’) f(R’/R’’) f(R’/R’’) Correction Factor 1 Assumptions: Uniform thickness Continuous (no holes) 0.2 1 10 100 1000 R’/R’’ Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 8

Van der pauw a implement a symmetric structure R′=R f(R/R)=1 Rc=4.53R ave Rave =12(R+R") Van der pauw test structure Arms should align to pins 7, 5, 21 and 15 Fall 2003-MA schmidt 3. 155J 6.152]-Lecture 6-Slide 9

Van der Pauw „ Implement a symmetric structure „ R’ = R’’ „ f(R’/R’’) = 1 „ RS = 4.53 Rave „ Rave = ½ (R’+R’’) Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 9

N Square resistor nask +△L W=Wn+△W mas N=Mask/Mask >>1 R=RS(L/W)=Rs [Mask/Mask +AW)I Used to determine the process bias(AW) Fall 2003-MA schmidt 3.155/6.152]- Lecture6-Side10

N Square Resistor L = Lmask + ∆L W = Wmask + ∆W N = Lmask/Wmask >>1 R = RS (L/W) = RS [Lmask/(Wmask + ∆W)] Used to determine the process ‘bias’ ( ∆W) Fall 2003 – M.A. Schmidt 3.155J/6.152J – Lecture 6 – Slide 10