Slide điện tử từ trường lecture 8 mosfet circuit

Lecture 08 MOSFET’s Circuit Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt topics Large-signal operation FET circuits at DC FET biasing schemes Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Large-signal operation ID = μCo w 2L (vGS − Vt ) Load line: VDD k n' w ID = (vGS − Vt ) 2L = iD RD + vDS Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt dvo Av ≡ dvI v I =VIQ Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.2 DC analysis ID = Given I = 0.4mA D VD = 0.5V Vt = 0.7V μ nCox = 100 μA / V L = 1μm w = 32 μm μC o w ⇒ vGS (vGS − Vt ) 2L = 1.2V VS = I D RS + VSS ⇒ VS = I D RS − 2.5V − 1.2 + 2.5 Rs = = 3.25k 0.4m VDD = I D RD + VD RD = ? ⇒ 2.5V = I D RD + VD RS = ? RD = 2.5 − 0.5 = 5k 0.4m Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.3 Given I D = 80 μA Vt = 0.6V μ nCox = 200 μA / V L = μ m w = μm VD = ? R=? ID = μC o w 2L ⇒ VD = 1V (VD − Vt ) VDD = 3V = I D R + VD −1 ⇒R= = 25k 0.08m Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Exercise 4.12 ID R = 25k VD ID = μC o w (VGS − Vt ) = 2L QVD = 1V ∴ I D = 80μA μC o w 2L (VD − Vt ) VDD = 3V = I D R2 + VD ⇒ VD = − 80 μ × 20k = 1.4V Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.4 let Vt = 1V K n′ (W / L) = 1mA / V find RD = ? vDS = VD = 0.1V ≤ VDD − Vt = − = vDS ≤ vGS − Vt → ohmic region ID = μ n Co w [2(vGS − Vt )vDS − vDS ] 2L → I D = 0.395mA − 0.1 RD = = 12.4k 0.395m Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.5 let Vt = 1V K n′ (W / L) = 1mA / V Assume FET in saturation mode since VG>0 10 M 10V = 5V 10 M + 10M W I D = k n' (VGS − Vt ) 2 L ⇒ I D = × 1× [(5 − I D × 6k ) − 1]2 ⇒ 18I D2 − 25I D + = VG = I D = 0.89mA, I D = 0.5mA if I D = 0.89mA VS = 6k × 0.89m = 5.34V → VGS < if contradiction I D = 0.5mA VS = 6k × 0.5m = 3V → VGS = 2V → VD = 10 − 6k × 0.5m = 7V → VD > VG − Vt 10 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.6 Design the circuit operate in saturation Find the range of RD 'W k n (VGS − Vt ) 2 L ⇒ 0.5 = ×1× [VGS − (−1)]2 ⇒ VGS = −2V ID = VDS ≥ VGS − Vt = −2 + = −1 ⇒ VD (max) = +4V = 8k 0.5m = = 6k 0.5m RD (max) = RD (min) 11 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.7 let Vnt = −V pt = 1V K n′ (W / L) = K ′p (W / L) = 1mA / V First case : vI=0V Q p on, Qn on VGS ( p ) = −2.5V VGS ( n ) = 2.5V vo = 0V I DP = ×1[−2.5 − (−1)]2 = 1.125mA I Dn = × 1[2.5 − 1]2 = 1.125mA 12 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Second case : vI=+2.5V Qp ID = μ n Co w 2L off , Qn [2(vGS − Vt )vDS − vDS ]≈ ohmic region μ n Co w L (vGS − Vt )vDS NOT gate VD = −2.44V = vo VGS = 5V I Dn = 1×1[5 − 1]VDS VDS = −2.44 + 2.5 = 0.06 = 1×1[5 − 1](− I Dn ×10k ) VGS − Vt = − = ⇒ I Dn = 0.244mA VDS ≤ VGS − Vt Ohmic region 13 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Second case : vI=-2.5V Qp ohmic region , Qn off NOT gate VSG = 5V I Dp = 1×1[5 − 1]VSD VD = vo = I Dp × 10k = 2.44V VSD = +2.5 − 2.44 = 0.06 = 1×1[5 − 1](2.5 − I Dp × 10k ) V − V = − = SG t ⇒ I Dp = 0.244mA VSD ≤ VSG − Vt Ohmic region 14 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Exercise 4.16 Amplifier Qn up Assume k n' (Wn / Ln ) = k p' (W p / L p ) = 1(mA / V ) Vtn = −Vtp = 1V λ =0 find iDN , iDP , vo for vI = 0V ,2.5V ,−2.5V First case : vI=0V Qn off Qp 2nd case : vI=2.5V 3rd case : vI=-2.5V Qn Qn on Q p off off Qp on off 15 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt 16 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Biasing method • • • • Fixed Bias by VGS Self Bias Biasing using Drain –to-Gate feedback resistor Biasing using constant current source Biasing circuit Small signal 17 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt ID = μ n Co w 2L (vGS − Vt ) Device change ( Vt , Cox ,W L ) Temperature change ( Vt , μ n ) Change ID 18 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.9 Self Bias find Vt = 1.5V Change MOSFET K n′ (W / L) = 1mA / V ΔI D = ? mA k n' w ID = (vGS − Vt ) = × 1× (vGS − 1.5) 2 2L vGS = − 10k × I D ⇒ I D = 0.455mA when Good biasing scheme Vt = 1V K n′ (W / L) = 1mA / V I D = 0.5mA I D ↑⇒ I D RS ↑⇒QVG fixed ⇒ VGS ↓ k n' W Q ID = (vGS ↓ −Vt ) ⇒ I D ↓ 2L 19 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Self Bias I D ↑⇒ VS ↑⇒ VGS ↓⇒ I D ↓ Good biasing scheme ' n k w ID = (vGS − Vt ) 2L VGS = − I D RS 20 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Exercise 4.21 Drain –to-Gate feedback bias k n' w (VGS − Vt ) ID = 2L 0.5m = × 1× (VGS − 1) 2 ⇒ VGS = 2V parameters VDD = 5V Vt = 1V K n′ (W / L) = 1mA / V requirement VGS = VDS = VDD − I D RD I D = 0.5mA VGS = VDS = − I D × RD find 2V = − 0.5 × RD RD , VD ⇒ RD = 6k VGS = VDS ⇒ saturation I D ↑⇒ VD = VDD − I D RD ⇒ VD ↓ ⇒ VGS ⇒ I D ↓ Good biasing scheme 21 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Constant current bias I REF = I D1 = VDD − VGS = I D2 R 22 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt 23 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt 24 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt ... n' (VGS − Vt ) 2 L ⇒ I D = × 1× [(5 − I D × 6k ) − 1]2 ⇒ 18I D2 − 25I D + = VG = I D = 0 .89 mA, I D = 0.5mA if I D = 0 .89 mA VS = 6k × 0 .89 m = 5.34V → VGS < if contradiction I D = 0.5mA VS = 6k... = 25k 0.08m Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Exercise 4.12 ID R = 25k VD ID = μC o w (VGS − Vt ) = 2L QVD = 1V ∴ I D = 80 μA μC o... change ( Vt , μ n ) Change ID 18 Microelectronic circuits by meiling CHEN CuuDuongThanCong.com https://fb.com/tailieudientucntt Example 4.9 Self Bias find Vt = 1.5V Change MOSFET K n′ (W / L) = 1mA