Explanation Given: V + = + 3 V V − = − 3 V V T N = 0.7 V λ = 0 K n 1 = 0.35 mA/V 2 K n 2 = 0.30 mA/V 2 K n 3 = 0.10 mA/V 2 Calculation: Consider the circuit shown below. The I REF for transistor M 3 I REF = K n 3 ( V G S 3 − V T N ) 2 = K n 3 ( V + − V S 3 − V T N ) 2 Substitute the given values in above expression, I REF = ( 0.10 × 10 − 3 ) ( 3 − V S 3 − 0.7 ) 2 I REF = ( 0.10 × 10 − 3 ) ( 2.3 − V S 3 ) 2 ( 1 ) Now I REF for transistor M 1 will be, I REF = K n 1 ( V G S 1 − V T N ) 2 I REF = K n 1 ( V S 3 − V − − V T N ) 2 Substitute the given values in above expression, I REF = ( 0.35 × 10 − 3 ) ( V S 3 + 3 − 0.7 ) 2 I REF = ( 0.35 × 10 − 3 ) ( 2.3 + V S 3 ) 2 ( 2 ) Now equating equation 1 and equation 2, ( 0

MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

4th Edition

ISBN: 9781266368622

Author: NEAMEN

Publisher: MCG

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Chapter 10, Problem 10.7TYU

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The value of IREF and IO for the given circuit.

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Chapter 10, Problem 10.6TYU

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MICROELECT. CIRCUIT ANALYSIS&DESIGN (LL)

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The value of I REF and I O for the given circuit.

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The value of IREF and IO for the given circuit.

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