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3/9/2011 Reducing the Effect of Input Bias Current lecture 1/7 Reducing the Effect of Input Bias Current R2 i2 vin R1 vIB 1 i1 v+ R3 IB 2 - We found that the input bias current will cause an offset in the output voltage. There is a solution to this problem— place a resistor (R3) on the noninverting input! vout + Q: Maria, why is this resistor here? I don’t see how it can do any good. Jim Stiles The Univ. of Kansas Dept. of EECS 3/9/2011 Reducing the Effect of Input Bias Current lecture 2/7 The voltage v+ is non-zero! A: Let’s analyze this circuit to determine how this new resistor helps. First, notice that the voltage at the non-inverting terminal is now non-zero! The bias current IB 2 means that, by virtue of KVL: v + = 0 − R3 IB 2 = −R3 IB 2 R2 Now, because of the virtual short: v − = v + = −R3 IB 2 And from KCL: vin i2 R1 vIB 1 i1 v+ i1 = i2 + IB 1 where from KCL and Ohm’s Law: i1 = R3 IB 2 - vout + vin − v − vin + R3IB 2 = R1 R1 Jim Stiles The Univ. of Kansas Dept. of EECS 3/9/2011 Reducing the Effect of Input Bias Current lecture 3/7 It seems like this just made the offset even larger R2 And also from KCL and Ohm’s Law: i2 = vin − vout − (R3IB 2 + vout ) = R2 R2 i2 vin Combining these results: R1 vIB 1 i1 vin + R3IB 2 − (R3IB 2 + vout ) = + IB 1 R1 R2 v+ R3 IB 2 - vout + Performing the usual algebraic gymnastics, we rearrange this result and find that the output voltage is: ⎛ R2 ⎞ ⎛ ⎞ R2R3 − + − v R I I R I ⎟ in ⎜ 3 B 2 2 B1 ⎟ B2 R1 ⎝ R1 ⎠ ⎝ ⎠ vout = ⎜ − Jim Stiles The Univ. of Kansas Dept. of EECS 3/9/2011 Reducing the Effect of Input Bias Current lecture 4/7 Awl be baak Again we find the output consists of two terms. The first term is the ideal inverting amplifier result: − R2 vin R1 and the second is an output D.C. offset: −(R3IB 2 + R2R3 IB 2 − R2IB 1 ) R1 Q: Resistor R3 was supposed to reduce the D.C. offset, but it seems to have made things even worse. Fix this or I shall be forced to pummel you. Jim Stiles The Univ. of Kansas Dept. of EECS 3/9/2011 Reducing the Effect of Input Bias Current lecture 5/7 We must choose the proper value of R3… A: Say we set the value of resistor R3 to equal R3 = R1 R2 , i.e.: R1R2 R1 + R2 R3 = In this case, the D.C. offset becomes: ⎛ R1R2 ⎞ R22 IB 2 + IB 2 − R2IB 1 ⎟ −⎜ R1 + R2 ⎝ R1 + R2 ⎠ ⎛ (R + R2 )R2 ⎞ = −⎜ 1 IB 2 − R2IB 1 ⎟ ⎝ R1 + R2 ⎠ = R2 (IB 1 − IB 2 ) = R2Ios Typically, the bias currents IB 1 and IB 2 are approximately equal, so that offset current IB 1 − IB 2 = Ios is very tiny. Therefore, the resulting output offset voltage is likewise very tiny! Jim Stiles The Univ. of Kansas Dept. of EECS 3/9/2011 Reducing the Effect of Input Bias Current lecture 6/7 …and this is that proper value Therefore, when designing an amplifier with real op-amps, always include a resistor R3 equal to the value: R3 = R1 R2 = R1R2 R1 + R2 This is true regardless of whether we use the inverting or non-inverting configurations! R2 i2 vin R1 i1 v- R2 - i2 IB 1 R3 = R1 R2 v+ IB 2 R1 + i1 vin vIB 1 R3 = R1 R2 v+ IB 2 Jim Stiles The Univ. of Kansas - + Dept. of EECS 3/9/2011 Reducing the Effect of Input Bias Current lecture 7/7 This is just the type of subtle point that shows up on an exam If the impedances are complex (i.e., Z 1 (ω ) and Z 2 (ω ) ), then set the resistor R3 based on the D.C. values of the impedances: R3 = Z 1 (ω = 0 ) Z 2 (ω = 0 ) In other words, set the capacitors to open circuits and inductors to short circuits. Jim Stiles The Univ. of Kansas Dept. of EECS