The op-amp circuit below is to be used at DC and very low frequencies. A closed-
ID: 2084532 • Letter: T
Question
The op-amp circuit below is to be used at DC and very low frequencies. A closed-loop gain of -200 is required. Specifications indicate that: (i) the error due to finite open-loop gain cannot exceed 0.1% (ii) DC output voltage due to input offset voltage lessthanorequalto 100 mV (iii) DC output voltage due to input offset current lessthanorequalto 5 mV Determine: (a) the minimum open-loop gain required of the op-amp (b) the maximum input offset voltage required of the op-amp (c) assuming that the op-amp's input offset current is 10 nA, and a suitable compensating resistor R_3 is used, calculate the maximum value of R_2 that can be permitted.Explanation / Answer
a) Ideal close loop given is Ai = -200
Exact close loop gain for inverting amplifier is fiven by
Af = -AR2 / (R1 + R2 + AR1 ) ( A = open loop gain )
Since exact gain is less than ideal gain and error due to open loop gain can not exceed 0.1% i.e. -0.2
Af = - 199.8
AR2 / (R1 + R2 + AR1 ) = 199.8
Ai = -R2/R1 = -200
R2 = 200R1
A200R1 / (R1 + 200R1 +AR1) = 199.8
200A /( 201 +A) = 199.8
200A = 40159.8 + 199.8A
0.2A = 40159.8
A = 200799 .......................1
This is the required open loop gain for op-amp
b) DC output voltage due to input offset voltage is Voo is less than or equal to 100 mV
Voo = ( 1 + [R2/R1] )*Vio (Vio = input offset voltage of op-amp )
100 = ( 1+ [200R1/R1] )*Vio
100 =201*Vio
Vio = 0.497 mV ................2
This is the required maximum input offset voltage for op-amp
c) When suitable compensating resistor R3 = (R1R2) / (R1 + R2 ) is used then DC output voltage due to input offset current Iio is given by
Voio = R2*Iio
5mV = R2*10nA
R2 = 5 mV / 10 nA
R2 = 500 K Ohm ......................3
This is the required maximum value of R2 that can be permitted
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