List down various parameters of Opamp along with their typical values for IC74 I. Also explain what the significance of CMRR and Slew Rate is?

Mumbai University > Computer Engineering > Sem 3 > Electronic Circuits and Communication Fundamentals

Marks: 10 Marks

Year: May 2016

Parameters IC741 Values
Differential input Resistance 2MΏ
Input capacitance 1-4 pF
Open Loop Voltage Gain 200,000
CMRR 90 dB
Output Voltage Swing ±13 to ±15 V
Output Resistance 75 Ώ
Input Voltage Range ±12 to ±13 V
Power Supply Rejection Ratio 30µV/V
Power Consumption 85 mW
Gain-Bandwidth Product 1MHz
Average Temperature Coefficient of Offset Parameters $12 pA/C^0$
Supply Current 2.8 mA
Slew Rate 15 µA


It is the ratio of differential voltage gain $A_d$ to the common mode voltage gain $A_C$.


Now $A_d$ is nothing but open loop voltage gain $A_{OL}$. And $A_c$ is measured by using the circuit as shown in figure

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The common mode input $V_C$ is applied to both the input terminals of OP-amp. Then the output $V_{OC}$ is measured. Then common mode gain $A_C$ can be obtained as,


It is generally very small and not specified in the data sheet. The CMRR is generally specified for the op-amp and is expressed in dB. For op-amp 741C it is 90dB.

Slew Rate

The slew rate is defined as the maximum rate of change of output voltage with time.

The slew rate is specified in V/μsec. Thus,

$$Slew \ \ Rate=S=\dfrac{dV_0}{dt}\bigg| \max$$

The slew rate is caused due to limited charging rate of the compensating capacitor and current limiting and saturation of the internal stages of an op-amp, when a high frequency, large amplitude signal is applied.

The internal capacitor voltage cannot change instantaneously. It is given by $(dV_c)/dt=I/C$. For large charging rate, the capacitor should be small or charging current should be large. Hence the slew rate for the op-amp whose maximum internal capacitor charging current is known, can be obtained as


For example, for IC 741 the charging current is 15 μA and the internal capacitor is 30 pF, hence its slew rate is

$$S=\dfrac{15 \times 10^{-6}}{30 \times 10^{-12}}=\dfrac{0.5}{10^{-6}}V/\sec \\ =0.5 V/\mu \sec$$

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