A. Average of two input bias current
B. Summing of two input bias current
C. Difference of two input bias current
D. Product of two input bias current

Answer: A

Input bias current is the average of two input bias currents flowing into the non-inverting and inverting input of an op-amp.

2. Although the value of input bias current is very small, it causes

A. Output voltage
B. Input offset voltage
C. Output offset voltage
D. All of the mentioned

Answer: C

Even a very small value of input bias current can cause a significant output offset voltage in circuits using relatively large feedback resistors.

3. The formula for output offset voltage of an op-amp due to input bias current

A. V_{OIB}= R_{F}*I_{B}
B. V_{OIB}= (R_{F}+R_{1})/I_{B}
C. V_{OIB}= (1+R_{F})*I_{B}
D. V_{OIB}= [1+(R_{F}/R_{1})]*I_{B}

Answer: A

The output offset voltage due to input bias current is V_{OIB} = R_{F}*I_{B}.

4. Find the input bias current for the circuit given below

A. 10mA
B. 2mA
C. 5mA
D. None of the mentioned

Answer: C

Input bias current

I_{B}=(I_{B1}+ I_{B2})/2

=> I_{B} =(4mA+6mA/2 = 5mA.

5. Mention a step to reduce the output offset voltage caused due to input bias current?

A. Use small feedback resistor and resistance at the input terminal
B. Use small feedback resistors
C. Reduce the value of load resistors
D. None of the mentioned

Answer: B

Since the output offset voltage is proportional to the feedback resistor and input bias current. The amount of V_{OIB} can be reduced by reducing the value of the feedback resistor.

6. Given below is a differential amplifier in which V_{1}=V_{2}. What happens to V_{OIB} at this condition?

A. V_{OIB}= 0
B. V_{OIB}= V_{OIB}×10^{-10}
C. V_{OIB}= V_{OIB}/2
D. V_{OIB}= -1

Answer: A

The voltage V_{1 }and V_{2 }are caused by the current I_{B1} and I_{B2}. Although this bias current is very small, if they are made equal, then there will be no output voltage V_{OIB}.

7. Name the resistor that is connected in the non-inverting terminal of the op-amp which is in a parallel combination of the resistor connected in inverting terminal and feedback resistor.

A. Random minimizing resistor
B. Offset minimizing resistor
C. Offset reducing resistors
D. Output minimizing resistors

Answer: B

The voltage is a product of resistors and input bias current. Therefore, the value of the resistors is adjusted such that the resistors are connected at the inverting input terminal and are made equal to the resistor connected in the non-inverting input terminal.

The use of these resistors minimizes the amount of output offset voltage and therefore, they are referred to as offset minimizing resistors.

8. Calculate R_{OM}, if the value of I_{B1} = I_{B2} in the given circuit.

A. 1173.11Ω
B. 171.31Ω
C. 1171.43Ω
D. 1071.43Ω

Answer: D

Offset minimizing resistor

R_{OM} =(R_{1}* R_{F})/( R_{1}+R_{F}).

=> R_{OM} = (1.2kΩ*10kΩ)/(1.2kΩ+10Ω) = 1071.43Ω.

9. Calculate the output voltage for the given circuit using the specification: R_{1} = 820Ω; R_{OM}=811.882Ω; V_{in}=10mVpp; V_{OIB}≅0.

A. 1.025Vpp
B. 1.8Vpp
C. 1Vpp
D. 2Vpp

Answer: C

Offset minimizing resistor

R_{OM} = (R_{1}*R_{F})/(R_{1}+ R_{F})

=> R_{F} = (R_{OM}* R_{1})/( R_{1}– R_{OM})

= (812Ω*811.882Ω)/(820Ω-811.882Ω) = 82kΩ.

∴ V_{o} = -(R_{F}/ R_{1})* V_{in}

= -(82kΩ/820Ω)*10mVpp = 1Vpp.

10. Analyse the given circuit and determine the correct option

A. V_{oo} ≥ V_{IOB}
B. V_{oo} = V_{IOB}
C. V_{oo} >> V_{IOB}
D. V_{oo}IOB

Answer: C

741op-amp has V_{io} = 6mvdc and I_{B} =500nA.

The output offset voltage due to input offset voltage is given as

V_{oo} =[1+(R_{F}/R_{1})]*V_{io}

= [1+(4.7kΩ/47Ω)]*6mv = 0.606v.

The output offset voltage due to input bias current is given as