# Linear Integrated Circuit Questions and Answers – Input Bias Current

This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Input Bias Current”.

1. Input bias current is defined as
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

Explanation: Input bias current is the average of two input bias current 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

Explanation: 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) VOIB= RF*IB
b) VOIB= (RF+R1)/IB
c) VOIB= (1+RF)*IB
d) VOIB= [1+(RF/R1)]*IB

Explanation: The output offset voltage due to input bias current is VOIB = RF*IB.

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

a) 10mA
b) 2mA
c) 5mA
d) None of the mentioned

Explanation: Input bias current, IB=(IB1+ IB2)/2
=> IB =(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

Explanation: Since the output offset voltage is proportional to feedback resistor and input bias current. The amount of VOIB can be reduced by reducing the value of feedback resistor.

6. Given below is a differential amplifier in which V1=V2. What happens to VOIB at this condition?

a) VOIB= 0
b) VOIB= VOIB×10-10
c) VOIB= VOIB/2
d) VOIB= -1

Explanation: The voltage V1 and V2 are caused by the current IB1 and IB2. Although this bias current are very small, if they are made equal, then there will be no output voltage VOIB.

7. Name the resistor that is connected in the non-inverting terminal of op-amp which is in parallel combination of resistor connected in inverting terminal and feedback resistor.
a) Random minimizing resistor
b) Offset minimizing resistor
c) Offset reducing resistors
d) Output minimizing resistors

Explanation: The voltage is product of resistors and input bias current. Therefore, the value of the resistors are adjusted such that the resistors are connected at the inverting input terminal is made equal to resistor connected in non-inverting input terminal. The use of this resistors minimize the amount of output offset voltage and therefore, they are referred to as offset minimizing resistors.

8. Calculate ROM, if the value of IB1 = IB2 in the given circuit.

a) 1173.11Ω
b) 171.31Ω
c) 1171.43Ω
d) 1071.43Ω

Explanation: Offset minimizing resistor, ROM =(R1* RF)/( R1+RF).
=> ROM = (1.2kΩ*10kΩ)/(1.2kΩ+10Ω) = 1071.43Ω.

9. Calculate the output voltage for the given circuit using the specification: R1 = 820Ω; ROM=811.882Ω; Vin=10mVpp; VOIB≅0.

a) 1.025Vpp
b) 1.8Vpp
c) 1Vpp
d) 2Vpp

Explanation: Offset minimizing resistor, ROM = (R1*RF)/(R1+ RF)
=> RF = (ROM* R1)/( R1– ROM) = (812Ω*811.882Ω)/(820Ω-811.882Ω) = 82kΩ.
∴ Vo = -(RF/ R1)* Vin = -(82kΩ/820Ω)*10mVpp = 1Vpp.

10. Analyse the given circuit and determine the correct option

a) Voo ≥ VIOB
b) Voo = VIOB
c) Voo >> VIOB
d) Voo << VIOB

Explanation: 741op-amp has Vio = 6mvdc and IB =500nA.
The output offset voltage due to input offset voltage is given as Voo =[1+(RF/R1)]*Vio = [1+(4.7kΩ/47Ω)]*6mv = 0.606v.
The output offset voltage due to input bias current is given as VIOB = RF*IB =4.7kΩ*500nA = 2.35mv.
=>∴ Voo >> VIOB.

11. The specification for LM101A op-amp is given as IB =75nA. Determine the value of VIOB– V1.

a) 0.112v
b) 0.750v
c) 0.374v
d) 0.634v

Explanation: The voltage at non-inverting terminal is given as V1 = ROM*IB1 = 148Ω*7.5nA = 1.11µv.
=> ∵ ROM = (R1*RF)/(R1+ RF) = (15kΩ*150Ω)/(15kΩ+150Ω) =148Ω
The output offset voltage is given as VIOB = RF*IB
=> VIOB = 15kΩ*7.5nA = 112.5mv
=> ∴ VIOB– V1 = 0.112v.

Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.

To practice all areas of Linear Integrated Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.