Operational Amplifier Fundamental MCQ [Free PDF] – Objective Question Answer for Operational Amplifier Fundamental Quiz

The input offset current from the circuit is I_io = |I_BA– I_BC|.

For a 741 op-amp

I_io = 200nA(Maximum).

V_OIio = R_F*I_io

= 27kΩ*200nA =5.4mv.

The gain of the inverting amplifier

A=-(R_F/R₁).

Therefore, the total output offset voltage is given as

V_ooT = -[(R_F/R₁)*V_io] + R_F*I_io.

The total output offset voltage due to input offset voltage,

V_ooT = -[(R_F/R₁)*V_io] + R_F*I_io

V_ooT = [1+(47kΩ/470Ω)]*7.5mv+(47kΩ*50nA.

= 0.7598 ≅ 0.76v.

The factor that affects the input offset voltage, bias current, and input offset current in an op-amp are

1. Change in temperature
2. Change in supply voltage
3. Change in time

Any change in the mentioned parameters affects the values of input offset voltage, bias current and input offset current from remaining constant.

The average rate of change of input offset voltage per unit change in temperature is called thermal voltage drift, i.e.

△V_io/△T.

When an amplifier is nulled at room temperature, the effect of input offset voltage and current is reduced to zero.

Change in the total output offset voltage occurs only if there is any change in the value of V_io and I_io. Therefore, the total output offset voltage will be zero at room temperature i.e at 25°C.

As the amplifier is used in inverting configuration, the effect of voltage and current drift is given as, the average change in total output offset voltage per unit change in temperature.

△V_ooT/△T = {[1+(R_F/R₁)]×(△V_io/△T)} + R_F×(△I_io/△t).

The maximum possible change in the total output offset voltage △V_ooT results from a change in temperature △t.

Therefore, the error voltage is obtained by multiplying △T by the average total output offset voltage.

E_v =( △V_ooT/△T)×△T = [1+(R_F/R₁)]×(△V_io/△T)×△T + R_F×(△I_io/△T)×△T.

The output voltage of inverting amplifier is

V_o= -(R_F/R₁)×V_in±E_v

=> E_v= ± V_o+(R_F/R₁)×V_in

= 3.99mv+(12kΩ/7.5kΩ)×1.33mv

= ±6.118 ≅ ±6mv.

Change in temperature △T = 50^oC-27^oC = 23^oC.

=> Error voltage

E_v =[1+(R_F/R₁)]×(△V_io/△T)×△T + R_F×(△I_io/△T)×△T

= [1+(100kΩ/1kΩ)] × (30µv/1^oC × 23^oC + 100kΩ × (300pA/1^oC × 23^oC = 0.06969 + 6.9×10^-9

=> E_v= 0.0704 = 70.4mv.

For an input voltage of 6.21mv dc, the output voltage,

V_o=-(R_F/R₁)×V_in±E_v

= -(100kΩ/1kΩ) × 6.21mv ± 70.4mv

= +0.69v or -0.55v.

A poorly regulated power supply gives different values depending on the size & type of load connected to it and a poorly filtered power supply has a ripple voltage riding on some specific dc level.

Even though the input bias currents change due to the change in supply voltages, the input offset current remain relatively constant because it is the absolute value of the difference between two input bias currents.

Supply voltage rejection ratio, SVRR in dB

= 20log(1/SVRR) =20log[1/(△V_io/△V)]

= 20log(1/150µV/V)

=20log(10⁶/150)

=20log(6666.67) = 76.48dB.

The total value of SVRR in µV/V should be zero. The lower the value of SVRR in µV/V, the better will be the op-amp performance.

The change in the input offset voltage is given as

△V_oo =[1+(R_F/R₁)]× [△V_io/△V] × △V

Where

△V = Change in supply voltage +V_cc & -V_ee,

△V_io/△V = Supply voltage rejection ratio (µV/V) and

[1+(R_F/R₁)] = Gain of the differential amplifier.

20log(1/SVRR) = 92dB

=> log(1/SVRR) = 92/20

=> 1/SVRR= 10^4.6

=> SVRR = 1/10^4.6 = 25.12µV/V.

△V value is the same regardless of whether it is computed from the change in low dc supply or change in +V_cc or -V_ee.

For example, suppose that -V_ee remains constant at -10v then the +V_cc has to vary from 8 to 12v as a result of a change in low dc voltage. This means that the change in △V in supply voltage +V_cc is 2v in either direction from 10v.

The SVRR = 15.85µV/V

for LM307 because of poor filtering and △V= 10mV_rms.

The change in output offset voltage

△V_oo =[1+(R_F/R₁)]× [△V_io/△V] ×△V

= [1+(950kΩ/1kΩ)] × (15.85µV/V) × 910mv = 8.1µV_rms.