This set of Tough Linear Integrated Circuit Questions and Answers focuses on “Voltage to Current Converter with Floating and Grounded Load – 2”.
1. Determine the full scale range for the input voltage if the resistance in series with meters are 1kΩ, 2kΩ, 47kΩ and full scale meter movement is 1mA in low voltage AC voltmeter?
a) 1.0 to 7.48 Vrms
b) 1.1 to 7.48 Vrms
c) 1.2 to 7.48 Vrms
d) 1.3 to 7.48 Vrms
Explanation: The minimum and maximum values of resistors are 1kΩ and 6.8kΩ.So, the range for the input voltages are
Vin(rms)|min = 1.1×R1/Io = (1.1×1kΩ)/1mA =1.1v.
Vin(rms)|max =1.1×R1/ Io =(1.1×6.8kΩ)/1mA = 7.48v.
Thus, the full scale input voltage ranges from 1.1 to 7.48 Vrms.
2. Determine the current through the diode, when the switch is in position 1, 2& 3. Assuming op-amps initially nulled.
a) Io (LED) =4.01mA; Io (Zener) =4.01mA; Io (rectifier) =8.33mA
b) Io (LED) =25mA; Io (Zener) =4.01mA; Io (rectifier) =4.01mA
c) Io (LED) =16.67mA; Io (Zener) =16.66mA; Io (rectifier) =4.01mA
d) Io (LED) =8.33mA; Io (Zener) =8.33mA; Io (rectifier) =8.33mA
Explanation: All the diodes are connected one after another in the feedback path. Therefore, current through the diode remains same. Io =Vin/R1 =1.5/180 =8.33mA.
3. A diode match finder circuit has input voltage of 2.6v and output voltage is 5.78v. Calculate the voltage drop across diode 1N4735
Explanation: The output voltage Vo =Vin+ VD.
∴ the voltage drop across 1N4735, VD =Vo – Vin = 5.78-2.6 =3.18v.
4. Find the voltage drop across the zener diode in the zener diode tester from the given specifications: IZk=1mA, VZ =6.2v, input voltage= 1.2v, output voltage =3.2v and resistance in series with meter =150Ω.
d) Cannot be determined
Explanation: Current through the zener Io=Vin/R1 =1.2v/150Ω =8mA.
Since, Io > IZk the voltage across the zero will be approximately equal to 6.2v. As the current is larger than the knee current (IZk) of the zener, it blocks VZ volts.
5. Which among the following is preferred to display device in digital application?
a) Matched zener diode
b) Matched LEDs
c) Matched rectifier diode
d) All of the mentioned
Explanation: Matched LEDs with equal brightness at a specific value of current are useful as indicators and display devices in digital applications.
6. The maximum current through the load in all application that uses voltage to current converter with floating is
Explanation: The maximum current through the load cannot exceed the short circuit current of the 741c op-amp which is 25mA.
7. For voltage to current converter with grounded load, establish a relation between the non-inverting input terminals and load current
a) V1 = [Vin+Vo-(IL×R)] /2
b) V1 = [Vin-Vo-(IL×R)] /2
c) V1 = [Vin+Vo-IL+R] /2
d) V1 = [Vin+Vo+(IL×R)] /2
Explanation: In the voltage to current converter circuit the relationship between the voltage v1 at the non-inverting input terminal and load is given as V1 = [Vin+Vo-(IL×R)] /2.
8. Find the gain of the voltage to current converter with grounded load?
Explanation: In voltage to current converter with grounded load all resistor must be equal in value.
∴ Gain = Vo/Vin = [1+(RF/R1)] = 1+R/R =1+1=2.
9. Find the output voltage and the load current for the circuit given below. Assume that the op-amp is initially nulled V1 =2.5v
a) IL=0.42mA, Vo =10v
b) IL=0.42mA, Vo =3.4v
c) IL=0.42mA, Vo =6.1v
d) IL=0.42mA, Vo =5v
Explanation: The load current IL =vin /R =5/12kΩ =0.42mA
vo =IL×R, can be obtained when Vo=2×V1 = 2×2.5 =5v.
Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.
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