250+ TOP MCQs on Basic DAC Techniques – 2 & Answers

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Advanced Linear Integrated Circuit on “Basic DAC Techniques – 2”.

1. For the given circuit find the output voltage?
A. -5.625v
B. -3.50v
C. -4.375v
D. -3.125v

Answer: D
Clarification: The given circuit is a 4-bit R-2R DAC.
So, the output voltage Vo=-RF [(b3/2R) +(b2/4R) +(b1/8R)+(b0/16R)]
From the circuit, b3=5v, b2=0v, b1=5v, bo=0v.
Vo= -20kΩ[(5/2×20kΩ)+0+(5/8×20kΩ)+0]= -3.125v.

2. Which type of switches are not preferable for a simple weighted resistor DAC?
A. Bipolar Transistor
B. Voltage switches
C. MOSFET
D. All of the mentioned

Answer: A
Clarification: Bipolar transistor does not perform well as voltage switches and MOSFET, due to the inherent offset voltage when in saturation.

3. The inverted R-2R ladder can also be operated in
A. Inverted mode
B. Current Mode
C. Voltage mode
D. Non inverted mode

Answer: B
Clarification: The inverted mode R-2R ladder circuit works on the principle of summing current. Therefore, it is said to operate in current mode.

4. Which of among the following circuit is considered to be linear?
A. Weighted Resistor type DAC
B. R-2R ladder type DAC
C. Inverter R-2R ladder DAC
D. All of the mentioned

Answer: C
Clarification: Current remain constant in each branch of ladder in inverted R-2R ladder DAC. So, the constant current implies constant voltage. The ladder mode voltage remain constant even when the input data changes. So, inverter R-2R ladder DAC is considered to be linear.

5. Multiplying DAC uses
A. Varying reference voltage
B. Varying input voltage
C. Constant reference voltage
D. Constant input voltage

Answer: A
Clarification: A digital to analog converter which uses a varying reference voltage is called a multiplying D-A converter.

6. Calculate the value of LSB and MSB of a 12-bit DAC for 10v?
A. LSB =7.8mv, MSB =5v
B. LSB =9.3mv, MSB =5v
C. LSB =14.3mv, MSB =5v
D. LSB =2.4mv, MSB =5v

Answer: D
Clarification: LSB=1/2n=1/212=1/4096.
For 10v range, LSB =10v/4096=2.4mv and MSB = (1/2)×full scale =(1/2)×10v =5v.

7. A multiplying DAC is given a reference voltage VR = Vom cos2πft. Determine the output voltage?
A. Vo(t) = Vom sin(2πft+180o)
B. Vo(t) = Vom cos(2πft+180o)
C. Vo(t) = Vom tan(2πft+180o)
D. Vo(t) = Vom sec(2πft+180o)

Answer: B
Clarification: Vo(t) = Vom cos2πft+180o. The 180ophase shift is added, since the VR is connected to inverting input terminal and Vom = 0v to (1-2-n)×Vim depends on the input code.

8. Multiplying digital to analog converters are used in
A. All of the mentioned
B. Digitally programmable filter
C. Digitally programmable oscillator
D. Digitally controlled audio attenuator

Answer: D
Clarification: In multiplying DAC, the output voltage is a fraction of the voltage representing the input digital code and the attenuator setting can be controlled by digital logic.

9. A 10-bit D/A converter have an output range from 0-9v. Calculate the output voltage produced when the input binary number is 1110001010.
A. ±7.96v
B. -7.96v
C. 7.96v
D. None of the mentioned

Answer: C
Clarification: Vo=9v[(1×1/2)+ (1×1/22) +(1×1/23)+(0×1/24)+(0×1/25)+ (0×1/26)+(1×1/27)+(0×1/28)+(1×1/29)+( 0×1/210)].
=9v×(0.5+0.25+0+0.125+7.8125×10-3+1.95 ×10-3) =9v×0.8547 =7.96v.

10. The basic step of a 8-bit DAC is 12.4mv.If the binary input 00000000 represents 0v. Determine the output, if the input is 101101111?
A. 1.36v
B. 2.27v
C. 5.45v
D. None of the mentioned

Answer: B
Clarification: The output voltage for input 10110111 = 12.4mv ×[(1×27)+(0×26) + (1×25) + (1×24) + (0×23)+ (1×22) +(1×21)+ (1×20)] = 12.4 × (128 + 32 + 16 + 4 + 2 + 1)
=12.4mv × 183 = 2.27v.