# A 6-bit ADC has a maximum precision supply voltage of 20 V. What are the voltage changes for each LSB present and voltage to be presented by (100110),

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A 6-bit ADC has a maximum precision supply voltage of 20 V. What are the voltage changes for each LSB present and voltage to be presented by (100110), respectively?

1. 0.317 V and 12.06 V
2. 3.17 V and 12.06 V
3. 0.317 V and 1.206 V
4. 3.17 V and 1.206 V

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Correct Answer - Option 1 : 0.317 V and 12.06 V

Concept:

Resolution

The ADC resolution is defined as the smallest incremental voltage that can be recognized and thus causes a change in the digital output.

Resolution = Step size

NOTE:

If only LSB is present in the input sequence applied then the output voltage is defined as the step size.

V0 = Step Size.

Calculation:

Given supply voltage is 20 V and 6 bit ADC

Output voltage when only LSB is present is Resolution itself.

${V_0} = \frac{{20}}{{{2^6} - 1}}$

${V_0} = \frac{{20}}{{63}}$

NOTE: In the ESE prelims calculator is not allowed so we will do the approximate calculation.

Calculate 20/64 we get the result as

$\frac{{20}}{{64}} = \frac{5}{{16}} = 0.3125$

But denominator less than assumed so, the fraction value increase.

According to options it will be 0.317 V

Converting into Decimal

(100110)2 = ()10

= 1 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20

= 32 + 4 + 2

= (38)10

Now the output voltage is

0 = 0.317 × 40

= 12.68 V

But the value is < 40 so answer also < 12.68 V

According to options, the answer is 12.06 V