A D And D A Converters Practice Questions

The input to digital circuuit is $X_2,X_1,X_0$ and output is $B_1,B_0$ $B_0=\bar{X_2}\bar{X_1}X_0+X_2X_1X_0$ $\;\;=X_0(\bar{X_2}\bar{X_1}+X_2X_1)$ $\;\;=X_0(\overline{X_2\oplus X_1})$

Temperature range of $-40^{\circ}C \; to \; 55^{\circ}C$ So. Total range in $95^{\circ}C$ Since, resolution is $0.1^{\circ}C$ So, number of steps will be 950 To have 950 steps, we need at least 10 bits.

The reference voltage is 5 V. The number of bits in ADC are 8. So, the resolution will be $=\frac{5}{2^{8-}1}=\frac{5}{255}$ The applied input is 3.5 V. The succesive approximation ADC start working from the MSB so, After one clock: SAR will toggle it's MSB from $0\rightarrow 1$ . So, output of SAR will be 1000 0000. After second clock: SAR will toggle its 7th bit from $0\rightarrow 1$ but 1100 0000 will result in value greater than 3.5. So, output of SAR after 2nd clock will be 1000 0000. After third clock: SAR will toggle it's 6th bit from $0\rightarrow 1$ and output will be 1010 0000.

Hex Number (AB.CD) $1010\;\;1011\;.\; 1100\;\;1101$ For finding its octal number, we can add one zero in both extreme and grouping. $010\;101\;011\;.\;110\;011\;010$ Its equavalent octal number is $(253.632)_8$

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Initial stage of the counter $=(111)_2=(7)_10$ So output will be equal to 7 V. Next state of counter $=(110)_2=(6)_10$ So output should be = 6 V But output is 3 V that meansLSB of counter is connected to MSB of DAC and MSB of counter is connected to LSB of DAC. Similarly next state oc counter $=(101)_2=(5)_10$ Input to DAC $=(101)_2=(5)_10$ So, output = 5V When counter goes to $(100)_2$ then input to DAC $=(001)_2=(1)_10$ So output = 1 V So connections are not proper.

Even when $V_1 \gt V_2$ , the (o/p) ' $V_0$ ' is high and for the next case ( $V_1 \lt V_2$ ) (o/p) is low.It is 1 bit quantizer. Since, it has two states which can be represented by 1 bit.

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Currently no explanation available