Electronics - Analog and Digital Converters - Discussion

Here Vref=+5 and the input is 0101(means 4 bit DAC).
In a 4-bit binary system, values should be 0000 to 1111 means 0 to 15 (or F).
As the R-2R ladder is connecting to inverting terminal of the Op-amp the output analog value will be ranging from 0V to -5V.

If the input is 0000 the output = 5 (0)/15 = 0 V.
If the input is 1111 the output = 5 (15)/15 = -5 V.

Here we have 0101 means 5.

So, output = -5 * 5/15 = -1.66 V.

The correct equation for 4 bit DAC should be,

Vo = -R(b3 * 2^3 + b2 * 2^2 + b1 * 2^1 + b0 * 2^0).
= -R (b3 * 8+b2 * 4+b1 * 2+b0 * 1)
Here R = Resolution = Vref/(2^n) - 1.
Correct me if I am wrong.

Analog output for the input code 0101.

Formula: Va = Vref / 2^n * input code bit (Start from MSB).
Given : Vref = 5V; Input code bit = 0101.

Va = (Vref / 2^0 * 0) + (Vref / 2^1) * 1) + (Vref / 2^2 * 0) + (Vref / 2^3 * 1).
Va = (5 / 1 * 0) + (5 / 2 * 1) + (5 / 2^2 * 0) + (5 / 2^3 * 1).
Va = (5 / 1 * 0) + (5 / 2 * 1) + (5 / 4 * 0) + (5 / 8 * 1).
Va = (0) + (2.5) + (0) + (0.625).
Va = 3.125 V.

Formula = [Vref/(2^n)] {msb bit * 2^0 + next bit * 2^1 + next bit * 2^2 + lsb bit * 2^3}.

Where n = no of bit,

Solution:

Vref = 5v;
n = 4;

=> (5/(2^4)) {0*0 + 1*2 + 0*4 + 1*8}

=> (5/16) {0 + 2 + 0 + 8}

=> (0.3125) {10}

=> 3.125.

Answer => 3.125.

My answer is option C.

Explanation: The main advantage of these converters is that is possible to transmit frequency even in a noisy environment or in an isolated form.

The dual slope is preferred over charge balancing ADC.

Yes,

Actual answer is 1.5625 volts.

Given in the sum is reference voltage = 5v, and resolution = 4.

So voltage per lsb = vref/2^n = 5/2^4 = 0.3125 v/lsb.

Therefore = 0101 in decimal is 5 = 5*0.3125 = 1.5625.

Formula= Vref/2^n(msb bit*2^3+nextbit*2^2+nextbit*2^1+lsb bit*2^0)
where, n= no. of bits

output= (5/2^4)(0*8+1*4+0*2+1*1)
= (5/16)(5)
= 1.5625 V

This should be the correct answer for any type of DACs.

@Nirja.

Formula = (Vref/2^N)(dec equivalent of i/p).

N- is number of significant bits (here 3 as msb is 0).

Here 0101 = 101 hence consider 3 bit.

Therefore,
Vo = 5(Vi/8) = 3.12.

Formula= Vref/2^n(msb bit*2^0+nextbit*2^1+nextbit*2^2+lsb bit*2^3)
where, n= no. of bits

output= (5/2^4)(0*1+1*2+0*4+1*8)
= (5/16)(10)
= 3.125 V

Vout = Resolution * Decimal equivalent of given No.
Where Resolution = Vref/(2^n)-1
Resolution = 5/(2^4)-1 = 5/7
And dec No = 5

So answer = 25/7 = 3.125.

Formula= Vref/{(2^n)-1}(msb bit*2^0+nextbit*2^1+nextbit*2^2+lsb bit*2^3)
where, n= no. of bits

output= (5/2^4)(0*1+1*2+0*4+1*8)
= (5/15)(10)
= 3.125 V