### Discussion :: Analog and Digital Converters - General Questions (Q.No.2)

Vimal said: (Oct 15, 2010) | |

Here most significant bit '0' then it's value can be 5 (0/2+1/4+0/8+1/16). |

Askani said: (Nov 17, 2010) | |

Formula= Vref/2^n(msb bit*2^0+nextbit*2^1+nextbit*2^2+lsb bit*2^3) |

Nirja said: (Mar 11, 2011) | |

Vo = 5(Vi/16) = 5(5/16) = 1.5625 |

Sundar said: (Mar 12, 2011) | |

A 4-bit R/2R digital-to-analog.... Here what about R/2R...? I think 1.5625 x 2 = 3.125V |

Anuradha Awasthi said: (Mar 23, 2011) | |

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 |

Nilima Sinha said: (Sep 10, 2011) | |

Vo=Vref(0*2^0+1*2^-1+0*2^-2+1*2^-3) =5(0.5+.25) =3.125V |

Thanihai said: (Oct 8, 2011) | |

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. |

Kiran said: (Oct 28, 2011) | |

It can b solved as =ref. voltage * binary equivalent/ total no of steps. |

Ankur said: (Dec 6, 2011) | |

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 |

Pushpinder Kansal said: (Dec 30, 2011) | |

v=5(5/16) =1.5625 now 1.5625*2=3.125ans |

Manjeet said: (Mar 25, 2012) | |

Vo = 5(Vi/16) = 5(5/16) = 1.5625 4-bit R/2R digital-to-analog.... Here what about R/2R...? I think 1.5625 x 2 = 3.125V |

Sumit Shah said: (Jun 6, 2012) | |

Main equation is, Vo= Vref * (b0*2^(-1)+b1*2^(-2)+....) Vo= 5 * (1*2^(-1)+0*2^(-2)+1*2^(-3)+0*2^(-4)) Vo=3.125V |

Ashu said: (Jul 31, 2012) | |

Vo = 5(0/16+1/8+0/4+1/2). Vo = 3.125. |

Vidya said: (Dec 5, 2012) | |

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. |

Vaishnavi Murugan said: (Jul 5, 2013) | |

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. |

Gopal said: (Jul 10, 2013) | |

@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. |

Shah said: (Oct 10, 2013) | |

MOST EASY METHOD: V(0ut) = (Vi/8)*digital input. |

Nitu said: (Dec 20, 2014) | |

Vo = (Rf/R)(Vref)(Do/16+D1/8+D2/4+D3/2). |

Nisha said: (Feb 25, 2015) | |

In case of inverting, Vo = -1.5625 V. In case of non-inverting, Vo = 3.125V. So the given answer is correct. |

Sarika Rajesh Banda said: (Sep 3, 2015) | |

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. |

Iuuu said: (Jul 20, 2016) | |

Formula = Vref/2^n(msb bit * 2^0 + nextbit * 2^1 + nextbit * 2^2 + lsb bit * 2^3). |

Mahesh R said: (Jul 22, 2016) | |

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. |

Akshay said: (Dec 25, 2016) | |

@Vidya. R2r ladder has resolution =Vr/2^n (only). |

Kiran said: (Apr 3, 2017) | |

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. |

Subodhunu said: (Aug 5, 2017) | |

Va= Vref(1/2 + 0/4 + 1/8 + 0/16) (because 0101). = 3.125 v. |

Mohith said: (Dec 31, 2017) | |

Vref/8=5/8=0.625, Va/0.625=5=0101, Va=3.125. |

Ravi said: (Mar 23, 2018) | |

Thanks @Vidya. |

Ajit Patil said: (Nov 12, 2018) | |

Thanks @Vidya. |

Chandrashekhar Pantina said: (Jun 19, 2019) | |

5/16*5 = 1.5625. As R/2R so 1.5635*2 = 5.125V. |

Chandrashekhar Pantina said: (Jun 19, 2019) | |

5/16*(5) = 1.5625V. As R/2R network so, 1.5625*2 = 3.125V. |

Pradeep Sharma said: (Aug 10, 2019) | |

It should be 1.5 because it is resolution * decimal equivalent And it is 4 bit dac so 2^n =16. |

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