For interfacing the digital circuit with the analog circuit, we must be able to convert the digital signal into an analog signal and analog signal into a digital signal.

If there is a digital output and we need analog output, we need to convert that digital signal into the analog signal. DAC (Digital to analog converter) takes digital inputs and generates the output which is equivalent to the analog signal.

There are many circuits to convert the digital signal into the analog signal. One of them is using the op-amp summing amplifier. Let us recall the Summing Amplifier.

The figure shown below is the connection for the summing amplifier.

Output of this circuit is given by:

$$ V_{out}=-\left\{ \frac {R_f}{R_0}V_0+\frac {R_f}{R_1}V_1+\frac {R_f}{R_2}V_2+\frac {R_f}{R_3}V_3 \right\} $$

In this circuit, if

$$R_f=R_0=R_1+R_2=R_3=R$$

Then output becomes

$$v_{out}=-(V_0+V_1+V_2+V_3)$$

It is the sum of input voltages.

**Please note that negative sign only indicates the inverting amplifier.**

We can use this circuit for converting the digital signal into the analog signal. Circuit diagram for DAC using op-amp is shown below:

In this circuit, V0, V1, V2 and V3 are for digital inputs.

We have replaced resistors of value given below :-

$$R_3 \rightarrow 1 K\Omega $$

$$R_2 \rightarrow 2 K\Omega $$

$$R_1 \rightarrow 4 K\Omega $$

$$R_0 \rightarrow 8 K\Omega $$

and also, feedback resistor

$$R_f \rightarrow 1 K\Omega $$

So that the output of the circuit becomes

$$V_{out}=-(V_3+\frac 12V_2+\frac14V_1+\frac18V_0)$$

(Negative sign only represents inverting apmlifier and we are not considering polarity).

Let us consider full-scale voltage of **binary 1 is 5V** and for **binary 0 is 0V**.

If binary input is “1001” then output voltage becomes:

$$V_{out}=-(5+0+0+0.625)V=5.625V$$

The output of this circuit for different digital inputs are given below:

A | B | C | D | Vout |
---|---|---|---|---|

0 | 0 | 0 | 0 | 0 |

0 | 0 | 0 | 1 | -0.625 |

0 | 0 | 1 | 0 | -1.250 |

0 | 0 | 1 | 1 | -1.875 |

0 | 1 | 0 | 0 | -2.500 |

0 | 1 | 0 | 1 | -3.125 |

0 | 1 | 1 | 0 | -3.750 |

0 | 1 | 1 | 1 | -4.375 |

1 | 0 | 0 | 0 | -5.000 |

1 | 0 | 0 | 1 | -5.625 |

1 | 0 | 1 | 0 | -6.250 |

1 | 0 | 1 | 1 | -6.875 |

1 | 1 | 0 | 0 | -7.500 |

1 | 1 | 0 | 1 | -8.125 |

1 | 1 | 1 | 0 | -8.750 |

1 | 1 | 1 | 1 | -9.375 |

Clearly, the output of this circuit is equal to the weighted sum of the digital inputs.

Full-scale value of this circuit is (-9.375V).

The output of this circuit depends on two factors, the first one is the value of the feedback resistor and the second is the precision of the input voltage.

Thank you Mr. Raj for posting this very helpful tutorial. I constructed this exact this circuit with the exact specifications and found that the output voltage responded as a logarithmic function of the input value, not linearly. Do you have any insight into this? Thank you. – Ben