Since the D/A converter has an internal output resistor, if a pull-down resister is added, resistance from both the internal and pull-down resistors will divide the output level. Accordingly, D/A conversion precision will be worse than if there were no pull-down resistor.
The D/A converter can be thought of as a type of power source. If R represents the impedance of the externally connected load, conversion occurs in an equivalent circuit such as shown in the figure below. In this example, the D/A converter includes output resister R0, so the D/A converter has a power source of output voltage V from internal resistor R0. According to the figure, D/A converter output current I becomes
I = V/(R + R0),
and the voltage to both ends of resistor R becomes
Vr = IR = VR/(R + R0).
It can be seen here that when output current I flows to the D/A converter, output voltage will be lower than the actual voltage V output from the D/A converter due to the voltage drop caused by internal resistor R0.
For example, to output 3V as the D/A output, if the op-amp input impedance is infinity and the internal output resistor R0 is 6kΩ, the DAi pin level will be as follows.
3[V] X 100k/(100k + 6k) = 2.830[V].
Note that the output current of D/A converter in the above example is 20 to 30μA, a level that will not result in MCU malfunction or program runaway.
Figure. When the D/A converter output pin is pulled down via a 100kΩ resistor