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Adding a 100kΩ resistor for pull-down to stabilize power supply?

Latest Updated:10/13/2010

Question:

I set the D/A output to be input to the op-amp. In order to stabilize power supply, I'm considering adding a 100kΩ resistor for pull-down. Would this be okay?

Answer:

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

Suitable Products
R8C/2A
R8C/2B
R8C/2C
R8C/2D
R8C/2E
R8C/2F
R8C/33C
R8C/34C
R8C/35C
R8C/36C
R8C/38C
R8C/3GC
R8C/3JC
R8C/33M
R8C/34M
R8C/35M
R8C/33G, R8C/33H
R8C/38A
R8C/L35C
R8C/L36C
R8C/L38C
R8C/L3AC
R8C/L35M
R8C/L36M
R8C/L38M
R8C/L3AM