How can I increase the power of the internal heater on the DL100 Distance sensors?

Knowledge article no. KA-02782 already describes how to protect the DL100 optics from fogging or icing.

If, in spite of the measures described there, fogging or even icing occurs in the device at extremely low ambient temperatures (for example, due to cold air flow), the solution may be to increase the heating power.

The heater built into the DL100-xxHxxxxx device variants is a flat heating element . The element is internally supplied directly by the supply voltage applied to the device and behaves like an ohmic resistor R (nominal 24 V, 15 W). The available heating power P can therefore be calculated using the following formula:

P = U x I = U² / R.

The formula shows that the heating power P has a quadratic relationship to the supply voltage. Therefore, an increase in the supply voltage has a significant effect on the heating power.

The following calculation shows the heating power at minimum and maximum specified supply voltage of the device (18 V and 30 V respectively) compared to the nominal supply voltage of 24 V.

P_{18 V} = (18 V)² / (24 V)² x P_{24 V}

P_{18 V} = 0.56 x P_{24 V}

P_{30 V} = (30 V)² / (24 V)² x P_{24 V}

P_{30 V} = 1.56 x P_{24 V}

The above results show that at 18 V the heating power is about 50 percent lower, at 30 V about 50 percent higher than at the nominal voltage of 24 V.

If you use a power supply unit with adjustable output voltage for the device supply, you may be able to increase the output voltage of the power supply unit and thus the heating power accordingly if necessary.

In this context, a possible voltage drop on the connecting cable of the DL100 must also be considered. It can have a noticeable effect on the input voltage at the device and thus on the heating power. This applies in particular to long cables and/or small cable cross-sections.