In the last log entry we discussed how to calculate the power rating of the programmable decade resistor. We assumed that the decade resistor is made of resistors with the same power rating – in reality this might not be the case. Also, it might be a good thing to derate the resistors a bit. So let’s see what the implications are.

## Resistor power rating and derating

In this particular implementation, the assumption of equal power ratings of all resistors is only true for the upper five decades-the first decade uses resistors with only *P _{00}* = 0.125 W instead of 0.4 W due to limited availability of higher rated ones. Now the equation looks like this:

That's why we'd have to check the first decade (*n* = 0) as well, whenever the second decade (*m* = 1) is the highest non-zero decade.

## Derating of the resistors

There is another aspect that might actually help us out: Derating. It's a good idea to derate the resistors. This not only limits the temperature and therefore improves (long- and shortterm-) stability, but also gives some additional headroom for an accidental overcurrent situation. (A warning: I hope that the comparatively wide traces help to cool the resistors, so that they don't suffer a little less. However, utilizing the maximum or even the mildly derated power dissipation capability will likely result in some resistance drift. If you want to minimize that effect, overspec the resistors like ... a lot.)

The initial requirement for the programmable resistor was a power rating of *P _{max,tot}* ≥ 0.5 W for all ranges. Since the power rating of a single decade is

*P*≥ 2

_{max}*P*(see table for a single decade in the previous log), we can derate the resistors from 0.4 W to 0.25 W (-37.5%) for all decades n ≥ 1 without compromising the goal.

_{0n}Not so great: The first decade has to use 0.125 W resistors (due to availability of precision resistors, at least at the time of building the unit), making it impossible to achieve anything close to the inital goal of 0.5 W, even without derating. So would it at least possible not to limit the maximum *current* even more than the second decade does (while still allowing for some derating)? Yes, with a derating to 0.1 W (-20%) we get:

So no special checks for the first decade required anymore. Summed up:

Decaden | Resistor power ratingP_{0n} / W | Derated resistor power rating P _{0n,d} / W | Decade power rating P _{max,n} / W |

0 | 0.125 | 0.1 | ≥ 0.2 |

1 .. 5 | 0.4 | 0.25 | ≥ 0.5 |

## Calculation of the limits

After all those considerations it's fairly simple to calculate the current, voltage and power limits:

- Determine the maximum allowable current: It's trivial to determine the highest decade
*h*that is not shorted. Therefore the maximum current Imax,tot can be determined easily (see table for single decade in the previous log). - Calculate the maximum allowable voltage:
*U*=_{max,tot}*R*⋅*I*_{max,tot} - Calculate the maximum allowable power dissipation:
*P*=_{max,tot}*R*⋅ (*I*)_{max,tot}^{2}

There might be other reasons to limit the current, voltage or power. A good example would be voltage limits imposed by the PCB's clearance/creepage distance between conductors of different potential.

## Next steps

In the next log we will dive into the topic of switching the relays.

## Discussions

## Become a Hackaday.io Member

Create an account to leave a comment. Already have an account? Log In.