Other than cost, is there a reason to not use higher rated components than called for by design?

(Apologies upfront if this question seems too vague, I'm trying to formulate the problem statement as I ask this)

I'm thinking about circuit assembly in a high component variability / low MOQ environment, from the standpoint of passive components. Assuming identical footprints, I'd like to be able to substitute:

Resistors

• tighter % tolerance for looser


• higher wattage for lower

Capacitors

• Higher voltage for lower

What I'm getting at is - if I'm prototyping lots of different designs, parts inventory management becomes a substantial issue. If I could aggregate by component value, and use the same parts where possible, inventory requirements go down, likely enough to offset the potentially higher cost of the individual items.

Does an approach like this work, or would I break designs by doing this? If this works, what other components can I try to standardize?

Answer

Yes, this is valid and commonly done for one-offs produced from local "lab" stock. There is cost in maintain more parts, which can easily dwarf the savings of using 5% resistors instead of 1% resistors in those cases where 5% is good enough.

There are also costs in production for each different part used. Even at high volumes, the pick and place machine has to be set up separately, different reels need to be bought, kept somewhere, etc. Unless you have a very high volume product, it makes no sense to use a 10 kΩ 5% resistor in one place when you need to use 1% tolerance of otherwise the same resistor elsewhere on the same board.

In other cases you have to be careful that the better spec in one dimension doesn't cause tradeoffs you care about in another dimension. For example, you mention higher voltage capacitors substituted where lower voltage is required. That's OK electrically, but higher voltage caps will be physically larger most of the time. The same is true for higher wattage resistors. Electrically, a 2 W 100 Ω resistor is a superset of a ¼ W 100 Ω resistor, but the 2 W resistor will be significantly bigger, which may incur other costs.