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I was working on designing a single-cell Li-Ion MPPT solar charger PCB. I am relatively new to PCB design and was struggling understanding why most PCBs, like the "Sunny Buddy" linked below, use resistors with really small power rating values (typically 1/4W) that seemingly don't correlate to the input voltage and current.

I wanted to test my PCB design with a breadboard before I finalized any component choices so I was wondering if anyone could explain why the resistors chosen have really small power ratings, because to my understanding based on the description of the Sunny Buddy with a 20V max input and a 450mA charge current that is around 9 watts of power going through the board which the resistors definitely would not support.

I feel as if the answer is right in front of me and very simple but it is just not clicking yet. Any clarification or help would be greatly appreciated. Thank you.

Sunny Buddy link : https://www.sparkfun.com/products/12885 Board

Board Schematic

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The only resistor in that circuit that is inline with the charge current is a 0.22 ohm resistor. Dissipation in a resistor is current squared times resistance, or about 55mW on average. Thus a 1/4W resistor is almost 5x larger than the maximum expected dissipation.

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    \$\begingroup\$ That resistor is for current sensing, and is needed to support some of the features of the LT3652 like programmable charge rates (constant current) and to terminate charging. \$\endgroup\$
    – SteveSh
    Jun 15, 2023 at 16:29
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Why does this circuit board use resistors with small power ratings that seemingly don't support its recommended input?

Because power is defined as Volts x Amps.

20 Volts and 0.5A sound like a lot (10 Watts if all those Volts and Amps could be used.) But what of that does the resistor actually "see?" To understand this, requires also considering its resistance value (Ohms.)

Let's assume there is 0.5A flowing through a resistor of 0.22Ω. How much power is that? The answer is in Ohm's Law:

$$W = \Omega \cdot A^2$$ $$W = 0.22 \cdot 0.5^2$$ $$W = 0.22 \cdot 0.25$$ $$W = 0.055$$

That is 55mW. This is such a low value due to the very low value of the resistor -- had it been a larger value like say 5Ω, it would have had significantly more power dissipation and needed to be physically larger.

When specifying resistors, it is always wise to power-rate them for double what is expected, so a 100mW or 0.1W resistor would be the smallest that could be put in there. No need to use a 1/4W (0.25W) or anything larger. Sometimes, if the board had a 0.22Ω 1/4W elsewhere, the same would have been used here, just to save on (unique) parts count.

P.S. When designing circuits, yes, the power-handling rating of all components should be determined. There are exceptions, but until those are known, it is better to verify and know that the design is robust.

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  • \$\begingroup\$ Of course! Thank you, that makes so much sense. I was so focused on current * voltage of the entire circuit I didn't consider the individual resistance to be substituted into the power formula. Thank you for the clarification. \$\endgroup\$
    – John G.
    Jun 15, 2023 at 18:11
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The job of the board is to pass as much power as possible from input to output.

In other words, the board itself must waste as little power as possible.

So even if the board passes 9W, there is very little power dissipated in the board so it does not require high wattage resistors.

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