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Requirement: During an external test 20W of power will be drawn from my supply circuitry for 1 minute. My mission is to implement a solution that limits the power draw to a level below 20W in under one minute.

Background to the requirement: it is somehow considered that 20W is a power at which enough heat could be generated that a danger of fire is at hand.

My system: I am designing a system containing a MCU and a couple of things to be measured and controlled, among them a DC motor. I have a safety requirement which says that when 20W is drawn from my supply circuit the power drawn from it must be limited (permanently by fuse or temporary by another measure) in order to prevent heating and hinder unintended short-circuits. My power supply is a battery pack of 10,8V which during short circuit test gives 25W and 4A in a small resistor for longer than a minute, during this lapse the battery voltage sinks a couple of volts.

Now to the problem: Under normal circumstances I would use a usual fuse, but there is one big quirk, which is that when the motor is on, it draws around 2,5A from the supply (might have higher inrush as well) (however the voltage sinks to a level so that the effect in the motor is below 20W). This makes it hard to find a reliable solution with fuse in the unclear region between 2,5A and 4A.

I have this far looked on thermistors, active current limiters, fuses and polyfuses. But none of the options look optimal, especially because the solution should preferable be reliable (as this is a safety measure), and active devices are not preferred.

And advice on design of a current limiting solution is appreciated, the more elaborated, the better.

Also, what percent of motor's nominal current draw should the fuse be in general case? From what I have read at least 150% - am I correct?

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closed as unclear what you're asking by Enric Blanco, uint128_t, Voltage Spike, Trevor_G, Wesley Lee Apr 29 '17 at 17:40

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    \$\begingroup\$ "active devices are not preferred" that does not leave many options..... \$\endgroup\$ – Trevor_G Apr 28 '17 at 17:43
  • \$\begingroup\$ I have a feeling this is an XY problem. \$\endgroup\$ – Trevor_G Apr 28 '17 at 17:44
  • \$\begingroup\$ Try and be clearer, do you require a current limit that rises as the battery voltage falls i.e. based on limiting power or, do you need a current limiter? Also, the reaction time isn't that clear nor the actual current limit if it is really a current limiter and not a power limiter. \$\endgroup\$ – Andy aka Apr 28 '17 at 17:44
  • \$\begingroup\$ So you have a 10.8 volt battery, and a motor that draws 25 watts and a current of 4 amps, right? Since power is voltage times current, and 10.8 times 4 equals about 43 watts, do you see that there is a small problem? \$\endgroup\$ – WhatRoughBeast Apr 28 '17 at 17:54
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    \$\begingroup\$ Active devices could be considered. This is not a school problem. I need to limit the current in order to limit the power. 4A and 25W is durng short circuit test, during it supply voltage falls to a lower level. \$\endgroup\$ – NoobPointerException Apr 28 '17 at 18:33
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Your available power is pretty close to the allowed power.

Look up polyfuses, they may be suitable. A well selected polyfuse may be convenient because they are slow blow devices. They are also self resetting.

A dropping resistor might move the power from the load into your battery box as a quick hack, perhaps a carefully selected filament light bulb as an overload indicator that has negative resistance characteristics that comes to your rescue.

An in-line LDO linear regulator with a current limit to 2.5A would be the active device solution.

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    \$\begingroup\$ Could you please explain why polyfuse is a better solution than a usual fuse? (Except that it doen't get destoyed in the proces). Your other ideas are also valid. \$\endgroup\$ – NoobPointerException Apr 29 '17 at 14:37
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    \$\begingroup\$ Edited answer to expand benefits. \$\endgroup\$ – KalleMP Apr 29 '17 at 14:40

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