ac-dcfusesmainspower supply
I am designing a system which will use two 12V AC/DC 60W modules (Meanwell IRM-60-12)
I would like to protect the system using a fuse. In the case of a parallel design like the one below, should I use one fuse for each supply or a global fuse at double the amp rating for both?
e.g (sorry for the bad drawing):
Left or right design?
Thanks!
Note the very very (lfe savingly) important aspect of HRC = "High rupture capacity" fuses, discussed at the end.
You have done a fairly good job of summarising both the reasons and the dilemmas involved.
Fast blow are used where possible, where the fuse can be sized such that typical faults will always cause it to blow but nuisance blowing is rare. Suh situations have little or no startup surges or large occasional current excursions.
Slow blow are used where large short term transients are known to occur and if sizing of the fuse to accommodate the transients will result in inadequate protectionm against typical faults.
Where neither fast or slow blow fuses offer adequate protection (transients are very high but faults may be relatively low compared to maximum usual) then a cicuit breaker can be used, whose characteristics can be mapped accurately to a desired time/current profile.
Fast blow is the "more ideal" where possible.
Circuit current is well defined within known limits,
Start up transients are not so large compared to typical current that allowing for them is going to cause problems.
Fault currents are liable to be much much larger normal operate current and much larger than expected transients.
Slow blow is a compromise that allows protection while accommodating expected transient behaviour.
Startup or other transients may occur which cause much higher than average currents but for short periods.
Sizing a fast-blow fuse to allow the transients would result in a fuse which may not provide protection during some expected fault conditions.
The ideal may be both a fast and slow blow fuse in series (very unusual and possibly also illegal for regulatory reasons) or a circuit breaker with a well defined current versus time "envelope".
Regulatory requirements often make it clear which sort of fuse must be used.
________________________-
HRC / High Rupture capacity.
In some situations fault conditions can develop which can result in fault currents vastly in excess of the normal operating current and so high that massive destruction to property or loss of life may occur. An excellent example is a multimeter intended or measuring AC mains voltages of 230 VAC or higher. A meter measuring nominal 230 VAC mains voltages may easily be exposed to over 330 VDC peak, and transients on the waveform may cause much higher voltages to occur. A domestic range/stove/oven may be supplied with two phases with phase to phase voltages of 400 VAC or approaching 600 VDC peak to peak.
In either case above, if these voltages break down circuitry in the meter, an arc may occur followed rapidly by carbonisation of components, PCB, nearby case etc and a relatively low resistance across mains short may occur. The mains may then be supplying a high energy load vastly in excess of what is expected or designed - at least kilowatts with ease and tens of kilowatts in some cases. The onset of arc formation and generation of heat can be so rapid as to cause an explosion of th equipment with debris being ejected violently and with electric shock hazard also increasing.
Standing in the gap to this happening is "the fuse".
Edit: Actually, the fuses in multimeters are used to protect the current measuring circuitry. The voltage measurement stuff is protected by MOVS and PTCs.
If the fuse is able to blow and stay functionally blown when such a fault occurs the meter etc 'just stops working". if the fuse holder arcs and the PCB carbonises or the fuse otherwise fails to interrupt current, then the above scenario can occur. And does.
People have died due to this scenario and will die in future
An answer is the use of an HRC fuse which is designed to "rupture" in suh a way that a damaging arc does not form and the circuit is cleanly broken.
HRC fuses are usually ceramic bodied, usually white.
Not all white or ceramic fuses are HRC.
Not all HRC fuses are white or ceramic.
Image below shows fuses said by makers to be HRC. note that one is glass bodied.
( From here.)
Many HRC fuse images and links here.
Test equipment intended for AC mains use will usually specify HRC fuses. DO NOT SUBSTITUTE inferior types.
I have only ever had one meter fail under high voltage high energy conditions.
That was on a 1000 VDC range with a 1200 ior so VDC transmitting power supply being measured.
Very impressive.
A good lesson.
Long long ago.
Cheap multimeters often have their high end ACV ranges marked "not for mains use" or similar. That's why.
If you use them on mains you usually won't die.
But if you do, you won't be able to say that you weren't warned.
Remember that before you can't !!!
Your question is more than clear enough.
I don't think there are 7A fuses on the market, closest will be 7.5A (used in cars), but I may be wrong.
If your load has nominal current of 7A - you should pick a bit higher fuse, 7A will blow up sooner or later at 7A current.
If inverter is "20A inverter" - this will be usually output current, but you may have to read some manual to make sure if this is input or output current.
Your supply current calculation looks wrong ok, but there is easier/more clear way.
Maximum input current if maximum output current is 20A:
Pout = 12V * 20A = 240W
Power including inverter loss (for 90% efficiency):
Pin = 240W/0.9 = ~266.67W
Input current:
Iin = 266.67W / 24V = ~11.11A
Typical inverter in real life at maximum power have efficiency worse than its maximum efficiency. If maximum efficiency is 90% - at maximum current you can expect 80% or less.
For your specyfic 7A load power demand is:
Pout = 12V * 7A = 84W
Power including inverter loss (for 90% efficiency):
Pin = 84W / 0.9 = ~93.33W
Input current at 24V inverter supply:
Iin = 93.33W/24V = ~3,89A
Best Answer
Two fuses, for two reasons.
First, I want to establish some arbitrary values. For the purposes of the below, I will assume the devices will draw 1 amp while idle, be connected to either one 10 amp fuse, or two 5 amp fuses, and become damaged at 7 amps.
Depending on how sensitive these devices are, limiting the current draw of the two devices may not be sufficient to prevent one from becoming damaged. Imagine that one or both devices are idle when one device suddenly malfunctions. A short circuit causes it to draw 8 amps of current, well beyond the 7 amp damage-point. With a single fuse, the total amp draw is 8 + 1 = 9 amps, so the fuse doesn't blow and the device is damaged. With two fuses, the five amp fuse on the malfunctioning device blows, preventing further damage. This isn't a contrived scenario, most devices are in standby most of the time, and, although some malfunctions can cause sudden power spikes, others can cause slow burns. I could answer the question more fully if I actually knew what the devices were.
Another reason is that, depending on the nature of the devices (From your question, maybe high power electronics?), I could imagine that one may internally malfunction and destroy the other. I really couldn't say whether this was a possibility without more information, but I would consider it. With a fuse on each device, each device has its own protection and you don't have to worry about hooking up another device to the circuit, removing one device from the circuit, or changing the nature of one device. You can rest assured that each is separately protected from a surge. It's just good practice.
If this device is a low-cost device, a low power device, or a temporary situation, it wouldn't be worth the bother. I would say that if it was worth the effort of asking the question, its worth a couple of bucks for a separate fuse.