



#129871  10/05/05 09:19 PM
Using OCPD at low voltages

Member

Registered: 06/16/04
Posts: 176
Loc: McKeesport, PA, USA


This may sound like a stupid question, but if you use a fuse or breaker that was rated for 250VAC on, say a 12.5VDC power supply, would the OCPD still be effective? For example, let's say you have a fuse with a resistance of 0.1Ω. That means at 250V 20A, there is 20 times as many watts than at 12.5V 20A, and therefore less heat to blow the element. Does it matter, or is it purely based on amperage? I see regular with the 250V rating protecting the secondary side of a class 2 transformer (not the wall wart type, usually the ones for higher end use... kinda like a laptop pwer supply.) is the 250V just a MAX rating, or a WORKING VOLTAGE rating? thanks, Josh.

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#129872  10/06/05 04:40 AM
Re: Using OCPD at low voltages

Member

Registered: 09/15/03
Posts: 649
Loc: boston, ma


Fuses blow based on the current flowing through them.
The heat produced in a resistive element is I^2R...the higher the current, the greater the heating.
Normally the load will limit the current flow, and the heating in the fuse is essentially constant for the same current flow. A 1 A fuse in a 12V circuit with 750mA flowing through it (9W dissipated in the load) will have essentially the same heating as a 1A fuse in a 240V circuit with 750mA flowing through it (180W dissipated in the load). In the event of an overload (say 1.5A flowing), the fuses will experience the same heating.
In the higher voltage system, more power is being delivered to the load, both during normal and overload conditions. The fuse will reach its trip point in the same way at both voltages.
I would expect a fuse to _open_ differently with different supply voltages. As the fuse opens, more and more of the supply voltage is dropping across the fuse, and less is being dropped across the load. As this happens, the heating in the fuse increases, and the supply voltage will determine the ultimate heating that is possible before the current flow actually stops. I expect that the details of the different operating characteristics are quite complex: high voltage would mean that the metal melts faster, but could mean that arcs get sustained for longer periods.
In any case, when a fuse is rated at a given voltage, it may be used at lower voltage, all other things being equal.
Of greater concern is the difference between DC and AC; a fuse that works just fine with 240V AC supply might not be able to function safely with a 60V DC supply. This is because once the metal melts you must also deal with quenching any arc that forms.
Jon

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#129873  10/06/05 07:54 AM
Re: Using OCPD at low voltages

Member

Registered: 06/16/04
Posts: 176
Loc: McKeesport, PA, USA


Thanks Jon, You're always helping me out with my little hypothetical questions. So does that mean (in reference to heat being I^2R) if you have 2 "toaster element space heaters", 1 rated for 12 Volts, and one rated for 80 Billion volts, and both draw 5 amps at their respective voltages, they produce the same heat? This kinda confuses me as heaters are rated in Watts (consumed). Does the watt rating just make it look better since the number is higher? Thanks, Josh

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#129875  10/06/05 08:42 AM
Re: Using OCPD at low voltages

Member

Registered: 09/15/03
Posts: 649
Loc: boston, ma


If you have two heating elements, one rated for 12V, and the other rated for 120V, _both_ rated for 5A, then the resistance _must_ be different. power dissipated is P = I^2*R (square of current times resistance) power dissipated is also P = I*E (voltage times current) voltage across a resistor is E = I * R (voltage = current times resistance)
notice that you can substitute E = I * R into P = I * E to get P = I * I * R, or the first equation.
In your example, the 12V 5A resistive heating element must have a resistance of 2.4 ohms, and must dissipate 60 watts. The 8*10^10V 5A heating element must have a resistance of 1.6*10^10 ohms, and must dissipate 4*10^11W (400 nuclear plants worth of output *grin* )
The confusing bit here is that E the voltage in question is _not_ the total voltage of the system, but instead the voltage dropped across the element being considered. With the fuses mentioned above, the supply voltage is being dropped across the _load_; and very little voltage is being lost in the fuse itself.
If you have a 250V 10A rated fuse, you could connect that fuse in a 120V circuit with a 500W lamp. The resistance of the fuse might be 0.01 ohms, and the voltage drop across the fuse might be 0.04V, with a total power dissipation in the fuse of 0.17W. The voltage drop across the lamp is 119.96V, and nearly 500W is delivered. The resistance of the lamp would be about 29 ohms.
Take this same fuse and put in in a 12V circuit with a 50W lamp. The current flow is still 4.2A, and the voltage drop across the fuse is still 0.04V with a power of 0.17W. But now the voltage drop across the lamp is 11.96V, and the resistance of the lamp is 2.9 ohms.
Jon

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