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I'm currently learning electronics and there's something I cannot seem to figure out regarding connectors.

Let's say I have a male + female connector which are rated up to 250 volts and 5 amperes. This should then be capable of handling 1250 watts as far as I know.

Now, when using lower voltages such as 12V, in order to reach 1250 watts, it should be around 104 amperes.

Let's say you make a circuit which uses 12V@10A, which is about double the current rating of the 220V@5A.

When using lower current like here, can you just use that connector because of the wattage conversion since it uses 120W instead of 1250W, or do I need to find a different connector which is rated for that amperage, even though the current is much lower?

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    \$\begingroup\$ Also see Using a 2 A, 48 VDC rated switch in a 8 A, 12 VDC circuit. \$\endgroup\$
    – ocrdu
    Dec 16, 2023 at 12:50
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    \$\begingroup\$ the connector does not have 250 V across it, so it is not dissipating 1250 W \$\endgroup\$
    – jsotola
    Dec 16, 2023 at 14:15
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    \$\begingroup\$ @jsotola The question states the connector would be "handling 1250 watts", not dissipating it. Does the number of upvotes suggest that people are not aware of the distinction? \$\endgroup\$
    – Graham Nye
    Dec 17, 2023 at 21:08
  • \$\begingroup\$ Isn't it a question of resistance and heat dissipation basically (ignoring the isolation when lowering the voltage)? I mean: Higher current will cause the resistance to heat up, and when it cannot dissipate the heat, it will melt (or become very hot at least). \$\endgroup\$
    – U. Windl
    Dec 17, 2023 at 22:27
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    \$\begingroup\$ "When using lower current like here, can you just use that connector because of the wattage conversion since it uses 120W instead of 1250W" - Do you think the connectors would be OK with 1250A at 1V? What about 1A at 1250V? \$\endgroup\$
    – marcelm
    Dec 18, 2023 at 14:25

5 Answers 5

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Let's say I have a male + female connector which are rated up to 250 volts and 5 amps. This should then be capable of handling 1250 watts as far as I know.

Now, when using lower voltages such as 12v, in order to reach 1250 watts, it should be around 104 amps.

Connectors do not have a constant power rating. The voltage and current limits apply independently of each other. So your connector can handle a maximum of 5 A at any voltage from 0 to 250 volts and it can handle a maximum of 250 V at any current from 0 to 5 amps. The only way it can handle 1250 W is to run it at both 250 V and 5 A. The current rating doesn't increase when you run it at lower voltages (and the voltage rating doesn't increase when you run it at lower currents).

I'm assuming here that the connectors stay mated, i.e. plugged together. The ratings for making (plugging together) or breaking (pulling apart) a connector while the load is active can be smaller. To find out what that is you'll need to look at the data sheet for your specific connector, if it's even allowed at all.

Let's say you make a circuit which uses 12v@10A, which is about double the current rating of the 220v@5A. ... Do I need to find a different connector.

Yes. You need to use a connector rated for at least 10 A at least 12 V. You can't use a 5 A connector.

Also, for safety, if a connector has an expected use you shouldn't use it for something else even if that's within the connector rating. For example, an AC grid/mains connector shouldn't be used for something else because at some point somebody, possibly yourself, will plug it into the AC supply and cause injury or damage.

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    \$\begingroup\$ Also note that connectors can have different carry and disconnect ratings; a connector that can carry 5 A safely might not be safe to unplug while it's carrying 5 A. Disconnect ratings usually differ between AC and DC usage, for the same reason switches and relays have different AC and DC ratings. \$\endgroup\$
    – Hearth
    Dec 16, 2023 at 13:53
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    \$\begingroup\$ Also also note that some connectors will have different single circuit ratings. I.e. while any one contact can handle up to, say, 5 amps, you cannot carry 5 amps on all contacts simultaneously. \$\endgroup\$
    – vir
    Dec 16, 2023 at 17:21
  • \$\begingroup\$ Another note: Some connectors have different ratings in the mated and unmated conditions. MHV is one such; it's rated for (if I remember right) 5 kV in the mated condition, but when unmated it's only suitable for much lower voltages. In MHV's case, this is mostly because it's possible to touch the center contact of an unmated connector, which would be dangerous if it had high voltage on it. (This is why MHV is considered obsolete and has been replaced with SHV, which is suitable for 5 kV in both conditions.) \$\endgroup\$
    – Hearth
    Dec 17, 2023 at 18:18
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    \$\begingroup\$ Yet another note: multi-contact connectors will often have a per-contact current rating. It is usually acceptable to use multiple contacts within such a connector for a single signal in order to handle larger currents. As this answer is basically saying, you need to look at all of the individual specifications, and the overall specifications, for the connector and be sure to comply with all of them. \$\endgroup\$
    – Makyen
    Dec 17, 2023 at 19:42
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The voltage and current ratings apply separately for connectors. The amperage is a rating for individual poles while the voltage rating applies between poles.

Such connector could fail at 300 V between poles and no current or at 10 A even with little voltage.

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The two ratings given for a connector apply to two different properties of the connector.

The voltage refers to the highest safe voltage between the conductors in the connector. As the voltage increases above this rated voltage, the risk increases that the insulation will breakdown and the current arcs from one contact to the other.

The other rating is current, and has nothing to do with voltage rating. The current rating comes from that fact that all real world conductors have some resistance, even if it is small. Current flowing through the resistance of the connector will generate heat. The heat energy generated is I^2 * R. The connector can only dissipate so much heat energy, and will consequently warm up. The current rating is the highest current that the manufacturer has calculated to produce a reasonably small temperature rise in the connector. Running more current that that through the connector risks melting the conductors, insulation, or both.

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To complement the other answers:

You have omitted two important parameters.
Realising that this is the case is a useful part of the learning process.

  1. Switches have an AC and a DC rating.
    A switch may be rated at eg 30VDC, 2A and at 230 VAC, 10A.
    Using the switch at the AC rated current on DC at or even below the rated DC voltage may result in switch destruction. This may be simply a reduction in switching cycle lifetime due to increased DC arcing on opening.
    Or, As a bonus, it may result in single operation destruction due to a non-extinguishing arc across the contacts when trying to interrupt DC current.

Alternating voltage drops to zero twice per cycle, which reduces the tendency for an arc to maintain when the current is interrupted. It does not completely stop arc formation and interruption of large AC currents may require very special methods to prevent arc continuation.

  1. AC switches may be rated at different currents for inductive and non inductive loads. Sometimes the worst case (inductive) is assumed in the ratings and a single figure is given.

When current in an inductive circuit is interrupted by eg a switch it MUST still continue to flow. The polarity of the voltage across the inductor will instantaneously reverse and the voltage across the inductor will rise (due to the collapsing magnetic field) and will assume whatever value is required to maintain the current initially. For a perfect switch and perfect inductor the voltage will rise to infinity. For an imperfect switch an arc may form to allow the current to cross the open contacts. If the switch opens adequately and there is no other current path the energy will transfer to the stray capacitance and then energy will flow to and fro in an oscillatory manner. Coil and stray resistance will dissipate the energy in time.

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    \$\begingroup\$ I know "MUST" is a strong word, but I think you could go even stronger: when the switch attempts to interrupt current, current WILL continue to flow somewhere. Every circuit has at least one meaningfully imperfect element(*), and given a perfect switch and perfect inductor, but imperfect cabling, the voltage would reach a level that allows the current to be rerouted through the insulation. \$\endgroup\$
    – supercat
    Dec 17, 2023 at 21:48
  • \$\begingroup\$ () In real circuits, some elements will be more imperfect than others, and it's not uncommon for a circuit to have some elements have enough uncertainty in their behavior that the overall circuit behavior would be *consistent with all other elements behaving in ideal fashion. In such cases, modeling the other elements as ideal may be more useful than trying to model their behavior more precisely. \$\endgroup\$
    – supercat
    Dec 17, 2023 at 21:50
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Generally speaking, when it comes to voltage the key is insulation and when it comes to current the key is wire size. Simple example with 120V and above (typical US values):

  • Most typical internal building wire has insulation rated at 600V.
  • 14 AWG wire up to 15A
  • 12 AWG wire up to 20A
  • 10 AWG wire up to 30A

If you want to install a 3000W heater, you can do it two different ways:

  • 120V at 25A on a 30A circuit with 10 AWG wire
  • 240V at 12.5A on a 20A circuit with 12 AWG wire (can't put it on a 15A circuit because...electrical code reasons)

The power output is the same. But higher current at lower voltage requires thicker wire than lower current at higher voltage.

The same rules apply to connectors and anything else in a circuit.

The same utility power will be at a higher voltage on wires attached to poles outside and even higher for longer distance transmission - 10,000V or more. That allows for thinner wires, but at a cost. That cost is either much thicker insulation (you will see that inside substations or in underground wiring vaults) or no insulation at all but the wires are kept way up in the air (quick search: 80 - 200 feet up) and kept relatively far apart to avoid danger. A Megawatt of power at 10,000V is 100A - that's comparable to the current (but not the voltage) going into a single home. That same power at 240V would require impractically large wires.

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