Converting AC amp ratings to DC amp ratings

I haven't found the answer yet, but I imagine it's easy and here somewhere.

If I have a simple on-off switch rated at 5 amps 125v AC, then how many amps will it be rated at 12v DC?

What's the formula?

jaygrant,

There is no formula, actually. Surprisingly, it is usually lower, because an AC voltage switches polarities, and thus if there is an arc-over when the contact is broken, it is quickly stopped from the reversing voltage, going through zero and into the opposite polarity.

So a switch rated at 5 amps, 125 volts AC, will probably be rated at something like 4 amps, 12 volts DC. Just a guess. It depends on the specific switch design. But, it certainly wouldn't be any better at 12 VDC. So, the best you could expect would be 5 amps at 12 VDC.

Thanks. From what I have found on the internet, I had read that the amp rating increased as the voltage decreased even from AC to DC, so I wanted to ask before taking what I found for granted.

1st: an amp is an amp is an amp. AC or DC isn't a factor.


That said, amps @ particular voltages corresponds to power. The ability of a switch to handle the power is related to amps @ voltages.

I'd say that your switch can handle 5 amps at voltages going up to 125v. Don't exceed 5 amps and you'll be fine.

kromeke said:

1st: an amp is an amp is an amp. AC or DC isn't a factor.

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Actually, it is. Whenever there is a current flowing and the circuit is opened, there is an arc of current across the newly opened gap until the gap becomes large enough to stop the arc. As was stated above, AC current is considered "self quenching" in that the return to zero voltage every half cycle tends to stop, or "quench" the arc over much larger gap distances. So that a switch design that will open a gap large enough to stop the arcing under AC current may not be adequate to stop an arc under DC current. That is why DC-rated switches tend to be larger and more robust than their AC equivalents. There is no way to "de-rate" a switch for DC, since so much depends on the design of the switch itself.

Timothybil said:

Actually, it is. Whenever there is a current flowing and the circuit is opened, there is an arc of current across the newly opened gap until the gap becomes large enough to stop the arc. As was stated above, AC current is considered "self quenching" in that the return to zero voltage every half cycle tends to stop, or "quench" the arc over much larger gap distances. So that a switch design that will open a gap large enough to stop the arcing under AC current may not be adequate to stop an arc under DC current. That is why DC-rated switches tend to be larger and more robust than their AC equivalents. There is no way to "de-rate" a switch for DC, since so much depends on the design of the switch itself.

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I get your point, but we are talking about 12v DC. I don't think that arc is a factor here.

Also, a lot of switches are de-rated at higher voltages, for example, a switch rated for 6A @ 125v might also be rated 3A @ 250v. The higher the voltages, the more likely arc becomes a factor.

I'll stand by my statement that an amp is an amp is an amp. The size of the conductors in the switch is most likely the limiting factor.

Edit: here is a link from a switch manufacturer that will shed a little more light on the subject:
http://www.carlingtech.com/products/switches/learn_more.asp?page=switches_amp-rating

In particular, look at the "DC rule of thumb"

Ok I read the "DC rule of thumb" and I read the whole article. I see what you are saying and I think I would agree. Of course the only hard part of that, is the other things I've read stating ideas that are different about DC. It's confusing to get different opinions, especially when the "opinions" supposedly come from "knowledgeable people". I see that happen a lot in electronics. I read a lot of times where people will argue about this circuit, or that and not agree on it, even thought they are knowledgeable. It sort of makes one like me feel that electricity is still the stuff of old world wizards and dragons.....

Sorry, not really trying to aggrivate, just voicing my confusion here. I have found many switches rated at one Amperage for AC and another amperage for DC and usually the DC amp rating is higher than the AC rating. My thought is that they are not considering the "arc" but only the amperage thru the contacts. It seems that way.

Suffice to say that if I just look for at least 20-30% higher amp rating than I need, I should be OK.

Which means I'm looking for a 20amp rotary 3 position switch that will fit in a Mag D body.... Simple......

jaygrant said:

Which means I'm looking for a 20amp rotary 3 position switch that will fit in a Mag D body.... Simple......

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Alternatively, you may want to consider a switch with a lower amp rating and FETs, which is explained for the mag's stock switch here. (explained for one position, you would need 3 of the explained setups.

No aggravations here. You have to understand that there is no hard and fast rule with a lot of these things because there are multiple ways of arriving at a particular solution.

When I say an amp is an amp is an amp, I mean that in the general sense. There are no AC amps or DC amps, just amps.

Switch ratings have to take a lot of variables into consideration. Switch lifetime is one. If you start to exceed the ratings, you might have no problems in general, but the switch lifetime will likely be reduced. How the switch fails may be important and a design consideration. My rule of thumb basically matches the one on the link I sent you (if it is less than 30v DC then the rating at 125v AC will work with no problems).

If you start working with high voltage DC then things my change a little. "High voltage" depends on who you're talking to, for me, it is anything above 50v. For a lineman, it may be much higher.

kromeke,

Saying that "an amp is an amp is an amp" is IN GENERAL, not correct.

Amperage ratings for AC circuits can be

1. root mean squared: take the square of the instantaneous current, then take the time average of this variable (over one cycle), then take the square root of that time average. This is what most AC "amps" are. They are RMS values. Using RMS values is useful because when you go to things like power or I^2R losses and what not, it just algebraically substitutes.

But, they could also be

2. Peak Amps at the highest points in the cycle.

or

3. Complex phaser values to represent the sinusoidal wave with all the information present. You would use a phaser, which can be represented as a complex number, for current and voltage, when doing a full calculation of a network. And you even get into complex impedances for capacitance and inductance, for example.

So, saying that "there are no AC amps or DC amps, just amps" is incorrect, even if within certain applications it may be applicable.

Say this to a microwave cavity engineer and see what sort of response you get, for example.

Sure, but this is candlepower forums, not Microwave cavity engineering forums.

You're starting to get into Relativity vs. Newtonian physics. Newtonian works fine for most things. (analogy, not literally)

I'm not an EE, nor do I claim to be. I've done technician work and know the business end of an oscilloscope when I see it. I'm pretty well schooled on the basics, I get a little out of my comfort zone when we get into more complicated things.

In the scope of these forums, I think I can safely say an amp is an amp is an amp.

Edit: and when it comes to motors and such, you have to worry about starting current. A hotwire will also have a start current.

kromeke,

Point taken.

But, the thing is that when it gets down to it, switches are "relativistic" (to use your analogy) and not "newtonian", so there is no "newtonian" rule-of-thumb that you can rely on. The switching situation is inherently NOT steady state. Thus, an amp is not an amp is not an amp.

If you want to be precise.

js said:

You would use a phaser

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Set on stun, perhaps?

Ooops.

I should have been spelled it "phasor", as in complex notation, amplitude and phase (or real and imaginary rectangular).

hehe. Now that's funny.

js said:

jaygrant,

There is no formula, actually. Surprisingly, it is usually lower, because an AC voltage switches polarities, and thus if there is an arc-over when the contact is broken, it is quickly stopped from the reversing voltage, going through zero and into the opposite polarity.

So a switch rated at 5 amps, 125 volts AC, will probably be rated at something like 4 amps, 12 volts DC. Just a guess. It depends on the specific switch design. But, it certainly wouldn't be any better at 12 VDC. So, the best you could expect would be 5 amps at 12 VDC.

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JS is entirely correct, there is no formula, it depends on the switch design, such as the gap (assuming air gap), speed of opening, etc.

(Also if the circuit has inductance upon opening but that is a different discussion.)

Do flashlights use scrubbers?

kromeke said:

I'm not an EE, nor do I claim to be. I've done technician work and know the business end of an oscilloscope when I see it. I'm pretty well schooled on the basics, I get a little out of my comfort zone when we get into more complicated things.

In the scope of these forums, I think I can safely say an amp is an amp is an amp.

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Power consumption in a simple resistive load, will be the same from 10 AC amps, as 10 DC amps. Same is true for AC voltage and DC voltage dropped across a simple resistor -- a filament bulb on 12VDC will draw similar power as it would from a a 12VAC source.

Now for a good counterexapmle that is well within the scope of CPF:

Try doing the same thing with an LED (nonlinear load) instead of a incan lamp (more or less a simple resistor), and the results will be very different.

There the RMS value of voltage or current are completely irrelevant, as an LED will tend to draw a huge current when the sinusoidal voltage is close to its peak value, then close to zero current when the value of the sine wave is less than the Vf of the LED, or negative.

Or, let me put it this way -- let's say you are driving some LED christmas lights directly from AC. You design your circuit so that over one full period, the root mean square current, or AC current, is 20mA. I design a circuit where my LEDs are driven by a constant DC current of 20mA. If you compare the two circuits, The AC circuit (with the same number of amps) will be overdriving the LEDs mercilessly for about ~20% of the cycle, and they will suffer from a highly annoying 60Hz flicker. The DC setup will be running within spec, and will substantially brighter, more power efficient, and longer-living.

jaygrant said:

Which means I'm looking for a 20amp rotary 3 position switch that will fit in a Mag D body.... Simple......

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Do you mind ellaborating on your proposed setup, eg. what are you trying to do -- incan light? LED? Soemthing with multiple levels? If I know this I might be able to offer some specific suggestions. A light which involves multiple levels, and also 20 amps (of any kind) sounds very interesting...

LukeA said:

Alternatively, you may want to consider a switch with a lower amp rating and FETs, which is explained for the mag's stock switch here. (explained for one position, you would need 3 of the explained setups.

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Good idea on the MOSFET. However, rather than using three mosfets, a better soltion if the idea is just to get different output levels, is to simple use a rotary switch or even a continuously variable potentiometer, to adjust the biasing of a MOSFET -- and make it more or less resistive. Note that this will require heatsinking the FET to the body of the light as the transistor will dissipate significant power if used this way.

A "lossless" way to dim an incan, would be to actually use a MOSFET as a switch, and control it using a PWM driver (in a D-cell maglite you could probably fit a homebrew circuit based on a 555 timer) -- which will basically switch the lamp off and on very rapidly. This will effectively reduce the RMS voltage delivered to the bulb, without wasting power.