# What happens if I run 1A through a 50mA rated switch for 5 seconds, one time only?

The picture is showing a typical, miniature tactile switch, rated for 50mA. This switch suits my application perfectely, the only problem is that it's only rated for 50mA.

My application contains 2xN batteries in series (N=1000mAh, 1.5V). What will happen if I run 1Amp through the switch for, let's say 5 seconds, one time only? The whole device is disposable, so the switch will only be used once.

Will the switch manage one time use with this current, or will it break down? Are there other switches of this dimension that are rated for 1amp? I haven't managed to find any.

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seandogue8 months ago

Point is, if you *expect it to fuse, you might be rewarded, or you more likely will not. It's hit or miss at 100x overcurrent, favoring miss. (5A = 5000mA => 5000/50 = 100).

Presumably, the idea is that the device is a one time, once on always on thing.

you could employ a latching circuit to route power through a "beefier" switch (ex: a fet) and still employ the low-v microswitch for actuation.

If your gizmo employs a "Computer" of any sort (arduino etc) you could employ a digital input for the sense, and a digital out to trigger a fet/opt-isolator, etc

karolina81 (author)  seandogue7 months ago

Thanks, but my thingy has only room for one type of switch, however I have found a quite small 1amp switch (posted in a comment further down).

Woodclaver8 months ago

I do have to ask if you understand that fifty miliamps is fifty one thousandths of an amp?

You are asking to put 1000 miliamps, one entire amp, (1000/1000ths). through a switch rated for 50/1000ths. That's a factor of twenty times over rating! One amp isn't exactly dangerous to you, but it could be deadly to your equipment. There have to be one amp switches out there.

karolina81 (author)  Woodclaver7 months ago

Hi and thanks for a good explaination. I understand that 50mA@12V is tiny but I had an (also tiny) idea that maybe I could gain a little current due to my 3V and very short lifetime. But I understand that it will not be a reliable solution.

However, I did find a 1amp switch, I have posted it in a comment below. Do you think that this setup might work?

Woodclaver8 months ago

Hi:

If you run extra current through any circuit you run the risk of burning up something.

But a switch is more suseptable susseptable Susceptable prone to damage because of the connecting and disconnecting parts allowing arcs to cross the gap and melting the contacts.

I once put a fifteen amp switch on an electric roto-tiller I was converting: (Story can be read at:)

http://www.backwoodshome.com/converting-a-gasoline...

The fifteen amp switch was to operate a fifteen amp motor running at 240 vac. I used a DPDT switch, which means that both hot wires were disconnected simultaneously.

Since I was operating at the limit of the capacity, the switch literally welded itself together, in the on position, when I had it in reverse to back out of a corner.

I had to run away from the run away tiller, so it eventually ran over its own power cord, wound it up on the tiller, and pulled the cord apart. It was quite scary.

I replaced the fifteen amp switch with a twenty amp, and it worked ok for many years, still in fact.

iceng8 months ago

Once a switch is closed it does not matter how much over_current flows through it or for how long.. Short of a fusing current or already high current damage..

What damage high current does cause on_opening is micro arc pitting of the contact points from smooth metal to jagged sharp spike like teeth..

On_closing metal contacts bounce multiple to dozens of times arcing at every momentary separation... Heavy current can arc heat the metal contact points and can micro weld a protruding spike locking the switch closed...

Additionally micro spikes increase the switch resistance which depending on the circuitry can be undesirable...

Here are some suppliers of switches ;

http://www.goldmine-elec-products.com/products.asp...

http://www.goldmine-elec-products.com/prodinfo.asp...

http://www.goldmine-elec-products.com/prodinfo.asp...

http://www.goldmine-elec-products.com/prodinfo.asp...

this rare price http://www.mpja.com/DPDT-Mini-Push-Button-Switch-P...

http://www.mpja.com/PCB-Mount-Push-Button-Switches...

good hunting.. -.-. . -. --.

karolina81 (author)  iceng8 months ago

That's interesting information. Probably better to size-up to a 1amp switch then.

Jack A Lopez8 months ago

I think the quantity to worry about is the actual, ohmic, resistance of the switch when closed, and the size of that resistance in comparison to the resistance of your heating element.

I mean a switch made out of carbon printed on rubber, like the kind found inside TV remotes, or computer keyboards, a switch like that is super cheap, and the actual current it switches is very small, like milliamperes or microamperes. Also for a switch like that, the resistance of the switch when closed can be quite high, like 10s or 100s of ohms, and the switch will still work well for the circuit it is in.

For example if the current being switched is 1 mA, and Ron is 100 ohm, that gives a voltage drop of I*R = 0.1 volts, which is pretty small compared to the supply voltage supply of the keyboard, e.g. 3 volts DC or 5 volts DC, so for the other circuits that actually do the voltage sensing, the 0.1 volts seen across the switch looks close enough to the ideal, zero, that the circuit thinks, (or feels?, senses?) the switch is working.

So what is the actual resistance of your switch when closed? If it is not zero, and physically it kind of has to be not absolutely zero, then some real amount of power, as heat, is being dissipated in this switch, and that is power you would rather have going to your heating element.

By the way, I don't know if there is anyway to infer Ron from the current rating, but to sort of take a wild guess at it, suppose there is some small, close-enough-to-zero voltage drop , Vlow, for which we call the switch closed.

If Vlow=0.1 V, then Ron = Vlow/Imax = (0.1V)/(50 mA) = 2.0 ohm?

But you really shouldn't trust that math, because Vlow is just a made-up guess.

It would be preferable, to get an actual spec for Ron, or if that's not possible, just buy a few, if they're cheap, and test them in your circuit. Put a voltmeter across that switch and see how low the voltage across it drops to when you stab the button.

Or maybe your criteria for success is simply whether or not heating element succeeds in lighting your fire. You know: Just hold down the button and see what happens.

karolina81 (author)  Jack A Lopez8 months ago

OK, so if I understand things right, you're basically saying that a small switch will heat up more and steal more power from my nichrome wire (which I don't want). So switching to a "larger" switch, with the correct rating will steal less power?

I have purchased some different switches already and will do some experiments and see!

8 months ago

What I'm saying is if you have a stack of resistors, all wired in series with a voltage source, the power dissipated by any particular resistor is directly proportional to the size of that resistor.

In other words the power goes proportionally where the resistance is.

An example with three resistors, {R1, R2, R3}, will make this clear.

Suppose R1=R2= 1 ohm, and R3 = 8 ohm, and these are connected in series with a voltage source Vs = 3.0 volts. To sort of tie these symbols to some real things, imagine that R3 is a heating element, and R2 is the resistance of a closed switch, and R1 is the internal resistance of a battery.

Next, I do some math to calculate where the power is going.

The current that flows through every resistor and the voltage source too (since they are all in series) is, I = Vs/(R1+R2+R3) = 3/(1+1+8)= 3/10 = 0.3 A

The voltage across each resistor is, V1=I*R1, V2=I*R2, V3=I*R3, respectively, and numerically this is,

V1 = 0.3*1=0.3 V, V2 = 0.3*1=0.3 V, V3 = 0.3*8=2.4 V

The power dissipated by each resistor is P1=(I^2)*R1, P2=(I^2)*R2, P3=(I^2)*R3, respectively, and

P1 = 0.09*1=0.09 W, P2 = 0.09*1=0.09 W, P3 = 0.09*8=0.72 W

Total power is the sum of those,

Ptotal = (I^2)*R1 + (I^2)*R2 + (I^2)*R3 = (I^2)*(R1+R2+R3)

Ptotal = 0.09*10 = 0.9 W

If you just want to know fractionally where the power is going, the (I^2) divide out,

P1/Ptotal = R1/(R1+R2+R3) = 1/(1+1+8) [10% of Ptotal goes to R1]
P2/Ptotal = R2/(R1+R2+R3) = 1/(1+1+8) [10% of Ptotal goes to R2]
P3/Ptotal = R3/(R1+R2+R3) = 8/(1+1+8) [80% of Ptotal goes to R3]

The moral of this story is that for a bunch of things wired in series, the power goes where the resistance is. For this reason you want the resistance of your heating element to be bigger than all the other little resistances in series with it, e.g. bigger than the internal resistance in the battery, bigger than the resistance of the closed switch, bigger than the resistance of the non-heating wires, etc.

karolina81 (author)  Jack A Lopez8 months ago

Very good explaination, thanks so much. You should become a teacher, if you're not one already.

that is the capacity of the battery. the question is that will you really be using 1 amp? even so, i think the switch will handle it for 5 seconds

OK, thanks. The batteries will be powering a nichrome wire and from experiments I know that the batteries run relatively flat in less than a minute. And after abt 5 seconds, the nichrome wire has done its job so it doesn't matter what happens to the switch.

8 months ago

+1

The value you've quoted is the batteries rated capacity. The real question here is what is the battery going to be powering?

Just because a battery is capable of outputting up to 1A does not mean that it will output that full current regardless (otherwise the battery would run-flat in under an hour).. it will only output the current that the connected devices draw.

For example, if you had a 1.5v LED with a current rating of 20mA, powered by a 1000mA 1.5v battery the LED would draw around 20mA from the battery, allowing it to run continuously for approximately 35 hours.

karolina81 (author) 8 months ago

I learned something new today, which is that tactile switches normally are momentary. For my application I will need a maintained function of the switch, which means that I will have to look into other pushbutton switches. I found a small latching flashlight switch that might work, and which is also rated for 1 amp.

But as I know very little about circuits, could anyone tell me if this setup might work? The 2 N batteries shall power the nichrome wire, with the 2 switches in a normally open state. When pressing both buttons the circuit closes and the nichrome wire starts to glow.

Am I missing something or does it look reasonable?

Downunder35m8 months ago

If it is a throw away circuit anyway use a metal strip, safety pin....
Otherwise these metal disk push buttons you find on some toys, remotes and machine panels work great for high currents, at least a few times before they fail from heat damage....

karolina81 (author)  Downunder35m8 months ago

That's agreat idea and I relly would have loved using a pull pin switch, but as the whole product will be finally covered in shrink plastic (waterproof), that makes it difficult I think.

Will have a look at the metal disk push buttons though! Thanks.