Too Many Amps for Application?

The battery can supply unlimited AMPS.

Chargers you plug into the wall may supply 1 AMP.

Your plug-in supplied 14 volts at about 150 AMPS if there was nothing to limit the amperage.

Next time, use the bells and whistles, you can weld metal with power from the battery.

Let's take a look at my statements recently voted OT.

The battery can supply unlimited AMPS. Seems on topic and correct to me.

Chargers you plug into the wall may supply 1 AMP. Not outrageous and to the point of the topic "Too many Amps for Application?" Maybe I should have said 1-2 AMPS?

Your plug-in supplied 14 volts at about 150 AMPS if there was nothing to limit the amperage. Well? What's wrong here? Should I have said "Your plug-in COULD supply............?

Next time, use the bells and whistles, you can weld metal with power from the battery. Seems like OK advice.

Since the OT was applied late in the life of the thread, it's pretty obvious the source.

But, it's a free world, and who cares anyway. The Animas River is still orange.

I never use the childish OT feature myself. all batteries have a limit of their output. no battery is unlimited.

Unlimited was an incorrect choice, but relative to a regulated 120V wall wart, not totally outrageous in relative terms.

The "unlimited" was qualified in statement #3.

So shoot me.

"So shoot me.

"

Too messy. You get a pass. I'm too lazy to clean up afterward.

I got the qualification. It made sense to me.

Here, take this.

Agreed.

Assuming the charging device you plug into

The cig lighter is rated for an output of Dc 5v 1a isn't

That as much as the using device will recieve, that was my belief.

I'm inclined to believe that polarity was the issue.

Don't make it more complicated then it is.

Careful with the statement "don't make it more complicated than it is."

It might come over condescending.

If you want to make some sense in this thread maybe turn your inclination into knowledge and let us know if the same happens with the replacement charger when you reverse the polarity.

This would make the solution to the question official!

Thanks for your understanding.

Car chargers that use the lighter socket always have a current conditioner and limiter in the plug....here is a way to make a universal usb charger....or you can just buy one...

"The input voltage supply for a car charger is typically 12 V, but can range from 6 V to 14.5 V with input surges of up to 40 V for multiple 16-ms durations. The power supply must be able to tolerate these surges, and regulate the output to a nominal 5 V with a tolerance of 4.75 V to 5.25 V. Since the USB cable may cause the output voltage to fall at heavy loads, it is desirable to have droop compensation to raise the output voltage during this condition. Short-circuit protection is required in case of a fault with the USB port. The average current consumption depends on the device connected to the USB port, but can be as high as 2.1 A continuous. The form factor of the design is an important consideration, allowing for the car charger to be inserted and removed easily, with little material extending beyond the socket. The form factor must be small enough to meet UL standard 2089 and ANSI/SAE J563 specification."

Creating a Universal Car Charger for USB Devices From the TPS54240 and TPS2511

http://www.ti.com/lit/an/slva464e/slva464e.pdf

"Since the USB cable may cause the output voltage to fall at heavy loads, it is desirable to have droop compensation to raise the output voltage during this condition."

How is that accomplished?

The USB cable consists of 4 wires plus a shield (Red = +5VDC, White = D+, Green = D-, Black = GND, plus the braided shield over the cable). USB cables come in various lengths from a few feet to a maximum of 16 feet for USB 2.0 as well as various gauges.

How would the regulator end of the charger that is at the one end of that USB cable know what the voltage is at the other end?

I know that you can use an extra pair of sense cables to do this, but there isn't that extra pair in a USB cable (I know because we build specialty USB cables for one of our products).

Is there another way for the power supply regulator to know this? I don't know of one myself.

Well it regulates output by voltage drop at the supply side of the device....this assumes the voltage is the same throughout the usb cable length....probably best to keep the cable at 3' or under...

That is a contradiction, isn't it?

If the cable is responsible for drop in your first post, but then ignored in the second - hows that work?

The USB charger is a "CURRENT controlled" device, rather than voltage controlled. That is why it can vary from 4.75 to 5.25 volts. The current delivered to the controlling load (the charger) will determine the voltage required. And since the charger will not pass more current than A) the design limit, or B) the load requirement, whichever is smaller, the charger does not need a +-V sense pair.

Yes according to a voltage drop it raises the amps to a max of 2.1 I think...and .01 on the low end...

I don't think so. It can only regulate the current based on the voltage at its own output. It has no way to compensate for cable loss and no way of knowing what the voltage is at the device.

Misnomer. SolarEagle said "voltage drop". It isn't, really, though the implicit drop is seen in the loss of, and demand for, more current.

I'm trying not to push this too far, lest I state something wrong in a big way, but the charger/control device (wall wart, brick, whatever) DOES change it's own effective impedance, thus changing the total load impedance, so that the supply side will deliver more current, thus boosting the delivered current back up to requirements, in spite of, and to correct for, line (and other) voltage losses.

As Del said earlier, OHMs law answers this. It shows that if current has to stay the same, and voltage is not under the distant end's (the load's) control, then the only way to cause current to stay the same is for the load to vary it's impedance. That will reduce the "R" part of the equation, resulting in the E/R part of the equation (which IS I, or current) to rise.

The net effect of lowering impedance at the load is NOT to raise supply voltage, but to keep supply current under control regardless of supply voltage.

So, in a way, to correct my earlier statements, this does NOT control for line voltage loss. But it does compensate, without having to know about that loss.

Well, the original USB 1.2 and 2.0 standards were for 100 ma, followed by the handshake between the charging and charged devices, and a rise to a maximum of 500. That wasn't designed to POWER devices, other than things like USB sticks. For powering USB external HD and such, a hub was required, with it's own power brick.

But manufacturers were quick to see the advantage of being able to offer more power, thus the ability to support devices the competition didn't, and the race was on.

Hence the NEW USB "standards" evolved to the currently common 1 Amp and 2.1 Amp, plus whatever the USB 3.0 standard allowed, at inception, and now allows.

But commonly, I see 1A and 2.1A. I did not know the minimum allowed after handshake, but that would have been .1A, rather than .01A, to start. It may have evolved down to .01A, for maintenance charging on "flea-power" devices.

It must be hard to keep up with "standards", when there don't seem to be any.

Right. And any losses in the cable are just that, losses.

The charger doesn't know if the cable is lossy or not. It just supplied the required current. How much of it actually gets to the device is not known to the charger.

Therefore, the charger doesn't compensate for line loss at all.

The charger does, incidentally, account for line loss. Remember that current delivered will always equal current consumed, since current at the charger represents the SUM of all consumption in all the parallel and complex connected loads, no matter how complex the circuit. So, if the supply is designed to deliver a specified current (within specified limits), the CHARGER will vary the voltage to drive the current into the load, regardless of the load impedance.

In effect, the charger doesn't have to know the line loss. It only has to know how much current to push out. THAT fact derives, by implication, the line (and any other) losses in the circuit.

In the old days (my old days, at least) all the power supply designs were for VOLTAGE controlled designs (I always was told current control was too hard or too expensive to implement. Maybe so, but apparently ICs solved that problem, since almost all power supplied now, at least on ICs, are current controlled. I don't know well enough how switch-mode supplies work, but I think they are actually cumulatively voltage controlled.)

So, with current controlled supplies, or what used to be called CONSTANT CURRENT GENERATOR CIRCUITS, voltage sense is by implication. In the old precision bench supplies, where Vsense was always implemented for really close control, they were CONSTANT VOLTAGE GENERATOR CIRCUITS, and Vsense was used to give EXPLICIT control of voltage, while the load impedance was relied on for implicit current level control. (Both were really Control, rather than Generator circuits, but as a peon in the employ of my masters, I didn't get to correct their English).

"Remember that current delivered will always equal current consumed"

Yes, I agree with that, but only as a total system. However, it's useless when you want to know how much power the device is getting over delta t.

I'm thinking about it this way. The cable is a resistor. So is the device. The resistance of the cable is unknown, but it will have to dissipate some energy (in the form of heat) and whatever is not dissipated or lost goes to the device.

The equation is simple: P = V • A.

If 20% is lost in the cable, then the device charges at a slower rate. It may still get to a full charge, eventually, because the source is constant current as you pointed out.

So, I guess the charger can't compensate for the loss in time or the charge rate.

First (to my mind) answer that answers the OPs question....

I actually think he simply screwed up the polarities because he either does not have a good meter (any working meter is good!) or he does not know how to use it!!

install something like this before you melt/weld your little wires

C'mon guys ...
I answered the question... where is my sack full of GA's?

I'd hope the plethora of confused answers will convince the OP that if we don't know, then he certainly doesn't and he should take my advice and stop messin'

Del

The grammar board has deducted points from your aggregate score for this, "Yes, I've made plenty of screw us too"

Screw us too?

Keep trying. There's a good kitty.

Cheers.

D'oh

I actually gave you one, didn't you see it yet?

Someone may have not agreed with me, you know how they are here.....

I did component level repair on TV's back in the 70's, also have an EE degree. My first thought is this should have worked. 2nd thought is it was hooked up backwards or it was just time for the port to fail or the 12vdc to 5volt sub converter failed. Mark

Anybody Explain why and how to achieve successful results in the future?

To answer your question, simple, you screwed up! And do use a car electrician!!

So go get a car electrician to rewire your cigarette lighter socket to be always "live" with a fuse, or even better, add a proper 12 volt power socket(s) for cars, not quite the same as a cigar lighter socket, but similar, including an inline fuse that you can easily replace of course....

Also, adding an electronic unit to warn you, or to switch off the socket, before the battery is too discharged to start the engine, could be cream on the cake of course!!!

Del got it right, don't meddle with what you don't understand, it can get VERY expensive......and very painful too.....

IMO the whole expression The battery can supply unlimited AMPS. Is subtlely wrong

It supplies Volts, the load draws amps... a subtle distinction but an important one.

A battery can't just pump out amps willy nilly... however it will always supply the volts (yeah ok there is volts drop on a really heavy load... but hopefully you know what I mean)

Amps depends on load... the Volts are always there.

The title of the thread is bonkers and so are half the replies.

You should all go sit on the naughty step with a crate of beer and think about it.
I'll join you

Del

(PS... wasn't me marked anything OT)

"The battery can supply unlimited AMPS. Is subtlely wrong"

OK, it's subtlely wrong. And subtlety wrong too.

Many times, this one included, the forum over-complicates things for the non-expert.

Chances are very good that Jtarleton is not an electrical expert, (If they were, they'd not be here this time) and the technical difference between voltage, amperage, resistance, conductance and OHM's law would not only be lost on them but, after trying to wade through 7 of these posts that do nothing to help the OP, but rather are (once again) technical discussions trying to prove points having nothing to do with why the damn thing burnt up they would have left the thread, eyes glazed over in disgust.

It burnt up because too much power or (insert favorite, technically correct term here) was supplied to the device.

So giving the OP a lesson on the various electrical terms and what they mean to the forum (not the OP) do nothing to help answer the basic question.

I freely concede that you are all very intelligent and far more advanced in your knowledge of electricity, and theory, than I am.

Jus sayin'

Come on over, I've got the beers waiting.

To all here present (and any who might have left in disgust, also)

Mea Culpa, Mea Culpa, Mea Maxima Culpa (and I'm neither Catholic nor Latinate).

I hereby confess, I am guilty as charged.

The others were right. You supplied to blamed much power, in the form of both unconstrained volts and unconstrained amps, which totals unconstrained watts, or, as stated, too much power.

Simple enough? Good. Find an electrician to fix it, and to get it right before you fry your ride.

Del, the funny thing is that a short will produce all the amps without a load.

So in a subtle way Lyn was right for a special case!

Funny you should say that! Its a very small load creating the highest current. Or is it considered a big load because it creates the highest current stream inclusive heat and light - undesirable but still a load.

I think current and load are proportional as follows:

I = E/R

As for the definition of BIG as far as a load goes I would classify that as either it requires two flushes or the impedance is very small compared to the potential. ;-)

The current caused by any "short" circuit is dependent on the impedance of the short.

The lower the impedance of the short, the higher the current load of the circuit until the circuit reaches "bolted fault" state and either fails or produces maximum heat.

Any/all short circuits occurring in any circuit adds to the total circuit load by decreasing total circuit impedance.

In the case of high resistance short circuits the circuit load is increased due to the fact that circuit impedances are being paralleled.

A short IS a LOAD.

Anything except an open circuit is a load.

A hot dog dropped across the terminals is a load

For pities sake... exits left sobbing

Del

(Blimey, I unsubscribe for a while and then peek back in to see what's happened while I've been away.

The naughty step for the lot of you!)

Peep!

Common give a none-EE a break.

For me a load is something useful. Its kind a backwards calling the shortest thing the biggest load. Its a load because it draws amps. Makes a leak in a water pipe looking really good.

And we were talking about loading USB devices from a battery didnt we? Is this resolved?