# Amperage vs Voltage - Which is more dangerous?

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#### Rick

##### Posts: 4,512   +66
So here's a question for those who are electronically inclined.

I deal with a lot of laptops which do not have their original power supplies. So we usually match up the connector, voltage and amperage. However, there's not always an exact match.

What are the dangers of power supplies rated for higher / lower voltages than recommended?

What are the consequences of higher / lower amperage than recommended?

And assuming Volts x Amps = Watts (From what I've figured out), do you suppose this would work?

Suppose I have a laptop designed for 15v / 3a. This is roughly 45w, correct? What if I were to use a 19v / 2.3a power supply? Does the total wattage dictate whether the laptop will get fried or not? Or do the amp/voltage ratings fry by theirselves?

The fellows at work tell me higher amps is the killer, but higher votlages are okay (within reason). From my own obverservations, this is not the case. It looks more like higher voltages kill, while higher amperages make no difference. I successfully used a number of lower voltage, higher amperage power supplies, but toasted some equipment (not laptops, thank goodness!) with higher voltage, correct amperage power supplies.

#### StormBringer

##### Posts: 2,218   +0
It depends, if you are speaking of dangers to humans then amps(current) is the concern. Components are a different story, in general, the voltage is usually the concern with most components because the current is drawn by the device. Just because your power source can deliver 15A doesn't mean you are delivering 15A, if the device only draws 5A, then only 5A will be used. On the other end of this, if I have a source that can deliver 16v, but my device only requires 12v, the 16v will likely kill the device. Some devices can handle it longer than others, some will die instantly.

In short, your source has a max current it can deliver, but the amount delivered is dictated by the load.

#### DraG

The simple answer to your question is that your screwed both way, but you have better chance at a lower voltage and higher amps battery pack. #### StormBringer

##### Posts: 2,218   +0
Originally posted by DraG

How do you figure that?
Amperage is drawn by the load, the power source doesn't supply more amps than the load draws.

I=E/R
I=Current(amps)
E=Voltage

For example, lets take a look at three conditions of the same circuit; first, we look at it in operating condition(no faults)
we have a 12v source that is rated at 15A max

E=12v
R=100ohms
I=0.12A

Next we look at the same circuit with a "short" condition.

E=12v
R=0ohms
I=0A

Now we look at it with an "open" condition.

E=12v
R=Infinity(since the circuit is broken)
I=MAX
In this condition, the source can supply more that the rated amout of current, although, since this circuit is broken, it is only supplying that current when you read it with a meter.

To further illustrate this, I'll use a fourth example of a "changed value" condition. This will be the same circuit, but the load will now be a little lighter, this can be caused by excessive wear or by components that are on there way of shorting.

E=12v
R=30ohms
I=0.40A

#### Rick

##### Posts: 4,512   +66
Excellent. Thanks Storm.

So being rated for a higher amperage determines the load of the power supply, not the output. The voltage on the other hand, delivers just as it advertises on the label.

I figured you of all people on here might have an answer to this. #### StormBringer

##### Posts: 2,218   +0
Yea Rick, the amperage rating means that (in theory) the power supply can delive that amount of current, if needed by the load. The voltage is the actuall value being output(+/-10%)
So, if we are talking about one of the +12v MOLEX connectors, it should measure anywhere between +10.8v and 13.2v. Measuring current on that connector would give you the MAX for that rail, unless you put a load on it and measure across the load. An excellent way to see an example of this would be to use a POT at the load, you can increase and decrease the resistence(load) and see how the current drawn by the load will vary. The source voltage stays the same.

#### oouter

##### Posts: 7   +0
to deal with the voltage/amp which is more dangerous thing, voltage doesnt kill you, the amps will.

#### DraG

That the thing we are not talking about a circuit. Talking about a supply to that circuit . If you have laptop that has 100ohms but a battery pack that supplys 18 to 19 volts. Then what is changing in your formula there? That right you are changing the amount current in the curcuit. So that why a lower voltage is better it the only battery pack that will not change the current used by the laptop.

#### Nic

##### Posts: 1,519   +1
A power supply having a higher rated output voltage will only kill a device if it can sustain the power demands of the load. If the power supply can only supply the power required by the load, and no more, then the output voltage will drop, as the current increases to supply the load. This is how unregulated power supplies work, and they only become dangerous when the power supply is more powerful than required to drive the load.

The power supplies of computers, laptops, and many external storage devices (e.g. cdrom drives etc.), are normally switched-mode units and their output voltages are thus regulated. Because of this regulation of output voltage, they will attempt to maintain a fixed output voltage according to their rating. If you use such a device to power your electronics equipment, and providing it can supply sufficient current, then it will almost certainly kill any attached device that normally runs on a much lower voltage.

In other words, always use a power supply of the correct output voltage for your chosen device. The current rating does not matter, provided that it is at least equal to that required by your device. If it can supply more current than that required, then that simply means that it is able to power more than one device simultaneously (up to the limit of its max current output).

#### StormBringer

##### Posts: 2,218   +0
Originally posted by DraG
That the thing we are not talking about a circuit. Talking about a supply to that circuit . If you have laptop that has 100ohms but a battery pack that supplys 18 to 19 volts. Then what is changing in your formula there? That right you are changing the amount current in the curcuit. So that why a lower voltage is better it the only battery pack that will not change the current used by the laptop.

Are you saying my explanation is incorrect?
Your statement makes absolutely no sense whatsoever

#### offamychain

A little late, but I just had to throw my input into this one. To give my answer to the actual title of the thread, Storm pretty much hit the nail square on the head. As an electrician, I AM concerned about the amperage I deal with, it kills quick. However, I usually don't go beyond "normal" safety precautions with simple high voltage. It can, & will, do damage. But is more likely to just singe your hair compared to high-amp output.
The statement about the "amp number" not really meaning much is pretty much correct also. A PS, or breaker, or pretty much anything else, rated at whatever amperage, will only deliver what the device needs (well, that's how it's supposed to work, but that's a different story), but a higher voltage supply WILL deliver it's total voltage into a device. (always bad news) Even a LOWER voltage supply has the potential to do damage to electronics. They're designed for EXACTLY what they say, with a usually very small safety margin.
You can do what you want, but I would suggest 1 of 2 things. Investing in a RELIABLE variable-setting PS, or simply using what the device calls for. Any other option is just begging for trouble.

P.S. Oh yeah, the body does not limit Amps like a computer does. It will take however many amps you grab ahold of. POOEY!

#### alphnumeric

##### Posts: 173   +0
Originally posted by StormBringer
How do you figure that?
Amperage is drawn by the load, the power source doesn't supply more amps than the load draws.

I=E/R
I=Current(amps)
E=Voltage

For example, lets take a look at three conditions of the same circuit; first, we look at it in operating condition(no faults)
we have a 12v source that is rated at 15A max

E=12v
R=100ohms
I=0.12A

Next we look at the same circuit with a "short" condition.

E=12v
R=0ohms
I=0A

Now we look at it with an "open" condition.

E=12v
R=Infinity(since the circuit is broken)
I=MAX
In this condition, the source can supply more that the rated amout of current, although, since this circuit is broken, it is only supplying that current when you read it with a meter.

To further illustrate this, I'll use a fourth example of a "changed value" condition. This will be the same circuit, but the load will now be a little lighter, this can be caused by excessive wear or by components that are on there way of shorting.

E=12v
R=30ohms
I=0.40A

If R=0 (a short) your current is max, it will be what ever the supply can dish out, but only for a very short instant, until the smoke comes out. The voltage will be almost zero across a dead short.

If R=infinity (an open circuit) their is no current flow, measured or other wise.

To answer the question, if the resistance stays constant, then increasing the voltage will result in an increase in current flow. Using a power supply with a higher current rating doesn't mean their will be more current flow. If you use one with a higher voltage then the current will be higher across the same load. Will it cause damage? It could if it's a big difference. The laptop may have some regulation built in since it has to charge the battery, so their probably is some leeway. How much is hard to say. The better question is, how much is the cost of a new laptop versus the cost of the correct power supply? If you match the voltage you should be OK. If power supply can't supply enough current it's output voltage will likely drop off. Just take the battery pack out of the laptop to lower the load on the supply.

#### StormBringer

##### Posts: 2,218   +0
you somehow misunderstood me alphanumeric. My statements were made based upon measuring accross a faulted component. If you understand current flow and voltage drops, then you should fully understand what I said and be able to relate to it without any problem. It was a fairly textbook example of basic AC/DC theory.

#### alphnumeric

##### Posts: 173   +0
I didn't misunderstand it, it's wrong. I'm an electronic technician. I've been working in electronics for over 20 years.
You said
E=12v
R=0ohms
I=0A

If R=0 ohms that is a short circuit. If R is the only resistance in the circuit you will have maximum current flow. I=E/R
Lets say R=0.1 ohm if E=12V then I=120 amps.
If R=0.01 ohm and E=12V then I=1200 amps
Lower the resistance and the current flow increases.

You also said
E=12v
R=Infinity(since the circuit is broken)
I=MAX
If R=1m ohm 1000000 ohms and E=12V then I=0.000012 amps
If R=10m ohm 10000000 ohms and E=12V then I=0.0000012 amps
Increase the resistance and the current flow decreases.
If the circuit is broken there is NO current flow in a series circuit. Any questions?

#### StormBringer

##### Posts: 2,218   +0
OK, last time I'll explain this, you do the math.
I=E/R
R=E/I
E=IxR
P=I(squared)xR
EDIT: I was trying to get them both in and ended up combining them in my haste)

BTW, If R=0, then I has to =0(do the math)

E=Voltage
I=current(Amps)
R=Resistance(Ohms)

In the "short" example, if you measure accross a shorted load(essentially a wire) you'll get no resistance or current reading because there is no longer a load there.

I think what is happening here is we are looking at the circuit in different ways, you are looking at the entire picture, while I'm looking at it on a component by component basis.

For the record, I am an Electronics technician as well

#### alphnumeric

##### Posts: 173   +0
Originally posted by StormBringer
OK, last time I'll explain this, you do the math.
I=E/R
R=E/I
E=I(squared)xR

BTW, If R=0, then I has to =0(do the math)

E=Voltage
I=current(Amps)
R=Resistance(Ohms)

In the "short" example, if you measure accross a shorted load(essentially a wire) you'll get no resistance or current reading because there is no longer a load there.

I think what is happening here is we are looking at the circuit in different ways, you are looking at the entire picture, while I'm looking at it on a component by component basis.

For the record, I am an Electronics technician as well

Ok looking at a single resistive load. To measure the current the amp meter has to be in series with the load. You said in your other post that if the R=Infinity(since the circuit is broken) then I=max. That's not true, if the circuit is open their is no current flow. Even with a very high load the current is low, not high.
Now leave your amp meter connected and short out the load. The amp meter will peg and you'll blow the fuse. If your lucky the power supply won't blow up in smoke. If you don't believe me take a dime and put it across the two terminals on a nine volt battery. The dime will get hot very quickly since it is drawing a high current. And by the way P=I (squared) R, E=IxR. Putting zero's into the equation makes it invalid. I=E/R you can't divide 0 into E. R can never be 0 anyway, unless you have it cooled to absolute zero. That equation will work with any value of R other than zero. Would you agree that 0.00000000001 is almost zero. E/R=I 12 divided by 0.00000000001 =120000000000 Amps. The closer you get to zero R, but not 0, the higher the A value gets.

#### StormBringer

##### Posts: 2,218   +0
Are you measuring across the load(one lead on one side, one on the other) to get your readings? If not, then thats why we are seeing this differently. If you measure across a short, you'll read NO drop in voltage and NO current(since there is no change from one side to the other(since it is essentially a wire and the potential is the same on both sides) If you have an open(broken circuit) and measure across the break(which will complete the circuit) you'll measure source voltage(source being whatever is measured before going into the load) and the current flow will be MAX(since your meter is now the load)

Your explanation is correct from the perspective of the circuit, but I was explaing it from the perspective of the measurements taken from a meter which are exactly opposite from that of the circuit in this respect.

#### alphnumeric

##### Posts: 173   +0
You measure voltage across the load as you say (one lead on one side, one on the other), but you don't measure current across it. You have to connect the meter in series with the load to measure current. Connect the amp meter on one side and open the circuit on the other. You will not measure any current flow. Closing the circuit with the meter defeats the whole purpose. If you do that your not measuring the open circuit current flow are you, you have closed the circuit again.
Current ceases to flow; therefore, there is no longer a voltage drop across the resistors. Each end of the open conducting path becomes an extension of the battery terminals and the voltage felt across the open is equal to the applied voltage (EA).
Taken from here open circuit
If you look at the top circuit in the blue box you will see a fuse with a meter across it. The voltage is zero volts across the fuse. The fuse is essentially a short. That diagram also shows normal current flow. That current flow is going though the fuse that has zero voltage across it. If I was to use your math Quote BTW, If R=0, then I has to =0(do the math) then no current would be flowing. That kind of defeats the purpose of using the fuse in the first place. A short is a short is a short, and when in a series circuit it can conduct current. If the short is the only load in the circuit you get maximum current flow.

#### StormBringer

##### Posts: 2,218   +0
I see what you are saying, and have from the beginning, but what you aren't seeing is that my measurements were all of the load itself, when you are taking measurements of the load, you have to measure the load, if you remove the load(whether it is shorted, open, or good) you defeat the purpose of testing the load.

Again we seem to be in the same book but on different pages.

#### Nic

##### Posts: 1,519   +1
I think you should both leave this argument here before it gets too far off track. You are both highly experienced and talented electronics engineers (as am I), but appear to be having a little trouble communicating your exact thoughts on the subject.

Originally posted by alphnumeric
... I=E/R you can't divide 0 into E. R can never be 0 anyway, unless you have it cooled to absolute zero. That equation will work with any value of R other than zero.
You just proved that the equation also works for R=0, as anything divided by zero tends towards infinity, as you say here ...
Originally posted by alphnumeric
... If the short is the only load in the circuit you get maximum current flow ...

Just to be fair, here are some anomalies from StormBringer's calculations (apologies for pointing this out) ...

Originally posted by StormBringer
... For example, lets take a look at three conditions of the same circuit; first, we look at it in operating condition(no faults)
we have a 12v source that is rated at 15A max

E=12v
R=0ohms
I=0A
In a *real* power supply this is correct (except E would drop to zero also), because the short-circuit protection will have kicked in, or the fuse would have blown. Real power supplies are also voltage regulated, and current limited. If we were talking purely theoretical, and we did maintain 12v across 0 ohms, then the current would be infinite.

Originally posted by StormBringer
... Now we look at it with an "open" condition.

E=12v
R=Infinity(since the circuit is broken)
I=MAX
It would appear that you've made a simple typo, and gotten your 'open-circuit' and 'short-circuit' currents mixed up (see previous example). That appears to have led to this discussion moving a little off track. Now that its cleared up, maybe we could all try doing something a little more productive. Edit: One final comment ...

Originally posted by StormBringer
... If you measure across a short, you'll read NO drop in voltage and NO current(since there is no change from one side to the other(since it is essentially a wire and the potential is the same on both sides) ...
This is correct, but then you wouldn't have 12v across the zero ohm load, which makes the example misleading. Also, current can flow through zero resistance, as that is what superconductivity is about. However, there would be no power dissapated. Any real power source has its own internal impedence (hence its finite max current), so zero ohms will never be a reality in any real power circuit.

Originally posted by StormBringer
... If you have an open(broken circuit) and measure across the break(which will complete the circuit) you'll measure source voltage(source being whatever is measured before going into the load) and the current flow will be MAX(since your meter is now the load) ...
With a perfect (infinite impedence) voltmeter, and no load, the current will be zero (through the meter).

With a perfect (zero impedence) current meter, and no load, then the current will max out (through the meter) and PSU fuse will blow.

#### alphnumeric

##### Posts: 173   +0
I'm happy, nuf said. #### StormBringer

##### Posts: 2,218   +0
Same here, though Nic, in my examples, I was using values that would be gathered by readings taken with a meter across the component(the meter would read what I put there) this means that while you would not read a voltage drop, this is because there is no "drop" across the shorted component, meaning that the applied voltage(or voltage going in to the component, comes out) I hope that clears things up with that a bit more.

I would agree that this thread has gotten a bit sidetracked(as usually happens when a couple of us electronics guys tend to miscommunicate). This is probably partly to do with the way we were taught to look at things way back in basic theory. this would explain why we are both saying the same thing in the end, yet have such different ways of getting there. I know Nic and I have gone round and round for weeks on occasion, just to eventually discover that we were saying the same thing, just looking at it from different angles.

#### Nic

##### Posts: 1,519   +1
Originally posted by StormBringer
... I know Nic and I have gone round and round for weeks on occasion, just to eventually discover that we were saying the same thing, just looking at it from different angles.
I've done my fair share of that too. I do enjoy a good disscussion now and then, but it can sometime be difficult to know when to stop. I'm still learning myself, and hopefully getting better at it. A very good read guys, so all credit to you both. #### Rick

##### Posts: 4,512   +66
This all sounds like an unknown language to me. #### MrX

i plugged a 16V 4.5A charger into a laptop that is only meant to take 16V 3.5A. You reckon i just killed the laptop? I did smell something funny could that be my laptop frying? Is it safe for me to use this charger with this laptop?

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