I have an apparently simple question, but I can't seem to figure it out.

I learned in school that the generators (source powers) are defined by their voltage, and that made a lot of sense to me. Afterall, we say 1.5v, 6v, 9v, batteries, right?

But I noticed that in my source (that converts from wall socket's 100-240V AC), it reads: Output: 5V, 1A. My question is, how can it possible have a constant voltage and current? That makes no sense!

That is, we have U = Ri, and R is fixed in the circuit (sum of the resistance of the wires and components), so the source has to specify either the voltage OR the intensity of the current, not both! The other value is calculated, given the resistance. So which one is it? And what does the other value in my source means? Is it the maximum voltage/current? That would make more sense.

Some Background:

My problem with this was that I was trying to make my Integrated Circuits Chips (SN74LS series) to work with this source, but I think they burned out, because they simple didn't work. I made I simple test with the NOT gate, and it did nothing. It's such a simple setup I think it's very likely it has burned out. The 1A really freaked me out when I read them (after testing), because I'm aware that the regular current in a circuit is around 20-50mA (way less). If the current specified is the maximum (as it was clear to me), it should have worked; if the current was fixed, and the voltage was the maximum, then it might had burnt out, as it did in reality - but that makes absolutely no sense to me. What I am missing?

  • 1
    \$\begingroup\$ You're correct, it's either constant voltage or constant current. The rating on the source is the maximum of both. Chances are, it's a voltage source that can output a maximum of 1A (if a device tries to draw more than 1A, the voltage will either drop, or the source will be damaged). \$\endgroup\$
    – Shamtam
    Commented Dec 1, 2013 at 0:34
  • 1
    \$\begingroup\$ See here. \$\endgroup\$ Commented Dec 1, 2013 at 0:38

2 Answers 2


Both walk together. Imagine voltage to be, as the name says, potential. In other words, is what your source can potentially causes in your circuit. It is its potential to generate a current. As the ohms law states (U=Ri), we can think of current being a consequence of a resistence connected to this source. This is almost a rule for almost every source of electricity in our days, but there are exceptions (below)! So, whenever you think of your wall outlet or a car battery or a cellphone battery, they have a voltage potential and the current will be calculated by which material or what are you connecting to the power source (by its resistance). Generally, the current specs tells us what is the MAXIMUM current that this power supply can handle. But it does not tell you that this source will be always +V volts and A amperes!

But note that this is the case for voltage sources. As the name says, it guarantees a constant voltage (ideally). So you calculate current because you know that the nominal voltage will remain the same.

So the excepetions will be current sources. Now everything is inverted! This kind of power supply guarantee's that the current will remain fixed. So you can calculate what voltage it is applying to your circuit so it can delivery that amount of current (also by ohms law) but generally we do not need that. Although a current source concept is very useful inside electronic devices, we are not used to see them in our days. But one good example is your telephone line comming from the wall. Those are current sources. Note that you cant damage the wires or the telephone company by making a short circuit to these wires. You can connect a multimeter and you will see that the current will be stable at some point (here in Brazil at 24mA). I can connect a 10Ohms resistance or a 300Ohms resistance and the current will be the same. Of course the voltage applied will be different, and that is how a current source works.

So it all depends on what type of source you are dealing with. If it guarantees a current fixed, you can calculate the voltage difference between terminals. If it guarantees a voltage value, you can calculate its current depending on which load you put there. In most cases, those power supplies from computers, cellphones, etc are all voltage sources and its specifications guarantee a nominal voltage and a maximum current. But don't expect to have that current regardless of the connected load!

  • \$\begingroup\$ Thanks for clearing it up! I guess it was something else wrong with the circuit I made, then... As don't have a Multimeter yet, I can't easily test things, but I will try to make some simpler circuits and try it out. \$\endgroup\$
    – Luan Nico
    Commented Dec 1, 2013 at 0:52
  • \$\begingroup\$ Go on and stay calm about specifications. For a given load, you simply can't guarantee that your source will have a V voltage and a I current. Depending on the load, probably one of them (r both) won't match the specs. In your case, you just have to match voltage specs and your IC will determine the current. Those IC specs (20-50mA) exists only to you know what is the minimum current capacity your source must have (and also to calculate power consumptions). Good luck then. \$\endgroup\$ Commented Dec 1, 2013 at 0:57
  • \$\begingroup\$ One thing to watch out for when you build circuits with logic gates is that any inputs that can be floating need pull-up or pull-down resistors. If you used some switches on the inputs of your gates for testing and you didn't add pull-up or pull-down resistors, this may explain why it didn't work. \$\endgroup\$ Commented Dec 1, 2013 at 5:07

This is a simple case of misinterpreting the rating.

In this context (voltage sources), it is understood that a power supply rated at 5V, 1A can (safely) produce 5V across the load while sourcing up to 1A of current to the load.

So, for example, if you connected a \$5 \Omega\$ resistor across this power supply, it should be able to keep 5V across the resistor and, thus, 1A through the resistor.

Also, if the resistance is more than \$5 \Omega\$, no problem. The power supply will keep 5V across the resistor and less than 1A through.

However, if the resistance if less than \$5 \Omega\$, there is likely to be a problem. This could manifest itself as:

  • current limiting - there is 1A through but less than 5V across
  • current trip - there is zero volts across and zero amps through until you reset the power supply or replace the fuse
  • thermal trip - the power supply overheats and trips a thermal fuse
  • hard failure - the power supply "lets the magic smoke out" and must be R&R'd

This list is not exhaustive.


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