11
\$\begingroup\$

In the circuit below:

enter image description here

I find it awkward that the direction of the current on the 2V power supply is running opposite to the direction of the current on the 5V power supply. In the 5V power supply the current flows from - to + but in the 2V power power supply the current flows from + to -.

Mathematically, it all works out and the numbers add up but I am having a little bit of a hard time gaining some intuition regarding what is going on with this "backwards" current.

Here are a couple of questions:

  • If the 2V power supply was a battery, would this mean that the battery would be charging? If that was the case, would the battery eventually explode due to overcharging?

  • If the 2V power supply was a regular power supply (connected to the wall) why is the power supply not breaking? The power supply is not a battery, right? So there should be no such thing as "charging" a power supply so why do I not see smoke coming out of the power supply given that I am going against the natural flow of electrons?

\$\endgroup\$
  • \$\begingroup\$ current flows in the direction of voltage differences \$\endgroup\$ – KyranF Feb 27 '15 at 3:46
  • 5
    \$\begingroup\$ @KyranF, through resistors it does. But from OP's example, that's obviously not true for voltage sources. \$\endgroup\$ – The Photon Feb 27 '15 at 5:31
  • 2
    \$\begingroup\$ Note that a "negative" current is just a current flowing in the opposite direction of your guess. \$\endgroup\$ – Greg d'Eon Feb 27 '15 at 14:36
31
\$\begingroup\$

As the other answer says, from a purely theoretical point of view, there's nothing wrong with current flowing either way through a voltage source. That's exactly what an ideal voltage source is: a circuit element that maintains the same potential on its pins, no matter what the current through it is.

In the real world, our voltage sources are not ideal.

A linear regulator will not usually maintain regulation when current is reversed.

A battery will charge (if its chemistry permits it) when current is reversed.

But we also use voltage sources to model things besides power supplies. For example, a voltage source can model the output of an op-amp. And an op-amp output typically can maintain its output voltage whether sourcing or sinking current (within limits).

Or a voltage source could model a shunt regulator or zener diode. These devices will only maintain regulation when they are sinking current.

Or a voltage source could model a forward-biased diode. Diodes also will only work as voltage sources when current flows in to the more positive terminal.

So from a pure theory point of view, you should be prepared to accept current flowing either way through a voltage source. When thinking about real circuits, you need to consider what actual device the voltage source is modeling and then whether the model is still valid with whichever current direction the model predicts.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ A typical "wall" power source is a rectifier (i.e. diodes) connected to an AC source, right? \$\endgroup\$ – Random832 Feb 27 '15 at 19:30
  • \$\begingroup\$ @Random832, that would be an unregulated wall wart. It's also common for a regulator circuit to be included to give a more stable output voltage. \$\endgroup\$ – The Photon Feb 27 '15 at 19:41
8
\$\begingroup\$

In pure circuit theory, power supplies can absorb or deliver power (they don't need to deliver power). If your circuit has only one source, then it will only deliver power. But if your circuit has two or more independent sources, it's possible for one/more of them to absorb power (as long as at least one more is delivering power).

Keep in mind that these circuits are purely mathematical constructs. An independent voltage source is not a battery, but can be part of a model for a battery.

For intuition's sake, though: think of a rechargeable battery. When you recharge it... which way do you suspect current flows? If you guessed out of the battery... think again =)

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Pedantically, when charging a battery, current flows both into and out of the battery. \$\endgroup\$ – user253751 Feb 28 '15 at 0:02
  • 1
    \$\begingroup\$ And also when discharging a battery! \$\endgroup\$ – Jamie Hanrahan Mar 1 '15 at 1:13
5
\$\begingroup\$

Addressing the bullet points:

1) Yes, that's what charging a battery looks like: pushing current back through it by connecting it to a larger voltage. What happens depends on the chemistry and size of the battery relative to the current. Some types (NiFe, larger lead-acid) can be kept on a float charge of a few miliamps forever. Other rechargeables will be slowly reduced in charge capacity by plating at the electrodes until they're useless.

2) It's a simulation, not a wall power supply. Again, what happens when you try to drive a voltage source with another voltage source depends on how it's implemented; it may turn itself off, lose regulation, sink current (as in your simulation), or something else.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

Q1: this would be a charging current, but it depends on the type of battery if it actually stores the charging energy or will explode.

Q2: a regular power supply is not designed for negative loads and may have a diode to prevent a negative output current. But power supplies may work in 2 or 4 quadrants, to be able to sink a negative output current.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Q1: as previously posted, to cause the battery to explode (or burst, or cause safety venting) you'd have to provide enough energy / power relative to the size of the battery to cause sufficient pressure buildup. \$\endgroup\$ – Technophile Feb 27 '15 at 22:25
  • 1
    \$\begingroup\$ @Technophile: The energy required for massive overheating need not come from the charging circuit. If a small area of the battery passes through more "charging" current than the rest, feeding enough power to overheat that small area could cause it to start passing through the battery's own stored energy, resulting in thermal runaway. \$\endgroup\$ – supercat Feb 27 '15 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.