What happens if I put too much current in a voltage regulator?

I want to wire a solar panel to a battery pack to charge it.

If I use a voltage regulator (L7805) which regulates to 5V and a maximum of 1.5a...but exceed that by plugging in a bit too many solar panels...what happens?

Does it prevent any more than 1.5A going through for a while before melting? Or does it immediately melt at 1.51A? I was just wondering what would happen if I had enough solar panels for 2.1A at peak and put it through this regulator.

Why do you think a voltage regulator EATS all the current available at its input?
The clock in a car runs all the time and its voltage regulator is fed from the car battery. The car battery has HUNDREDS of Amps available so it can start a very cold engine but the voltage regulator in the clock does not draw all that current, instead it draws only as much current as it needs which is about 0.002A (2mA).

The circuit in the voltage regulator uses a few mA in addition to its output current. It limits the output current if it is higher than 1.5A to 3.3A.

A voltage regulator or a transistor is heated by the current through it times the voltage across it. If its input is 12V and its output is 5V then it has 7V across it. If its current is 1.5A then it heats with 7V x 1.5A= 10.5W. It needs a heatsink to feed that heat into the air so it does not get too hot. If it gets too hot then it senses it and shuts down without melting.

EDIT: You want to use a 5V regulator to charge a battery pack? NO WAY! The battery will probably explode and/or catch on fire!
A battery must be charged by a charger circuit made for the type of battery. The charger circuit limits the charging current and limits the charging voltage and stops charging when it detects that the battery is fully charged.

Are you sure? Those battery packs you buy to charge your cell phone can be charged by the 5V 1-2A AC to USB plugs that you use to charge your cell phone.

Why would 5V blow them up? I was going to take a microusb to usb cable, cut off the usb end and solder a voltage regulator in a circuit between that end and some solar panels. I was thinking that as long as I could get 5V coming in and not more than a couple amps, it should be the same thing as plugging it into a wall and having the AC to DC transformer (or whatever it's called) convert it to 5V DC.

I'm new to this so please let me know if I'm making a huge error.

I was thinking of using 18650 Li-ion batteries...I think there are two types and the ones I am thinking of using are 3.7V and require 4.2V to charge. So if I were to directly charge it with 5V that would probably blow it up -- but I am running it through the circuit in the battery pack that takes a 5V microUSB DC input and does some kind of circuit magic that then charges the 18650 batteries.

I was thinking of using the 18650 battery because they seem plentiful, cheap and hold a lot of mAh. I read somewhere that they are the #1 battery type used in laptop battery packs and so on.

Here's an example of a cheap 18650 5V battery pack on Ebay that I might use.
https://www.ebay.ca/itm/Portable-Re...l_Phone_PDA_Chargers&var=&hash=item5d4ed16f49
https://www.ebay.ca/itm/5V-2A-Dual-..._Phone_PDA_Batteries&var=&hash=item1e94286944

Also, 5V is a magical amount of voltage. Not only is it compatible with every cell phone out there, but it also works for the raspberry pi. I was hoping to build a solar powered raspberry pi for weather/environment sensing.

Aro2220 said:

So if I were to directly charge it with 5V that would probably blow it up...

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Yes, that's what Audioguru was worried about.

Aro2220 said:

... but I am running it through the circuit in the battery pack that takes a 5V microUSB DC input and does some kind of circuit magic that then charges the 18650 batteries.

Click to expand...

This should work fine.

Don't worry about too much current from the solar cell. The "magic" charging circuit decides how much current to draw from the 5V regulator to give to the battery. The 5V regulator only draws as much current from the solar cells as it needs to feed the charging circuit.

Be careful with the voltage from the solar cell though. The maximum input voltage for the L7805 is 35V, so if the solar cell produces more than that, the L7805 may be damaged.

Also bear in mind that if the voltage from the solar cell drops below about 6 or 7 volts, then the voltage at the output of the L7805 will drop below 5V.

godfreyl said:

Also bear in mind that if the voltage from the solar cell drops below about 6 or 7 volts, then the voltage at the output of the L7805 will drop below 5V.

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So if I had 5V coming from the solar panel there would be a voltage drop across the regulator and I would end up supplying less than 5V?

Aro2220 said:

So if I had 5V coming from the solar panel there would be a voltage drop across the regulator and I would end up supplying less than 5V?

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Yes, that's right. There's always some voltage drop with a linear regulator. If you want to minimise that, look for an LDO regulator (short for Low Drop Out).

With switch-mode regulators you can get output voltage higher than the input voltage but the circuitry is a bit more complicated. They're also more efficient i.e. waste less power as heat.

What you call a "Lithium battery pack" is really called a "Lithium juice pack" or "Power bank". A battery pack has just a battery that is 3.2V when it should have its load disconnected and is 4.2V when it is fully charged.
A juice pack also has the same battery but it also has a Lithium battery charger circuit that works from a 5V USB port, and it has a voltage stepup and regulator circuit so its output is always 5V.
Simply read the datasheet for the juice pack.

The datasheet for the L7805 regulator shows that its MINIMUM input voltage is 7.5V. An input voltage less than 7.5V might result in no voltage regulation so the output voltage will be lower than 5V.
Most of them fail to regulate when the input voltage is less than 7V.

Guess what? You might buy cheap Chinese FAKE 18650 batteries from ebay: https://www.youtube.com/watch?v=eOshOXcSkDA