Voltage regulator

hey, i got 5v voltage regulator 7805 and i got a problem. when i connect it (as all tutorial say) i get 5v on the output BUT the input voltage decrease. When the voltage regulator isnt and i measure the voltage of the battry its 8.88v. When i connect the voltage regulator it fall to about 7.6v. Why is this happening? the others legs are 0v (middle, ground) and 5v (output voltage). In addition, when i left it like that for couple hours i found out when i came back that the battery was 2.2v although i didnt connect something (the circuit was open).

As i said this happen even if i DONT connect the capacitors. Picture of my circuit with my capacitors:

Become the regulator warm?
The regulator needs 5-10 mA so i think you have a bad battery.

Pelle

MyNick:
when i connect it (as all tutorial say) i get 5v on the output BUT the input voltage decrease.
When the voltage regulator isnt and i measure the voltage of the battry its 8.88v.
When i connect the voltage regulator it fall to about 7.6v.
Why is this happening? the others legs are 0v (middle, ground) and 5v (output voltage).

Because the battery is being loaded and, since it has it’s own internal resistance, the combination of the load (from the regulator) and the internal resistance form a voltage divider that makes the actual voltage at the terminals of the battery lower. See the attached picture for details. The formula for the actual voltage at the terminal of the battery is: Vterm = Vbattery[Rregulator/(Rbattery+Rregulator)]. This will make the voltage at the terminals less.

MyNick:
In addition, when i left it like that for couple hours i found out when i came back that the battery was 2.2v although i didnt connect something (the circuit was open).
As i said this happen even if i DONT connect the capacitors.

Was the voltage regulator connected to the battery during this time? i.e., Vin → battery, Vout → unconnected, GND → Ground? I think that the regulator will draw a certain current even though the output is not connected, thus depleting the battery. You can measure this current by connecting an ammeter between the terminal of the battery and Vin of the regulator.

MyNick:
hey,
i got 5v voltage regulator 7805 and i got a problem.
when i connect it (as all tutorial say) i get 5v on the output BUT the input voltage decrease.
When the voltage regulator isnt and i measure the voltage of the battry its 8.88v.
When i connect the voltage regulator it fall to about 7.6v.
Why is this happening? the others legs are 0v (middle, ground) and 5v (output voltage).
In addition, when i left it like that for couple hours i found out when i came back that the battery was 2.2v although i didnt connect something (the circuit was open).

As i said this happen even if i DONT connect the capacitors.
Picture of my circuit with my capacitors:

First of all, did you connect the 7508 correctly? It should be like this:

±----+
| O | ← metal tab (is also ground)
+=====+
| LM |
|7805 | ← top view
+=====+
| | | ← pins
| | |
1 2 3

Where:
1 = INPUT
2 = GROUND/COMMON
3 = +5.0 DC OUT

Note that the 7805 needs at least a 2 to 3 volt differential across it. That is, since it’s putting out 5.0V, it needs 7.0 to 8.0V minimum as an input.

Also note that even with no load, a 7805 will draw 0.004 to 0.008 amperes (4 to 8 milliamps) quiescent current. If you assume 6 milliamps quiescent current and connect it to a 9 volt alkaline battery (nominally 300 milliamp/hours), you can expect the battery to go dead in about 50 hours.

If you had it connected correctly and it didn’t get warm or hot, and you had no load connected to it, I would guess your battery was defective or weak. (You did have those electrolytic caps connected the proper polarity I assume?)

Lastly, note that in most cases, those input and output bypass capacitors are not needed. Build your regulator circuit without them and put a scope on the output. If it’s clean, forget the caps. The 7805 (I’ve found) is stable without them.

Good luck!

Because one of those little 9V transistor radio batteries won't supply power for very long.