# Battery current reduction [duplicate]

Now I am trying to use this circuit in my bike. My bike has a 12V 9amp battery and while charging (bike running), the voltage fluctuates from 12V to 20V. To power this circuit with no fluctuation, I have made a voltage regulator circuit with LM7812 and capacitors (0.33uf, 0.1uf) as given in its datasheet.

I checked with multimeter, the output is good (around 11.9V to 12.1V), the problem is that the 7812 heats up (so much that I can't touch it) within 20 secs after switching on the supply. I believe that as the output current of 7812 is 1amp, it needs to reduce much current, causing so much heat. Do I need to change caps or is there something else I should try?

• 4.7k resistors (R3, R4) for LEDs powered from 9V? This schematic is wrong. R3 and R4 probably should be 470ohm, like R5 and R6. – Kamil Jul 17 '14 at 20:28

Heat production is caused by voltage drop on your LM7812 regulator.

When your voltage source produces 20V your regulator has to drop 8V to get 12V.

You schematic has two BC547 transistors, they have maximum current 100mA. That circuit on schematic can't consume not more than 250mA. It probably consumes about 100mA total.

Power lost on regulator at 250mA is 250mA * 8V = 2000mW = 2W of heat. Thats not a lot, but enough to heat up regulator (it's just few grams of metal) to "untouchable" temperature very fast.

You just have to add some heatsink.

Something like this should be enough for 2W:

(I took image from here)

but you can add bigger heatsink to keep temperature lower.

Keep in mind that LM7812 metal part of casing is electrically connected with GND pin inside device..

LM7812 operating temperature range is 0-125°C, so it can handle much more than your finger (I think "untouchable" temperature starts at 50-60°C).

Your capacitors are fine. 1A current is just maximum operating current for LM7812.

• I forgot to mention that i have already added the heat sink about the same size as suggested by Kamil (picture added). And the 0.33uf cap is of 100v, as i could get that only. I tested the 7812 ckt (picture added) when bike was not running, which means the 7812 got hot at 12v only which led me to think that the high current is the heating factor. – user2717098 Jul 18 '14 at 6:05

According to your schematic, the maximum number of LEDs you can have ON at the same time is two.

Then, assuming white LEDs with a forward voltage of 3 volts, about 0.3 volts Vce(sat) for the transistors, and a 12 volt output from the 7812 means that the maximum current that will be taken from the supply is:

$$I_s = \frac{12V - (3V + 0.3V)}{4700R / 2} \approx 3.7 \; mA$$

In addition, there are the transistors' base resistors, so if we consider a Vbe(sat) of about 0.7V for the transistors, then they'll pull about:

$$I_b = \frac{12V - 0.7V}{470R} \approx 24mA$$

from the supply, which is way more current than is needed to turn on the LEDs.

Then there's the 555, which can draw up to 15mA from the supply, so that's a total of about 44 milliamperes for everything.

If, then, the input to the 7812 is 20 volts and its output is 12 volts, the power it'll dissipate will be:

$$P_d = (V_{in} - V_{out}) \cdot I_s = 8V \cdot 44mA \approx 350 mW,$$

which should make the 7812 warm, but not uncomfortably so, so there's something definitely wrong in there.

I suggest you check all your wiring, especially the pinouts of the transistors and the 7812 in order to make sure there are no mistakes there.

Additionally, if you have the kind of charger / dynamo which gets disconnected from the battery once the battery gets to a certain voltage, but continues to put out, you probably have the 7812 input connected to the dynamo output instead of to the battery, which would account for the wild voltage swings you're experiencing.

In reality, if your battery isn't cooked and the charger is working, the battery voltage should stay between 12 and about 14V, which means that - if you hooked your circuit directly across the battery - you could get rid of the 7812.

If you want 12V from 20V at 1A, you are dropping 8V. 8V x 1A = 8W. That's a lot of power!

I recommend if you need to have the wide input voltage range, you change to a switching/buck regulator for your 12V supply, as a linear regulator cannot dissipate that much heat.

• Outdated answer. Circuit from schematic (added after question edit) can't draw 1A. – Kamil Jul 17 '14 at 20:27