Can LM7805 take 30 V max as input if the load current doesn't exceed 50 mA?

Question

Can I use a simple LM7805 for my beeper hobby circuit that monitors a 30 V battery?
The load current doesn't exceed 50 mA. It seems the IC will dissipate (30-5)*50 mA = 1.25 W worst case.
Will the IC survive?

(As of now I'm using a buck converter and it's taking lot of space; since the load is just 50 mA, I realized I could simply use a linear regulator instead)

Appreciate simple alternatives to convert 30 V to 5 V at 50 mA...

One idea:
I have a lot of 16 V 47 uF capacitors.
So can I put two 16 V capacitors in series and connect the combination to 30 V supply in parallel.
Then take the output 15 V from the middle and feed it to LM7805 input.
Will this work?

Min voltage of battery is 22.5 V and max is 29.2 V. This circuit operates on my scooter battery.

Answer 1

Can I put two 16 V capacitors in series and connect the combination to 30 V supply in parallel. Then take the output 15 V from the middle and feed it to LM7805 input.

Not capacitors: they'll charge as you describe, but then you'll just immediately drain the charge from the bottom capacitor, and the 15V will collapse to 0V and that's that.

If the load on the 7805 is limited to 50mA, you'd can just add a hefty dropper resistor on the input. It'll keep most dissipation away from the 7805.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

A TO-220 7805 will survive fine, especially given a small heat sink.

Without a heatsink, the thermal resistance junction to ambient is given as 23.9°C/W so you'll have 30V*5mA + 25V*50mA = 1.4W dissipation, leading to ~35°C rise, so at 50°C ambient your junction will be at 85°C, still relatively safe, if my assumption of maximum ambient and the datasheet assumptions of TO-220 mounting are valid. However, the 30V nominal input is really a bit high for comfort compared to the absolute maximum 35V of the 7805. Adding a 5.6V zener and a polarity protection diode helps with that:

^{simulate this circuit – Schematic created using CircuitLab}

It also reduces the power dissipation (of the regulator) by 0.3W (which, of course, is just moved to the Zener diode).

You could also use a small switching regulator module such as the XP Power SRH05S05 which will survive 72V input transients and is roughly a 10mm x 11mm x 7.5mm cube. There are others from different manufacturers, that is just an example, but your 30V is a bit high for comfort compared to many of them.

Answer 3

No, or maybe, but it is not recommended.

Datasheet for LM7805 says recommended input voltage is up to 25V and absolute maximum is 35V, and so anything above 25V is not recommended, but might work and might shorten the regulator life.

By itself the 1.25W dissipation is a lot but it does not sound too much as it should be handled by a heat sink. Which I guess you have no room for. Without a heatsink the Tja of the TO220 case is about 20 degrees per watt, so regulator internals should be 25 degrees higher than ambient.

Your idea of using capacitive voltage divider won't work because capacitors don't pass DC.

So you can simply put a resistor in series which will drop the supply voltage to be at least 8V at regulator input when 50mA flows through it.

Answer 4

The best solution is a step-down (buck) switcher. But if you want a low-parts-count easy-build linear solution, then Spehro's zener diode or Kuba's resistor (above) are good suggestions. But once you need that part to dissipate over 0.5W or so, then I would consider a series NPN pre-regulator. The NPN transistor in a TO-220 can dissipate 1 watt or so without a heatsink in room temperature environment. Or you can strap it to the same heatsink as the LM317 (both with heat sink insulators) to operate at higher temperatures.

^{simulate this circuit – Schematic created using CircuitLab}

There are more suggestions like this in "7805" and other linear-regulator datasheets.