# Best way to drop 10volts at 0.25A with linear regulator?

What would be a minimalistic circuit design to convert 14V to 3.3V at 0.25A with a linear regulator? I'm thinking about two linear regulators.

I've tried LM7805(14V-5V) to LD1117(5V-3.3V.) LM7805 gets hot even at 0.1A. What would be best solution without switching power supply? I'm trying to power efficiently ESP32 module from a car's battery.

• Do you prefer power FETs or Power Darlingtons? – Tony Stewart Sunnyskyguy EE75 Jan 17 at 4:04
• Darlingtons for sure – TheAfrizzz Jan 17 at 4:17
• Whatever you do, it's going to get hot. Choosing a heat sink will be an important part of this design. – The Photon Jan 17 at 4:25
• Linear regulators will all have about the same miserable efficiency as the 7805 + LD1117- less than 24% efficient. Total power is a bit more than (Vin - Vout)*Iout = 2.7W, which will require a substantial heatsink if the 0.25A is continuous. You could spread it among more or fewer regulators but the inefficiency will remain. – Spehro Pefhany Jan 17 at 4:30
• BTW, where does the 0.25A figure come from? If you're using WiFi on the ESP32, it may require more than 0.25A (I've covered that in my answer, but I'm curious as well) – anrieff Jan 17 at 7:53

you're planning to burn up 2.5 watts a TO220 part is going to need a heatsink at that power level, but you can put a resistor in series with the input and make the heat there where it's not damanging anything

simulate this circuit – Schematic created using CircuitLab

at low current the reguilator sees near full voltage, but the current is low so there's not much heat made. at 250mA R1 is dropping about 8V so the regulator is only working with 6V in and so making less that 1W of heat

• Is it possible to calculate r1 discipated heat somehow? Your solution looks good, thanks. – TheAfrizzz Jan 19 at 2:13
• As a first approximation assume all the current out of the regulator and 5mA more pass through the resistor. $P= I^2 R$ – Jasen Jan 19 at 5:16

Best? no , Simple Yes

Get a junkyard Festoon lamp holder but instead of a 12V bulb, buy a 24V 10W bulb designed for 250mA constant current with just enough voltage drop to run any 3.3V LDO and only dump < 0.5W in the LDO and 2.2W out of 10W in the Lamp, so it will be dim.

If the bulb is 10W rated at 24V it will be 3 Ohms cold and if rated at 28V, it will be 4 Ohms cold and you may need to bypass it with two 470 Ohm resistors in parallel across bulb.

• Came here to put the same suggestion then saw this :) – Rohat Kılıç Jan 17 at 5:43
• I wanted to do this in my 1st job out of school, my mentor said, sorry not rugged enough for 15g vibration spec for aerospace – Tony Stewart Sunnyskyguy EE75 Jan 17 at 5:56

## In a word: don't

The best way to drop 10 volts at 0.25A is already answered by other people, so I will not cover that.

Your question, however, aims at powering an ESP32 module specifically, and you don't disclose more details about the intended usage. However, if you are targeting an IoT scenario with WiFi communication, you need to consider that the ESP32 does NOT consume 0.25A. It's power usage varies with the distance to the AP, and the required increase of transmit power when RSSI falls down due to distance.

ESP32 supplies are typically rated for 500 or 600mA, for this reason. Personally the highest I've seen is 450mA, but your mileage may vary, and you may have other things on board, so... do the math.

If I were you, I'd use a cheap car USB supply (12-to-5V) with 600mA rating or more (which likely features a switching regulator inside), and use a LDO to drop the remaining voltage to 3.3V. You can go with 12 to 3.3V supply as well, but the 5V supply gives you the option to add a Li-Ion battery later, and the 5V supplies are more common.

I'd use a circuit similar to one given in Tony's answer.

simulate this circuit – Schematic created using CircuitLab

$$\D1\$$ (ES2J - but can be anything suitable) protects the circuit against reverse polarity. Power rating of both $$\R2\$$ (470R) and $$\D1\$$ (3V9 zener) should be at least 1W because their power dissipations increase under no- or light-load situations. $$\R3\$$ (1k) ensures the collector current of $$\Q1\$$ to be non-zero and thus the unloaded voltage to be 3.3VDC.

$$\DROPPER\$$ can be a resistor with enough wattage, or a 12V car lamp or even an ATX chassis fan.