1
\$\begingroup\$

I am kind of curious about this because for my project I have three power lines (12,5,3V3) but each one needs it own special voltage regulation designed for the current consumption.

I know there are three kind of voltage regulation:

  • Voltage divider
  • linear regulators
  • switching regulators

But none of these seem to care what the input current allows them to draw and "generates" their own limit.

Example of what I mean: I regulate a power adapter Vin with 12V/5A to 5V. My 5V power lines needs about 2A so I put in a small switching regulator like an LCW78. Last but not least I need to power my MCUs which need 3V3 and about 500mA - 1A, so I use a cheap AMS1117 linear regulator for that.

Here is the thing, ESP. With the linear regulator, that surprises me. If I don't use an AMS1117 but a cheaper Chinese 4B2x regulator that only supplys up to 150mA my MCU has a problem.

Why, though? In theory my 5V power lines has enough "power" to support the MCU as well but the regulator stops it.

\$\endgroup\$
4
  • \$\begingroup\$ It seems reasonable that a 10A regulator might be bigger and more expensive than a 2A regulator. If your load only required 1A, which one would you choose? Note: your esp32 requires more than 150mA. \$\endgroup\$
    – Kartman
    Commented May 31, 2021 at 7:31
  • \$\begingroup\$ I'd say it doesn't draw up to 150 mA, but it can supply up to 150 mA. \$\endgroup\$
    – Arsenal
    Commented May 31, 2021 at 7:34
  • \$\begingroup\$ @kartman obviously the cheaper one. But that isn't exactly the question. I am wondering why if I already have a 2A (5V) regulator on the board why does it not matter for the esp because it has it own regulator. And yes I know the esp needs lot more that is the problem but the board I bought had this cheap regulator on it. But that has nothing to do with the question itself.@arsenal true, changed it. \$\endgroup\$
    – Daniel Do
    Commented May 31, 2021 at 7:44
  • 5
    \$\begingroup\$ A voltage divider is not a regulator. \$\endgroup\$
    – Andy aka
    Commented May 31, 2021 at 8:14

3 Answers 3

1
\$\begingroup\$

Your voltage regulators draw the current they need to draw in order to perform their functions.

Let's take a 3v3 linear regulator for example: Its job is to maintain a fixed output voltage. But by design of such regulators (and omitting the small little intricated details that are not the point here) the input current must be equal to the output current. Thus, if your 3v3 MCU draw 500mA at some point in time, the regulator will output the 500mA and also draw 500mA at its input.

Now what happens if the regulator is not able to give the 500mA, because you choose one that in only rated for 150mA for instance ?

If you have 12V at the input and 5V at the output and the input current must be equal to the output current. Then, when some current flow through the regulator (let's say 500mA), the input power is 12V x 500mA = 6W and the output power is 5V x 500mA = 2.5W. Because of the conservation on energy principle, in a system, Pin = Pout. It's not the case here, thus there is a additional Pout : Power dissipation through heat. Here our regulator must dissipate an excess power of 6W - 2.5W = 3.5W

The regulator will heat up. Eventually to a point where is will fry. Or if it's a model with a thermal shutdown feature, it will power itself and its load off until it cools down and then restart, to re-heat up and shutdown etc..

\$\endgroup\$
1
\$\begingroup\$

Here is the thing, ESP. With the linear regulator, that surprises me. If I don't use an AMS1117 but a cheaper Chinese 4B2x regulator that only supplys up to 150mA my MCU has a problem.

It's a linear regulator so its output current is limited both by the size of the transistor inside the chip that actually passes the current, and by maximum thermal dissipation. Also these regulators usually have a current limiter to protect the internal pass transistor.

enter image description here

(LDO internal schematic from Diodes)

If the regulator has a maximum current of 150mA and the ESP draws current above that, output voltage will drop, which is the normal and expected behavior.

\$\endgroup\$
1
\$\begingroup\$

Each regulator (I do not count voltage dividers as regulators) must handle the current required by the device. That is, it must pass that current.

Take a look at your linear regulator. If it has to supply 5A at 3.3V from 5V, then it has to waste ((5-3.3)*5)= 8.5 watts of power as heat. That requires a large heat sink to get rid of the waste heat. The small linear regulators limit the current (or are only rated for low current) so that they don't burn up from waste heat or require a huge heat sink.

Take a look at a switching regulator. It uses an inductor to convert the voltage. Large output current would mean large currents through the inductor. To handle that properly, the inductor must be larger and/or made of different materials. The result is a regulator that is too large for convenient use, or too expensive, or both.


Those are just a couple of examples of reasons why a regulator will only be capable of supplying limited current. There are other reasons as well.

In the end, though, it comes down to cost, efficiency, and size.

Regulators should be designed for the expected load so that they are efficient, as small as reasonable, and reasonably priced.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.