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# LDO heating up, neutral switcher state

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#### makowka

##### Newbie level 6
Hi

I want to build a circuit like that shown in the picture. I haven't assembled it yet, instead I have made a small test circuit for the LDO to measure its correct operation. I have noticed one thing, if I don't apply the voltmeter terminals to the circuit, the LDO gets very hot with the danger of burning up. As you can see on my circuit, in the neutral switcher state the LDO doesn't have any current consumer to convey its current to. The batteries will be all the time inserted, of course. Do I take a risk of burning the LDO?

With no load the LE33 should be cold. I use them all the time and they seem very reliable. Check you have it's connections the right way around, with the flat side facing you and the pins pointing downward, the legs are OUT, GROUND, IN.

Bear in mind the heat produced is: output current * (input voltage - 3.3) so be careful if the input voltage is high.

Brian.

I think, that the capacitors make sure, that in the switcherless state the LDO doesn't burn out, some kind of oscillation. Am I right here?
On my test circuit, I didn't apply the capacitors, maybe that is why the LDO was immense warm.
Another relevant question is, if I have 3 AAA batteries and the LDO immediately after them, what current would flow to the LDO input? Maybe I would need a current limiting diode behind the batteries to limit the input current? I mean, the input current is unregulated, the LDO passes 100mA on the output. But that contradicts the traditional formula for the heat dissipation of an LDO: P=deltaU*I, in my case: (3,6V-3,3V)*0.1A = 0,03W.

There is no such thing as a "current limiting diode" so forget that! The LE33 has a 0.2V drop-out voltage so as long as you put at least 3.5V into it, it should work.
The regulator itself draws only 0.5mA so it will not cause any significant heating. The data sheet does state you must use at least 2uF across it's output so it probably was oscillating and that could account for the extra current.
I'm not sure what you mean by "input current is unregulated" as these are voltage regulators and the input current will be the output current + 0.5mA while in regulating state.

Brian.

Ok, I understand, that the input current will be output current + quiescent current. I must say, in the test circuit I didn't employ the capacitors, so I can't say, how the LDO would behave in a correct setup. Also, I am not sure, if the LE33CZ max output current is 100 mA (look at the title page) or 150 mA (page 9, specs for LE33C).

https://www.hobbytronics.co.uk/datasheets/LE33CZ.pdf

Below you will see my test board, the resistor in the board is supposed to resemble a voltage meter.

#### Attachments

• testcircuit.png
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They are designed to work at up to 100mA and have a current limit of 150mA, provided the power dissipation isn't exceeded.
You MUST fit the capacitors for it to work properly.
Your schematic shows the input and output pins reversed. The pin numbers are correct but they are shown backwards on the LE33 body. Check you have them the right way around!

I use LE33s on radio transmitter units which are solar powered and the input can be anything from zero to about 10V. Even at the highest voltage they run cold.

Brian.

makowka

Points: 2