LDO heating up

[makowka]

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?

 

[ronsimpson]

I can't see what you are doing. I think that regulator is oscillating with no load. Are the capacitors close to the regulator. 1 inch or less?

Normally we try not to parallel LEDs. You can get away with it some times. LEDs do not share current well. Use a resistor for every LED.

 

[makowka]

Hence the PCB is small (diameter 1,3 inch), the capacitors are close to the LDO. So, in switcherless state, the LDO is "discharging" through the capacitors, in order not to heat up til burnout?

I wanted to arrange the LEDs in parallel, since my power supply is only 3,6V, one led consumes 3,2V. Am I right here? If I arrange two leds in series, I would need a voltage of 6,4V for each branch of two leds. So, that would question the portability of my device, since I would need some sort of external battery pack.

 

[ronsimpson]

I wanted to arrange the LEDs in parallel, since my power supply is only 3,6V, one led consumes 3,2V. Am I right here? If I arrange two leds in series, I would need a voltage of 6,4V for each branch of two leds. So, that would question the portability of my device, since I would need some sort of external battery pack.

What you have is 1 resistor and two LEDs.
What you need is (1 resistor & LED)+(1 resistor & LED)

If you only make a couple it may not matter. In production you will find some LEDs are 3.0V and some 3.1 and some 3.2, 3.3, 3.4. If you parallel a 3.1 and a 3.3 LED the 3.1 will take the great majority of the current and the 3.3 will only get 1/4 the current.

 

[makowka]

OK, I already have changed my setup to have one resistor before each led. I think, with the capacitors the ldo won't heat up that much, since the dropout is 0.3V, max current 40mA. In my test setup I didn't use the capacitors, that may be the reason for overheating.

 

[ronsimpson]

In my test setup I didn't use the capacitors, that may be the reason for overheating.

The data sheet indicates that capacitors must be used.

 

[makowka]

Yes, that have must been the cause. I hope with the capacitors the LDO won't overheat. Thank you for the quick reply.

 

[MikeMl]

If all you need to do is light the LEDs, why bother with the LDO at all? Just size the resistors as required for the battery voltage and dispense with the regulator...

 

[makowka]

I want to protect my rechargeables from a damaging discharge - when the voltage in one of the batteries falls down below 3,1V, the LDO should shut the whole circuit off.

 

[ronsimpson]

With a 3.1V supply and a 3.1V LED the LED will be very dim.

 

[makowka]

Excuse me, it should read: "when the voltage in one of the batteries falls down below 1.1V, the LDO should cut the whole circuit off. " So, the batteries will be protected from over-discharging. We have still 3,3V output from the LDO, until the dropout level isn't reached.

 

[ronsimpson]

Excuse me, it should read: "when the voltage in one of the batteries falls down below 1.1V, the LDO should cut the whole circuit off. " So, the batteries will be protected from over-discharging. We have still 3,3V output from the LDO, until the dropout level isn't reached.

How many batteries are in series?

 

[MikeMl]

A regulator will just drop out of regulation, but keep delivering the battery voltage to the load, minus a few millivolts. You need a Voltage sensing switch...

 

[makowka]

There rare 3xAAA batteries in series. Isn't the LDO supposed to switch to the GND if the dropout is being reached?

 

[alec_t]

In your schematic the battery is the wrong way round! Unless the regulator has reverse-voltage protection (I haven't checked its datasheet) if you've actually wired up your circuit as shown then it's not surprising it overheats

 

[makowka]

Yes, I noticed, but in my test setup I have connected the batteries the right way. Just a typo.

 

[MikeMl]

There rare 3xAAA batteries in series. Isn't the LDO supposed to switch to the GND if the dropout is being reached?

Post the data sheet (or a link) of your LDO and we will look at it. Most do not.

 

[makowka]

Here it the datasheet:

https://www.electro-tech-online.com/custompdfs/2013/02/LE33CZ.pdf

I wonder, what is then the function of the dropout voltage? Isn't it supposed to be, that when the voltage falls down below the dropout level (in my case, 3,3V-0,2V=3,1V), the LDO shuts the circuit down, switching to the GND output?

 

[MikeMl]

The DropOut voltage of a regulator is the minimum differential voltage between the input pin and the output pin for the regulator to work properly (i.e regulate). The name comes from the regulator "dropping out" meaning to stop regulating.

For a 3.3V regulator in this series, the max Vdo is 0.4V, so that means to get 3.3V at the output, the minimum input voltage has to be 3.3+0.4 = 3.7V. Not stated on the data sheet, but based on experience, if the input voltage falls below 3.7V, the output will always be ~0.4V lower than the input, sort of like connecting a series Schottky diode between input and output. As the input voltage drops, the output will always be ~0.4V lower. There is no magic input voltage at which point the regulator shuts off the load... It might happen, but the manufacturer doesn't test for it, or specify it on the data sheet.

You might be able to use the ShutDown pin (with some external circuitry) to shut off the load based on the voltage of the discharging battery.

 

[makowka]

At least I will have a stable voltage supply to the leds. My former assumption was wrong, it should read: 3,3V+0,2V=3,5V, which is 0,1V below the supply voltage of 3,6V. So, when the voltage of one of the batteries in series falls to 1,1V, the LDO will cease to regulate? What voltage will I have then on the output, full 3,5V?

Another question, can I place the via like shown in the picture, direct on the wire?