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I'm currently designing a simple PCB that uses a MAX3221 to send/receive RS232 over an ethernet cable along with an optional 1-wire thermocouple reading (The thermocouple breakout can be soldered in depending on a users needs).

enter image description here Note: I forgot to add the Cap values to the diagram but they are all 0.1 μF

The end device i'm targeting has 4 lines that i'm trying to tap into (24 VDC, GND, Tx, Rx). My idea was to push the 24 V through a LDO and use that to provide a source of 3.3V to the parts where it's needed (E.g. MAX 3221 & Thermocouple breakout). What i'm finding though is that the LDO I'm using (LM1086 3.3V) is getting incredibly hot. My continuity checker doesn't show a connection between output and GND when unplugged, but the moment I plug the device in it bridges.

The system works fine when sourcing a 3.3 V line from elsewhere (E.g. cheating it from the raspberry pi I have attached on the other side of a different transceiver). Strangely, I have managed to solder up a couple of PCB's with an LM1086 & BA033CC0T with bent pins that are working.

Is there an issue in my circuit design? I have a feeling that I'm fundamentally missing something with the LDO that's causing this issue. Or could it simply be a matter of faulty parts or poor soldering on my behalf?

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4 Answers 4

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If you make 3.3V from 24V with a linear regulator, you will dissipate about (24V-3.3V) * drawn current = 20.7 V * drawn current. So, for e.g. 100 mA, this is 2.07W.

Although the LM1086 has a thermal protection and won't get easily damaged, the thermal protection will reduce the output current or even just shut down the regulator. In both cases, it is undesirable.

You'd better use a switching regulator.

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  • \$\begingroup\$ Thanks for the help. I believe the max Iout is ~22mA if i've not messed up reading the various datasheets. With that value it's only 0.5W, which leads me to believe it shouldn't be too hot to touch? I've taken a DMM to the Vin and it looks like it can vary up to ~27V which might be the source of the issue? I'll be sure to take a look at switching regulators. I wasn't tied to LDO, they just seemed cheaper and easier to implement (My electronics knowledge isn't the best so the extra pins were daunting haha). \$\endgroup\$
    – Cinfra
    Feb 17, 2020 at 14:04
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Are you sure you're using the 3.3V version, and not the adjustable version? This is not clear/mentioned in your schematic.

You are using a linear regulator. The dissipated heat will be (24V - 3.3V) * Iout, so the output current multiplied by 20,7 equals the dissipated heat.

Last but not least, a 0.1uF capacitor is not suitable to be used as filter capcaitors for the regulator. The datasheet recommends 10uF capacitor.

See the datasheet for more info about adjusting the output voltage, and for information about choosing suitable capacitors.

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  • \$\begingroup\$ My apologies for the lack of info. I don't think i'm using the ADJ version. My packaging shows LM1086IT-3.3. As far as the Iout goes, if I'm correct in my estimations, it should relatively small at a max of 22mA. My guess would be that I wrongly assumed the max output from Vin? The datasheet says 24V, but taking my DMM to it, i've just seen it hovering close to 27V. Could the proximity to the LDO tolerance also explain why I've had a couple successful PCB's? \$\endgroup\$
    – Cinfra
    Feb 17, 2020 at 14:01
  • \$\begingroup\$ Also, thanks for outlining the cap values. I think i'd mistakenly read it wrong in the datasheet. \$\endgroup\$
    – Cinfra
    Feb 17, 2020 at 14:02
  • \$\begingroup\$ I wouldn't know if the proximity to the tolerance explains your couple succesfull PCB's. You could search for switching regulators. Take a look at TI's products. Also, in most datasheets you can find a chapter called typical applications. There you can find some good info about which additional components you need, and how to choose those components. \$\endgroup\$
    – David
    Feb 17, 2020 at 14:30
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Possibilities:

  1. LM1086 has abs max Vin of 27V. Be sure 24V supply is always < 27V

  2. You do not say what max current is at 3.3V or LDO regulator package or heatsinking.
    24V to 3V3 is a LARGE drop. You dissipate 100 mW per 4.6 mA of current drawn.

  3. Minimum capacitance values are specified in the datasheet. They may be larger than expected.

  4. LM1086 needs minimum ESR in Vin and Vout caps for stability. With ceramic, & probably the evil tantalum you will need series resistors!. Maybe also with Al electrolytics. See datasheet 7.4.2
    Instability can lead to wholly unpredictable operation - and destruction of powered devices, worst case.


"7.4.2 Stability Consideration Stability consideration primarily concerns the phase response of the feedback loop. In order for stable operation, the loop must maintain negative feedback. The LM1086 requires a certain amount series resistance with capacitive loads. This series resistance introduces a zero within the loop to increase phase margin and thus increase stability. The equivalent series resistance (ESR) of solid tantalum or aluminum electrolytic capacitors is used to provide the appropriate zero (approximately 500 kHz).

Aluminum electrolytics are less expensive than tantalums, but their ESR varies exponentially at cold temperatures; therefore requiring close examination when choosing the desired transient response over temperature. Tantalums are a convenient choice because their ESR varies less than 2:1 over temperature.

The recommended load/decoupling capacitance is a 10 uF tantalum or a 50 uF aluminum. These values will assure stability for the majority of applications.

The adjustable versions allows an additional capacitor to be used at the ADJ pin to increase ripple rejection. If this is done the output capacitor should be increased to 22uF for tantalum or to 150 uF for aluminum.

Capacitors other than tantalum or aluminum can be used at the adjust pin and the input pin. A 10 uF capacitor is a reasonable value at the input. See Ripple Rejection section regarding the value for the adjust pin capacitor.

It is desirable to have large output capacitance for applications that entail large changes in load current (microprocessors for example). The higher the capacitance, the larger the available charge per demand. It is also desirable to provide low ESR to reduce the change in output voltage: ΔV = ΔI x ESR (2)

It is common practice to use several tantalum and ceramic capacitors in parallel to reduce this change in the output voltage by reducing the overall ESR. Output capacitance can be increased indefinitely to improve transient response and stability."

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Thermal considerations are very very important for the proper implementation of Low-Dropout regulator designs. Texas Instruments dedicates a full section about this topic in their datasheet page 19 (https://www.ti.com/lit/ds/symlink/lm1086.pdf). Do you know how much current is your circuit pulling from 3.3V?

Let's take a simple example and assume the LDO is loaded with 50mA output and that it has no ground current (Ig = 0). The power dissipated by the LDO would be:

PD = (24 - 3.3) x 0.05 ~ 1W

Now let's check the table on page 5: enter image description here

The "junction-to-ambient thermal resistance" which depicts the quality of the package thermals in specific conditions (ground plane size, mounting, etc.. see link below the table) tells you that the temperature of the case will be:

Tcase = Tamb + Roja x PD = 20 + 40 * 1 = 60C

(assuming Tambient = 20C and that you are using the KTT package)

60C is pretty hot to the touch already!

You can do your own calculation to determine what is the expected case temperature in your case, I was just giving an example. Also do read the "10.3 Thermal Considerations" section, there are tons more information there :)

There is a second thing I wanted to mention as you said you are exclusively using 0.1uF capacitors and I imagine they are all Ceramic/MLCC capacitors. In the LM1086 datasheet, they speak about the LDO stability ("7.4.2 Stability Consideration"), make sure to take a look at it, it is possible your LDO output isn't completely flat (have you looked on an oscilloscope?). Texas Instruments recommends using either Aluminum electrolytic or Tantalum capacitors to keep the stability in check, and you may also want to bump the input capacitor value a bit.

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  • \$\begingroup\$ It is also useful to note that the OP appears to have simply soldered the pins to wires; the thermal information in the datasheet is based on the device being thermally mounted on a PCB, so the actual thermal resistance is likely to be quite a bit higher. TI has a report on how they measure this: ti.com/lit/an/spra953c/spra953c.pdf \$\endgroup\$ Feb 17, 2020 at 12:49
  • \$\begingroup\$ Thanks @PeterSmith, yes it is the link I was referring to, below the thermal specifications, now added as a hyperlink. I did not know that person is mounting the LDO using wires, I can only assume what I read in the question :) \$\endgroup\$
    – eeintech
    Feb 17, 2020 at 12:55

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