0
\$\begingroup\$

Newbie question. I have a MCP1700-3302E LDO voltage regulator to power ESP8266 using a 3.7V battery.

However, as soon as I connect the negative lead to the ground pin and the positive to Vin the voltage regulator overheats and starts to smoke.

I tried 2 different MCP1700s thinking maybe the first one was defective. Tried using a different, stable power supply at 5V, same result. Tried with 1uF input capacitor, same.

What am I doing wrong? I just wanted to test the output voltage to be 3.3V.

Thank you in advance.

EDIT 3: Burned a few more, without seemingly any consistency. Some are stable at 4.6V and live, some go up in smoke right away, some burn out slowly at 2.6-3V. One was in expected range of 3.4V but was very hot.

I added a 2K resistor before Vin and input capacitors and it doesn't heat up any more but the output voltage is 0.002V. With a 100 Ohm resistor it's 0.016V, with a 47 Ohm one it's 0.03V (and resistor got hot). So looks like amount of current affects Vout. I'm close to giving up.

EDIT 2: In the setup below, it doesn't overheat anymore. Vout is a steady 4.61V, instead of expected 3.3V

EDIT: Latest setup, fresh LDO, 0.1uF ceramic + 47uF aluminum caps on input, 1uF ceramic on output.

Vin = 5.1V When it on, I'm measuring Vout = 4.6V until the problems start. Vout is supposed to be 3.3V.

enter image description here

\$\endgroup\$
  • 1
    \$\begingroup\$ 1- post a schematic. 2- Make sure that you didn't apply the voltage in reverse (i.e. positive voltage to GND, GND to VIN). \$\endgroup\$ – Rohat Kılıç Jan 10 at 18:22
  • \$\begingroup\$ When it has been smoking it might well be already damaged beyond repairs and no longer suitable to use. \$\endgroup\$ – Oldfart Jan 10 at 18:25
  • 1
    \$\begingroup\$ @MarcusMüller Could I be a victim of AliExpress for the first time? I've ordered most parts from there and never had an issue, but as you are saying, I don't see what could possibly explain what is being observed except those are fake/defective/mislabeled parts... \$\endgroup\$ – Alex Chumak Jan 10 at 19:22
  • 4
    \$\begingroup\$ @SpehroPefhany Despite very positive feedback 4.8 stars, I just saw someone post 2 days ago this comment on the seller's page: "They don't seem to be LDO regulators. I bought 20 and they don't work, they overheat with only 4V even without charge." I think it's time to stop wasting everyone's time. Thank you all! \$\endgroup\$ – Alex Chumak Jan 10 at 21:00
  • 1
    \$\begingroup\$ I think it might be good, from a book-keeping perspective, to write up a quick answer to the question and accept the answer. The answer could basically be that you are presuming faulty parts from seller. That way it is not yet another open question on EESE. \$\endgroup\$ – mkeith Jan 11 at 5:25
4
\$\begingroup\$

It appears that the purchased MCP1700 chips were faulty/damaged/fake/mislabeled. They did not function as they were supposed to and there was no rhyme or reason for it.

|improve this answer|||||
\$\endgroup\$
0
\$\begingroup\$

Needs output capacitor. Put it close to the output + gnd pins of the regulator.

"A minimum output capacitance of 1.0 μF is required forsmall signal stability in applications that have up to250 mA output current capability. The capacitor typecan be ceramic, tantalum or aluminum electrolytic. TheESR range on the output capacitor can range from 0to 2.0."

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ My original attempt did have one. But do you even need it if there is no load? Either way, i tried originally, and just tried adding it again. Same result. Turning power on results in it heating up in a couple of seconds. \$\endgroup\$ – Alex Chumak Jan 10 at 18:42
  • \$\begingroup\$ Was this with a new chip? \$\endgroup\$ – Calum Nicoll Jan 10 at 18:45
  • \$\begingroup\$ Also - it's not written in your schematic. Yes, you do need one on output. What is voltage on Vin and Vout? - i.e measure them. \$\endgroup\$ – Calum Nicoll Jan 10 at 18:47
  • \$\begingroup\$ First of all, adding 1uF cap to output does seem to help. It's hot, and i can smell it but it doesn't start to smoke like before. Vin = 5.1V, Vout at the beginning is bouncing around 2.6-2.7V, after like 5 seconds quickly goes up to 3V, then 4V, then to 0 and back to 2.6ish. Vout is supposed to be 3.3V \$\endgroup\$ – Alex Chumak Jan 10 at 19:00
0
\$\begingroup\$

Strange.

Try testing the component with a multimeter in diode mode between Vin and GND. If you get ~0.7V, that would suggest you've got the pins the wrong way round (somehow the datasheet could have a mistake? Very unlikely, especially since I have a similar working chip - MCP1703 - with the same pinout). Then try some different pin combinations with a 1k resistor in series with the power supply (to avoid blowing up another one) and see if anything will give you 3.3V output. For each combination, you could also measure Vin to get an indication of the impedance of the input (and hence whether it would blow up the chip). If Vin is close to the power supply voltage, it would imply a high impedance and so it should be safe to short out the resistor.

Alternatively, if the multimeter detects a total short (<0.3V) then the chip is probably damaged. It could be a bad batch, or there could have been moisture/ESD damage.

Finally, if there is no diode drop detected, try using the chip with a 1k resistor in series with Vin and see if there is a) any voltage on Vin (if not then your multimeter has lied to you) and b) any voltage on Vout.

Perhaps you could also check for a short in the breadboard; it always turns out to be something silly like that.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ I moved LDOs around the breadboard 3 times while testing. I did get 0.71V between Vin and GND! I think i tried all possible permutations after that and could not get any sensible Vout. Just saw someone post this comment on seller's page, 2 days ago: "They don't seem to be LDO regulators. I bought 20 and they don't work, they overheat with only 4V even without charge." I think it's time to stop wasting everyone's time. Thank you for yours \$\endgroup\$ – Alex Chumak Jan 10 at 20:55
0
\$\begingroup\$

LDO's with very low Ron and high loop gain have very low output impedance until current limiting. The only way you can burn these out when wired correctly is by turning it into an oscillator charging a battery or large C load. If the protection thresholds are oscillating out of phase with the voltage (OCP , UVP) it can be unprotected.

The photo looks like the connections are correct, but the long jumpers are 10nH/cm so with perhaps 100nH of series inductance , with I see 47uF on paper and 100uF+ load on card. This looks like a high Q resonator.

Possible Solutions to overdampen the response add <=1 Ohm between output and load until surge load on startup is reduced.

If you want to use a larger Cap, use up to 5 Ohms in series as allowed in datasheet. This may cause low f load regulation errors or static DC drop.

Twist the wires to lower the impedance of wire directly to output. use AWG 24 > 5 turns/cm** for 120 Ohm impedance roughly estimated from incremental sqrt(L/C) ratio.

  • In future consider dynamics of surge currents on startup and LC resonance and if rated , max load capacitance.

  • This design is not stable.

  • use current limiting R's as a means to safely turn on experiments and keep an eye on hot spots.

  • Add a good adjustable current limiter to your lab supply without bulk C, until you get more experience. on ESR and reactive impedance with high Q factors.

|improve this answer|||||
\$\endgroup\$

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