2
\$\begingroup\$

Update :-(:

I tested the pinouts of the IC with a multi meter and the connections are fine. Then I discovered I had done a very stupid mistake and connected Vout where Vin should have been connected. I corrected it and now there is no difference between input and output.
So problem discovered, and is my board toast?


I bought this. Its supposed to give out 5V DC. My input is 7.13V DC. Why is it giving out 6.34V?

Power is coming from a 12V adaptor, stepped down to 7.13V via a buck convertor before the linear regulator so that the AMS1117 doesn't have to do too heavy lifting. I'd expect it to then give 5V.

The images say on the PCB its a YL-46, whatever that means, but the thing which came had HW-122 printed instead.

Here's a pic of what I've got. Actual part

\$\endgroup\$
17
  • 1
    \$\begingroup\$ @Justme question edited. I have soldered everything, no wrong connections done, checked again. \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:00
  • 1
    \$\begingroup\$ More info added. \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:13
  • 2
    \$\begingroup\$ @SpehroPefhany yes the MM is working ok, I use it everyday and the 12V from the adaptor is showing fine. \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:45
  • 1
    \$\begingroup\$ @RussellMcMahon I only connected the AMS1117 to the buck convertor after setting the BC to 7.13V. I have pasted an image of the actual part in the question. There is nothing on the back side, its completely blank, no traces even. \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:47
  • 2
    \$\begingroup\$ @JohnSnow I had nothing to do with your question being closed. || So far you have failed to address my major important question despite it being the basis of my original answer and my having specifically asked it twice. it MAY be what is wrong. It may not. You can check it in a minute and report back. || I provided the pin outs for the regulator. Using an ohm-meter: Does regulator Vout Vin Ground - connect to PCB Vin Vout Ground correctly? || Also - ensure MINIMUM 10 mA load as others have noted. ||Please report back. \$\endgroup\$
    – Russell McMahon
    Sep 3, 2023 at 22:15

3 Answers 3

2
\$\begingroup\$
  1. Absolute maximum Vin.

The AMS1117 Vin specification varies with supplier.
Some have 18V and 30V Vin abs max ratings (see data sheet refs in my answer) BUT some have 12V abs-max Vin ratings.

The one in your cited ad says 12.5V in large red letters.
If you have allowed even a sniff of >= 12V on Vin the regulator may be dead.

I've seen the 12V max spec cited on low cost Arduino clones from China. Interestingly the Chinese LCSC supplied data sheet says 30V.


  1. PCB fault?

The PCB MAY have been incorrectly made.
Looking at the two versions in the ad that you cited it's not clear that the regulator connects correctly. This seems unlikely but is easily checked. Check viaually and/or use an ohm-meter to see if the IC pins connect to Vin, Vout, Gnd as expcted. see diagrams below for probably AMS1117 pinout.

Is the IC on your board an AMS1117?

This is the pinout shown in several datasheets that I found online.
Left to right, tab up, Ground Vout Vin - as shown in red at the bottom of the image below.

The two images are from the page you cited BUT show different PCBs.
What are the actual connection to Vout and Vin on your PCB.

If these are in fact in erro a little 'surgery' with a scalpel or similar will allow correction.

enter image description here

AMS1117 datasheet here

enter image description here


and here

enter image description here

  1. Minimum load

It has been suggested that the AMS1117 requires a minimum load of 2 mA typical, 10 mA max. This is true of the adjustable versions. It may not be true of the fixed voltage versions. Adding a 10 mA load when testing would do no harm.


REVERSE POLARITY PROTECTION:

The circuit below uses one P Channel MOSFET and one resistor to provide protction against reverse power supply polarity.
The PFET has the drain and source swapped compared to more normal use. This means that the FET's body diode conducts when the polarity is correct - not something that is wanted when the FET is used as an on/off switch.

The FET is turned fully on when the gate is pulled adequately negative relative to the source by R1. This bypasses the body diode and FET on resistances in the milliohm range are achievable at modest cost.

When the supply is reversed the body diode does not conduct, and R1 now pulls the gate positive relative to the source. As it is a P Channel FET this turns it off.

If Vin is near or above Vgate-source-max an additional resistor can be added to reduce the gate-source voltage to a safe level.

I added this system to a battery powered product I designed of which several hundred thousand were built. The Chinese manufacturers preferred this method over the use of mechanical polarity protection.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
16
  • \$\begingroup\$ @winny I suggest that your comment better belongs on your answer or on the question. || 106 means 10 uF and 104 means 0.1 uF in this context. Thos values suggest that the designer was at least half competent so my "wrong PCB": answer may be wrong :-) \$\endgroup\$
    – Russell McMahon
    Sep 3, 2023 at 12:12
  • \$\begingroup\$ I don't see this being incorrect. Can you describe or show which part you think is incorrect? To be fair, the tracks are hidden under the component so I can't verify it being correct either. \$\endgroup\$
    – Justme
    Sep 3, 2023 at 12:17
  • 1
    \$\begingroup\$ You're going to have to be more explicit about where you think the error is. The PCB looks fine to me, at least as far as not being able to see what's under the components or the other side (which I'm guessing is just all ground plane). \$\endgroup\$
    – Dave Tweed
    Sep 3, 2023 at 12:17
  • \$\begingroup\$ @DaveTweed I may well be wrong. Looking at the PCB I cannot be sure that some tracks join where they might. As two different PCB layouts are shown in the ad it is not certain which if either is his. I added a comment on his question re testing this - I'll add it to my answer. \$\endgroup\$
    – Russell McMahon
    Sep 3, 2023 at 12:25
  • 1
    \$\begingroup\$ @JohnSnow 35 countries so far. Id love to spend more time in India but quite likely never will (and many other places). My wife worked in Pune for 3 months. \$\endgroup\$
    – Russell McMahon
    Sep 5, 2023 at 13:00
5
\$\begingroup\$

It all looks okay to me. The AMS1117 is not guaranteed to be stable with a ceramic capacitor load. They recommend a 22uF tantalum capacitor in the datasheet.

That 'rule' is broken in many low-cost modules such as the one you have, and I've not heard of it causing an issue, however I don't think we can rule it out.

If you have a 22uF or thereabouts tantalum capacitor you could try replacing the output '106' capacitor (taking care as to polarity- tantalum caps usually have the + side marked rather than the negative as on aluminum electrolytic caps). Or replace it with an aluminum electrolytic of 20uF or greater.


Another (perhaps more likely) possibility is that you might have damaged the regulator accidentally by connecting the input in reverse, even for an instant. Using two red wires for both + and - makes that an easier accident to happen. There is no reverse voltage protection on this chip. The upstream buck regulator would limit the current so the damage would not be as visually (or olfactibly, or audibly) evident as if the module was (say) connected the wrong way across a car battery.

\$\endgroup\$
2
\$\begingroup\$

1117 requires a minimum load of 10 mA to regulate. It’s unclear if your module provides that or not.

enter image description here

Try hooking up a 470 ohm resistor on the output and check again.

\$\endgroup\$
7
  • \$\begingroup\$ Can I try 1K? Might not have 470. And this is to go between +ve and -ve, right? I sound like a novice because I am. \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:01
  • \$\begingroup\$ Sure. Or two in series. Or stack on top of the 1k to the LED to increase the current. \$\endgroup\$
    – winny
    Sep 3, 2023 at 12:07
  • \$\begingroup\$ @winny Minimum load usually refers to the adjustable part to allow for reference pin current. I do not know which datasheet you are using but eg this one notes that the current is for the ...-ADJ part. What does your note 5 say? \$\endgroup\$
    – Russell McMahon
    Sep 3, 2023 at 12:23
  • 1
    \$\begingroup\$ soldered 1K between output +ve and output GND, nothing changed \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:30
  • \$\begingroup\$ check out my most recent edit to the question also \$\endgroup\$
    – John Snow
    Sep 3, 2023 at 12:31

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.