I have the following connection:

R1= 300 \$ \Omega \$.

The issue is, when I measure the output at the out of AMS1117-3.3, I am getting 10 V instead of 3.3 V.

What can be the problem? AMS1117 can handle up to 15 V. So 12.5 V shouldn't be a problem.

Is it because I have put parallel connection to the LED and resistor? I don't understand.

I have made the changes in the circuit. Now I assume it should work. R1 = 294 \$ \Omega \$ C1 = 100 nF C2 = 10 uF Regulator is LD1117v33.

enter image description here

  • 5
    \$\begingroup\$ You must have capacitors on the input and output, as in the data sheet. \$\endgroup\$ Dec 2, 2016 at 11:28
  • 4
    \$\begingroup\$ You didn't include any of the capacitors the datasheet specifies?! \$\endgroup\$ Dec 2, 2016 at 11:30
  • 4
    \$\begingroup\$ Minimum load current is 5 - 10 mA. Was there any load during the measurement? \$\endgroup\$ Dec 2, 2016 at 11:33
  • 10
    \$\begingroup\$ The LED is the wrong way round. \$\endgroup\$
    – Andy aka
    Dec 2, 2016 at 11:37
  • 11
    \$\begingroup\$ @shantanu please don't argue with me - either the LED is the wrong way as depicted in your circuit or you are powering the circuit with the supply reversed. \$\endgroup\$
    – Andy aka
    Dec 2, 2016 at 12:16

2 Answers 2


To be clear and to protect against changes to the question, here is your schematic:

There are a number of problems here, both with the schematic and the circuit:

  1. The schematic is missing junction dots. It's a convention we expect everyone here to follow. Don't be so lazy.

  2. The logical flow in the schematic is right to left. It took me a little while to realize that the 12 V was being fed into the AMS thing.

  3. Text is overlapping connections and other drawings. Don't be so sloppy. You wouldn't hand in homework like this, so it certainly doesn't belong here. Your teachers get paid to look at your homework. We don't.

  4. You don't say what the purpose of the LED is, but if it's supposed to light when the 12 V power is on, then it's backwards. If not, then it is almost certainly being abused by too much reverse voltage. LEDs are usually only good for about 5 V in reverse. If backwards, then it may be getting abused due to excessive current. The current is over 30 mA. Some LEDs can handle that, but typical indicator LEDs are good to only 20 mA.

  5. There is no input capacitor, assuming this ASM thing is a linear regulator. Not having a cap close to the input can lead to problems in many regulators. What does the datasheet say?

  6. There is no output capacitor, assuming this ASM thing is a linear regulator. That is almost certainly violating minimum requirements that are surely mentioned in the datasheet.

In short, this is a sloppy mess. Pay attention, read the datasheet, and stop trying to cut corners. Neatness and attention to detail matter, both in design and in presenting to others. Until you take your work seriously, you can't expect others to.

  • \$\begingroup\$ There are two problems here, one with the text of the answer, another with the content. 1) That "AMS thing" or "ASM thing" is actually an "AMS1117-3.3". It reads right there in the image you quoted, and I've no idea why you refer to it with two distinct incorrect names. Don't be so sloppy. 2) AMS1117-3.3 is a "1 A Low dropout voltage regulator", which I found out with a trivial google search. You could have done that yourself. Don't be so lazy. \$\endgroup\$
    – ilkkachu
    Jan 11, 2018 at 22:17
  • \$\begingroup\$ @ilk: No, I shouldn't have to Google a part number just to find out what it is. It's the OP's job to provide a link to the datasheet, and to give rough description right in the question. Remember that the people answering are all volunteers here. \$\endgroup\$ Jan 11, 2018 at 22:22

You certainly need a capacitor at the output. From dataheet.


The circuit design used in the AMS1117 series requires the use of an output capacitor as part of the device frequency compensation. The addition of 22 μF solid tantalum on the output will ensure stability for all operating conditions.

When the adjustment terminal is bypassed with a capacitor to improve the ripple rejection, the requirement for an output capacitor increases. The value of 22 μF tantalum covers all cases of bypassing the adjustment terminal. Without bypassing the adjustment terminal smaller capacitors can be used with equally good results.

To further improve stability and transient response of these devices larger values of output capacitor can be used.

Also there is minimum current requirement of 5 - 10 mA.

enter image description here

Note 5: Minimum load current is defined as the minimum output current required to maintain regulation. When 1.5V ≤ (VIN - VOUT) ≤ 12V the device is guaranteed to regulate if the output current is greater than 10 mA.

To achieve a simple load at the output use a LED and a resistor. Modify your circuit as follows:


simulate this circuit – Schematic created using CircuitLab

  • 4
    \$\begingroup\$ This is the right answer. Capacitor(s) + minimum current. The AMS1117 is very similar to the ubiquitous LM1117, and very popular in China. Moving the LED to the output and adjusting the resistor value (and LED orientation) could provide the necessary minimum load current to keep the output from rising out of regulation. \$\endgroup\$ Dec 2, 2016 at 19:41

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