1
\$\begingroup\$

I have several MCP1700-3300E LDO regulators that I plan to use. The datasheet claims that this LDO has:

Low Dropout (LDO) Voltage- 178 mV Typical @ 250 mA for VOUT=2.8V

And the dropout plot for 2.8V:

enter image description here

Actually, I am using 3.3V variant, not the 2.8V one, but unfortunately the datasheet doesn't provide dropout plot for 3.3V variant. Nevertheless, dropout should be similar, right?

So, I rigged up a simple schematic. My PSU feeding 3.7V to the MCP, and a 22 ohm resistor between the output of the MCP and the ground to give it a load. Running, my PSU says the current is 128 mA, around half of the maximum this LDO can handle. But measuring the voltage between the output and the ground of the MCP gives 2.97V, which is a 0.73V drop - far more than the datasheet suggests. This is a schematic of the exact setup I have for testing:

enter image description here

So, what am I missing here? According to the datasheet, I should be getting ~ 0.1V dropout, so the output should still be 3.3V. But the actual dropout is way more than that. I tried several different MCP's, thinking this one might be defected, but all of them show the same behavior. Can anyone help?

\$\endgroup\$
6
  • \$\begingroup\$ Are you using a solderless breadboard? The contact resistance is significant. \$\endgroup\$
    – Mattman944
    Oct 24, 2020 at 22:28
  • \$\begingroup\$ I tried both soldered and solderless. Same result. \$\endgroup\$ Oct 24, 2020 at 22:29
  • \$\begingroup\$ Have you tried higher input voltages? This might be a regulation issue, not a dropout issue. Is there any chance that this is a 3.0V part? \$\endgroup\$
    – Mattman944
    Oct 24, 2020 at 22:35
  • 2
    \$\begingroup\$ Can you share a photo of your physical setup? \$\endgroup\$
    – The Photon
    Oct 25, 2020 at 0:09
  • 1
    \$\begingroup\$ Are you able to check both the IN and the OUT nodes with an oscilloscope? \$\endgroup\$
    – Ste Kulov
    Oct 25, 2020 at 4:36

2 Answers 2

2
\$\begingroup\$

The first thing I would check is how it works if you use the recommended decoupling:

enter image description here

This from the datasheet link you posted.

Without the decoupling, the LDO may oscillate and not give the correct DC voltages.

Also note they use ceramic capacitors - it may not work if the ESR does not meet the requirements - see sections 5.1 and 5.2 in the datasheet.

Most voltage regulators are critical about the decoupling - LDOs with low drop out are particularly demanding - always read the datasheet.

\$\endgroup\$
4
  • \$\begingroup\$ Good point on the decouplers. I added 1uF ceramic capacitors both to the input and output as you've shown. Unfortunately, I am still getting the exact same voltage reading. \$\endgroup\$ Oct 24, 2020 at 22:12
  • \$\begingroup\$ @JustinasRubinovas - that's weird. It's not getting too hot is it? Does the current change with time, or when you touch it? \$\endgroup\$ Oct 24, 2020 at 22:45
  • \$\begingroup\$ No, it's cool. Current does not change, touching doesn't seem to do anything either. \$\endgroup\$ Oct 24, 2020 at 23:00
  • \$\begingroup\$ @JustinasRubinovas - Have you tried changing the input voltage to see if the output tracks the input or maybe you have actually got parts with 3.0 V output? \$\endgroup\$ Oct 25, 2020 at 15:46
0
\$\begingroup\$

Dropout does not refer to the output, the output should be within the stated tolerance of the regulator, at the voltage it is meant to supply. Dropout means the amount over that voltage that you have to supply at the input, in order to reliably deliver the output voltage promised. For example, a 7805 regulator has a minimum input voltage of 7 volts to deliver 5 volts at the output. So the dropout is 2 volts. A low dropout regulator can operate reliably with just a little more than the output voltage specified. So your 1700 will be happy with about .2 volts more than the 3.0 or 3.3 or whatever you plan to deliver. These regulators can be more efficient, they don't have to waste as much energy as heat, because they aren't (depending on what you give them) having to drop so much voltage from input to output. And they may also be more convenient or give you more margin in your circuits, stepping down from a moderate voltage to a lower one, you aren't required to have such a high voltage input. For example, I use a 7805 to get 5 volts from a 9 volt wall adapter, but I use the 1700 to go from 5 V to 3.3 V elsewhere in the system. As an aside, related to other comments, you DO want to follow the manufacturer's recommendations on input and output capacitors, this is necessary for the regulators to be stable and respond adequately to changes in load.

If you're seeing too low a voltage at the output, you are probably current limiting or thermally limiting, and the regulator is not able to keep the voltage up at that current load. Since you measured the current, my guess is that the regulator is not able to transfer the heat out effectively, and is limiting because it is getting too hot.

Or perhaps you mistakenly bought the 3 volt variant.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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