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I am using MCP16312-E/MS buck regulator and need the component to always be on. I am therefore connecting the enable pin to the Vin supply. Do you need a current limiting resistor? Whilst the resistor I have in series with the enable pin will limit the current, it will also 'take' all of the supply voltage so the voltage at the enable pin itself will be 0V? Is this correct? If so, what should be done instead?

Buck regulator with current limiting resistor on enable pin

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    \$\begingroup\$ Usually no for every IC I have used so far. What does the datasheet say? \$\endgroup\$
    – winny
    Commented Nov 16, 2021 at 15:30
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    \$\begingroup\$ it's very likely that EN is a high impedance input and "essentially" zero currently will flow into the pin under any circumstances. \$\endgroup\$
    – vicatcu
    Commented Nov 16, 2021 at 15:37
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    \$\begingroup\$ (a) a current-limiting resistor is generally not necessary (b) 1 Megohm is far too large for a series current-limiting resistor. (c) It is sometimes nice to have a series element during the development/prototyping stage of a project, which gives a place to disconnect EN from the supply and jumper it to either the ground rail (during troubleshooting) or to a digital control, if one finds that controlling it might be valuable after all \$\endgroup\$
    – Ben Voigt
    Commented Nov 16, 2021 at 15:51

3 Answers 3

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I absolutely agree with @DirkBruere - you should read the datasheet.

I read the datasheet and it says nothing

Let's see if that is true.

First, the very first picture in it has EN pin connected to Vin. This tells you that for "always ON" applications you don't need the resistor at all.

enter image description here

Second, in the Absolute Maximum Ratings you can see that maximum input voltage is 32V and EN pin can be 0.3V higher, i.e. 32.3V maximum. This confirms that you can connect EN pin directly to input voltage.

enter image description here

Third, in DC Characteristics you can see that maximum EN Input Leakage Current is 1 microampere at 5V. It won't be too much higher than that at higher input voltages.

enter image description here

1 uA current through 1 MOhm resistor allows you to calculate voltage drop on resistor. In this case 1uA * 1MOhm = 1V. So, if your input voltage is 5V then the voltage on EN pin would be 5V - 1V = 4V.

Finally, in the same table you look for EN input voltage characteristics, and find that EN Input Logic High minimum is 1.85V, so anything above it will switch device ON. Calculated above 4V voltage is certainly sufficient. However depending on input capacitance the start-up time can be quite high. For this reason 1k-10k resistors are more typical for this pin, just as @DirkBruere suggested.

So, from the datasheet it is clear that 1) you don't need resistor at all, but 2) if you add it then 1 MOhm resistor will work, although smaller value is more appropriate.

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  • \$\begingroup\$ Thank you @Maple. your explanation regarding how much voltage will appear across the resistor answers my question. I am using 1MOhm because originally I had the EN pin connected to Vin as shown in the first diagram in your answer. When looking at the recommended layout, they all have a 1MOhm resistor (even though there is no mention of this throughout the datasheet- hence my confusion). Because the datasheet has a 1MOhm resistor in the recommended layout, should I keep this high value resistor, or should it be reduced in size or even removed? \$\endgroup\$
    – MRB
    Commented Nov 17, 2021 at 13:00
  • \$\begingroup\$ @MRB: From a inventory and production perspective, choose a value that's already used somewhere else in the design. So if you've got a 100kOhm elsewhere, use that here as well - the production line will already have a tape with 100kOhm resistors. \$\endgroup\$
    – MSalters
    Commented Nov 17, 2021 at 13:30
  • \$\begingroup\$ @MRB: I suspect the suggested resistor is to prevent accidents. A short overvoltage transient won't kill the part; that takes both voltage and current (=energy). But if you connect EN directly to Vin, with its associated Cin, then there's no need for additional protection. \$\endgroup\$
    – MSalters
    Commented Nov 17, 2021 at 13:35
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    \$\begingroup\$ @MSalters I think that resistor in suggested layouts could be a simple "everybody does it" mistake. Another possibility is a holdover from similar part with undervoltage control function. A lot of DC-DC regulators use EN pin to provide user-selectable startup threshold. Somebody who drew a dozen schematics for those could simply copy-paste without much thinking. \$\endgroup\$
    – Maple
    Commented Nov 17, 2021 at 15:11
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No, everything in the datasheet says there is no need for a resistor, EN can be directly connected to Vin.

Datasheet says EN pin has no internal pull up or pull down resistor, it must not be left floating, and it can handle whatever voltage there is on Vin.

However, it is not wrong to connect it via a resistor, for some purposes it might be even beneficial to do so.

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Put in a pullup resistor. 10k is usually nice. It's easier to short one out or replace with zero ohm resistor than try to add one later. Also read the data sheet!

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  • \$\begingroup\$ I read the datasheet and it says nothing. If I put a pull up resistor (10k, 100k, etc) the resistor (in series with the enable pin) will 'take up' all of the supply meaning at the other end of the resistor (i.e. at the enable pin) there will be 0V? Is this correct? \$\endgroup\$
    – MRB
    Commented Nov 16, 2021 at 15:46
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    \$\begingroup\$ @MRB datasheet does say a lot of things. \$\endgroup\$
    – Justme
    Commented Nov 16, 2021 at 16:04
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    \$\begingroup\$ @MRB No, that's not correct. The pin has a high input imepdance and draws essentially no current; why would the entire supply voltage drop across the pullup resistor? \$\endgroup\$
    – nanofarad
    Commented Nov 16, 2021 at 18:56

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