So I have a module that supposedly has a P-Channel Open Drain Output (INH on Infineon TLE6251-2G) and I am trying to interface that with the EN (Enable) pin of the Texas Instruments TPS57160-Q1 Step Down Converter.

The TPS57160-Q1 activates if the EN pin is at a voltage above 1.25V and has a "pullup current source" (whatever that means) in order for it to be activated when the EN pin is left floating.

Now the TLE6251-2G's INH active output (means the Step Down converter is to be activated) is VS=36V in my case. When the output is "disabled", the INH is left floating.

Now the question is: How can I interface the two pins? I only have the 36V rail and GND available and it should be ultra low power in the disabled state.

If I add a pulldown resistor at the INH pin and connect that to the EN pin of the TPS57160-Q1, I should have it properly disabled but I exceed the absolute maximum voltage rating of the EN pin by far. (36V vs. 5V) To get around this, I thought of a simple voltage divider to get the voltage to a healthy range. Furthermore, I don't really know how the TPS57160-Q1's EN pin reacts to a pulldown resistor with its "pullup current source" (This is my main problem in creating a proper interface)

Thank you in advance!

  • 2
    \$\begingroup\$ Page 4: Absolute Maximum Ratings, Isource, EN: 100uA --> No more than 12k Pull Down. Use no more than 10k to be sure. i.e. 8.6k pull down, 1nF capacitor, 72k "pull up" from INH. Gives no more than 0.86V when no voltage, no more than 3.84V when INH 36V. Whether that works depends on your supply stability. \$\endgroup\$
    – Asmyldof
    Mar 5, 2016 at 14:18
  • \$\begingroup\$ @Asmyldof Oh, I get it now! Thanks for the comment. Would you consider writing that as an answer so I can accept it? Is the capacitor necessary in that setup? \$\endgroup\$
    – fscheidl
    Mar 5, 2016 at 14:38
  • 3
    \$\begingroup\$ The capacitor (parallel to the 10K resistor) helps stabilize the voltage "seen" by the input pin in case of "supply ripple," to keep your TPS57160 from oscillating on/off/on/off/on/off... It may not be "necessary," but it's sure less of a headache to solder in one capacitor than it is to figure out why your circuit's "having convulsions." ;) \$\endgroup\$ Mar 5, 2016 at 14:52

1 Answer 1


As you request, I'll flesh it out to an answer close enough to being acceptable.

If you look at page 4 of the datasheet there's a table of absolute maximum ratings, it also lists the maximum Enable Pin source current as being 100μA.

This means to stay under 1.2V you should certainly keep any resistor to ground below 12k. I'd prefer a smaller one, to be certain, such as 10k.

With a bit more headroom even (and sub 1mA current at "ON" state) you could calculate for 8k2 lower resistor. This would give you: 100μA * 8200 Ohm = 0.82V if the EN pin is the only one sourcing.

This means, that you want, let's say 4V on your EN pin when INH is turned on, because you never dimension for ideal case, you may have power spikes. So even 4V might be too high, but that would depend on your power supply stability and reliability. You can take the Absolute Worst Case Maximum supply voltage and use that for a 5V calculation and then check if at lowest supply the EN will still go a few tenths of volts above 1.2V

which means you calculate for a current of:

4V / 8200 Ohm =~ 487μA

Now, you'd be tempted to subtract the full 100μA from that, and it'll most likely be perfectly good if you do, but since we don't know if that is honest in this case, as we don't know if the current might already be dropping off at that point. It is however safe to assume the EN pin will not bias itself above its own safe point and that the source will turn off at or before 5V is reached, since the Datasheet allows floating pin.

For the sake of argument I'll go compromise here and take away 50μA, which makes 437μA.

The upper resistor now needs to take up 32V at 437μA with your supply of 36V and 4V across the bottom resistor, so the top resistor should be:

Rtop = 32V / 437μA = 73091 Ohm.

Since we took away 50μA at a guess we can round up or down, but 75k Ohm standard value is pretty close.

Add a small capacitor for noise and supply glitch problems, since not much is specified for either chip at switch on or off in sense of input/output glitch suppression, and you get:


simulate this circuit – Schematic created using CircuitLab

Depending on your turn-on-time requirement versus expected noise and glitches the capacitor can be anything between 1nF and 1μF. Maybe even 2μ2 if a dozen or more ms turn-on isn't a problem. Point is, if you design it in, you can easily change it to a higher value or a lower value if the behaviour isn't what you want.

Fun point on the side, a regulator like this, that clearly specifies an internal current source on the EN pin, or even just a weak pull-up, has a very nifty start-delay option. If you connect only a capacitor on its EN pin the internal source will charge it up a while, causing a turn-on delay. Also very convenient if you need supply sequencing and you know how much the production tolerance likely is. (i.e. if it's 50% to 200% of stated value, step your caps in factor-4 at least.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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