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There is a NUD3160 relay driver on the board that I need to set ON at power-up. According to datasheet, it has internal 100k pull-down and 2V threshold voltage. So, if I add external 910k pull-up to 24V power line it should form a voltage divider sufficient to open the driver.

What bothers me though, is how reliable the datasheet is, especially in the long run. Technically, the manufacturer does not have to use these exact values, as long as the electrical characteristics remain the same.

I don't know if this question is "opinion-based" or not, but what would you recommend?

enter image description here

Update

After reading all the comments and answers I've come to the conclusion that relying on internal workings of the driver was a bad idea. So, here is what I am going to do:

  1. Add second resistor to get normal external divider;
  2. Adjust divider to produce around 5V gate voltage. This will ensure full opening of the driver and will reverse-bias diode so that this circuit would not load 3.3V power control signal coming from TC7PZ17 buffer (see full circuit here).
  3. Use lower values so that driver's input impedance won't matter much, but not too low to draw much current; Something like 24k + 8.2k seems to be a good compromise, wasting only 0.8mA / 0.01W per node.

Critique?

Maple
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    It's worth noting, by the way, that this will probably not be sufficient to turn on the FET to any significant extent. The threshold voltage is the voltage at which it *just barely* begins to conduct, not the voltage at which it's fully on. – Hearth Oct 05 '19 at 21:30
  • You must consider the effect of the TVS in your divider , it's meant to be a weak pull for the purposes of floating pin stability. Especially when divider currents are on par with leakage, it may not be very stable with this threshold as @Hearth says. – crasic Oct 05 '19 at 21:51
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    If the datasheet doesn't specify a value for the resistor, with minimum and maximum limits, then you should not count on the resistor having a specific value. – Elliot Alderson Oct 05 '19 at 21:55
  • @Hearth, actually, 2V is absolute maximum when it MUST switch on fully, per datasheet. So, 2.4V should be sufficient... providing, of course the internal resistor is close to 100k, which was the gist of the question – Maple Oct 05 '19 at 22:17
  • @crasic After reading your comment and le_top answer I now realize that other components in the driver should be considered as well, making the whole idea of getting by with single resistor rather dubious :( – Maple Oct 05 '19 at 22:24
  • @ElliotAlderson unfortunately, it does not. And as was mentioned in one of the answers, the actual input impedance seems to be quite different. Hmm... too bad. I tried to avoid extra power drain via strong external divider. I guess I can go with something like external 20k+2k – Maple Oct 05 '19 at 22:32
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    @Maple Where does it say that? All I see is that 2V is the maximum threshold voltage, which does *not* mean the voltage at which it's fully on... – Hearth Oct 05 '19 at 22:48
  • @Hearth Figure 7 shows 30mA for 2V and 150mA (maximum rating) at 2.5V. So, you are correct, it is not fully ON at 2V. Fortunately, getting 2.5V should not be a problem – Maple Oct 05 '19 at 23:54
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    "threshold voltage" is a well-defined figure of merit for a FET, and it does not mean the voltage at which it's fully on. Be aware that figure 10 shows it at a drain current of 250 μA, and figure 8 is looking at a 0.8V Vds. – Hearth Oct 05 '19 at 23:55

2 Answers2

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Schematics like the one that you show should be regarded as simplified schematics - the actual implementation is likely more complex.

Anything that is not specified in the tables, graphs or text --- generally with tolerances --- is not a useable design property.

When reading the NUD3160 specification, I do find a property related to the value of the resistor, the \$I_{GSS}\$ or Gate Leakage Current: Extract from the NUD3160 specification There is also a graph showing the dependancy on the junction temperature for two voltage levels.

At 5V, nominal temperature you get 90uA, which means an effective impedance of about 55k . At 125°C, that becomes 45k. The variance is at least 10%, and the observed value is half of the 100k value shown in the schematic.

So as far I can can see, there is an indirect specification of the 100k value, and it shows that there is more to it than the schematic shows. I am not going to guess where the other half of the current flow is going.

Regarding the pull-up to enable the driver.

As your core issue is to get the driver "ON" at startup, you basically need to use this leakage current specification. The maximum rating for the Gate-Source voltage is 12V, and it is best kept close to a value not exceeding 5V. You get between 40 and 110uA at 5V. You are powering from 24V, so your target is 19V accross the pull-up resistor - so it would take a value between 170 kOhm and 475 kOhm. You do not want to exceed 10V on the gate - at that voltage with the minimum current (40uA), the resistor has to be at least 350kOhm. The actual minimum current is not specified (the table indicates "maximum current"), but the 40uA at 5V is surely below your practical usage - at 0°C it is about 47uA.

Checking the 350kOhm resistor, at 40uA, the gate voltage will be 10V (due to the resistive voltage drop). With a gate voltage of 2V, the current would be 63uA. That is just above the worst case current at 3V for a nominal temperature.

Figure 5 of the specification (the typical \$I_{GSS}\$ current versus temperature) shows that the typical current is lower. And implicitally - due to the variance in the production process - some devices can even have a lower current (just like some devices have a higher than typical current). The fact that the lower bound is not given is an issue.

As a result, with the available information, it seems impossible to ensure proper operation under all conditions with a resistor alone. The current might be lower than 40uA, which means that the gate voltage might exceed the absolute maximum rating. Increasing the resistor to lower the voltage does not ensure a high enough gate voltage at higher currents.

To conclude with a solution, it seems best to add a 5.1V zener diode across the Gate/Source, and use a 150kOhm pull-up resistor. With low currents, the zener diode ensures that the voltage remains within the maximum ratings, and with high currents, the voltage drop of the resistor is low enough to still provide 110uA.

le_top
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    MCUs often include the pull resistance in datasheet explicitly, and the range is usually quite wide! Like you say there is likely an internal pull down plus some leakage current from other sources. But it would be non trivial to rely on this as a precise voltage divider ! However, if this is for sampling purposes, and you can tolerate gross errors (e.g. simple supply detection), it should be usable given the analysis you provided. +1 – crasic Oct 05 '19 at 21:49
  • You guessed it right, this is simplified schematics. The actual one is close to second schematics [here](https://electronics.stackexchange.com/q/456549/187920). The most reliable solution would be to have an actual voltage divider with much smaller resistors, somewhere in 10k range, so that input impedance can be ignored. Unfortunately that will also increase power consumption, and in case of pull-up disconnected by SW1 (all boards but one master) this will create multiple pull-downs in parallel, increasing the load on enable line. That was the reason I tried to use internal R. – Maple Oct 05 '19 at 22:12
  • so from that data-sheet it looks like 60uA into a 3V load will be sufficient so for a 24V supply a 330K series resistor looks good? – Jasen Слава Україні Oct 05 '19 at 23:31
  • I've analysed the pull-up resistor - 330k isn't a good value. At 24V, the gate voltage would be 2V for 67uA and not guarantee that it would work for higher leakage currents. For typical devices, under typical conditions (Figure 5 of the specification), the leakage current is lower than 60uA for all temperatures, so "generally" it will work. – le_top Oct 06 '19 at 00:06
  • @Maple, generally you should use low impedance into high impedance, your current arrangement is a high impedance output into high impedance input. Note that a divider has a lot of queiscent current since it is a basic linear device , using an opamp buffered divider would provide a low impedance output with low quiescent current, and with a high impedance load, the drive current will be low as well. – crasic Oct 06 '19 at 17:35
  • @crasic You are right, I am trying to turn things around now. See the update in the original question. I'd use unity-gain buffer if it wasn't a cost issue. The design requirements call for lowest possible cost boards, that is why I was trying to save even on one $0.02 resistor. – Maple Oct 06 '19 at 21:19
  • @Maple Now that you decide to use two resistors, you just need to check that the voltage of the divider is within limits (minimum 3 or 5V and maximum 12V) for 110uA and the lower uA limit which you could test at 0uA (strict worst case) or 110k (10% variation of suggested value int he specification) in parallel with your lower leg of the divider. – le_top Oct 06 '19 at 22:21
  • @le_top I found one more option, which seems to beat the others at power consumption and layout simplicity. The voltage reference MAX6035 uses only 73uA supply current and does not require external resistor. Unfortunately it is 100 times more expensive than resistor divider. I guess I have to look at the requirements again, to see if cost takes priority over power. – Maple Oct 07 '19 at 19:48
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    Ok, you can also add a zener in reverse mode instead of the resistor. As long as your zener's value is between 13V and 19V you should be safe. The imprecise pull-down will pull the voltage down and only draw the current that it needs at 24V minus the zener voltage. Your voltage level is pretty much guaranteed and the current adapts to the actual "gate current". – le_top Oct 08 '19 at 11:58
  • The BZX84C15VLFHT116 costs less than 4cts per 1000. I guess it isn't going to break the budget. – le_top Oct 08 '19 at 12:01
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If you can't trust the datasheet in anything it writes, get a different part. It's really as simple as that.

If the datasheet specifies the pull-down to be 100 kΩ, then the manufacturer shouldn't simply change that. It's a specified property, just like the current capability or voltage rating of that relay.

Now, ONsemi isn't "some" manufacturer, but one of the largest and most reputable power semiconductor producers in existence. They will most definitely give out a specification change document if they decide to change the part – which I doubt would ever happen; these resistors are almost certainly part of the same silicon die, and rolling a new mask, even at the gigantic structure sizes they use, is simply something that costs a lot of money.

Just make sure you buy your relays from somewhere you trust to actually get the original part.

Marcus Müller
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    OK, that's reassuring, thanks! Buying everything from digikey, that's as close to trusted source as I can get. – Maple Oct 05 '19 at 20:25
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    I didn't see a tolerance rating for the pull-downs in the datasheet though. Can one assume a certain tolerance (e.g. 10%? or 20%?) when this is not stated? – Dampmaskin Oct 05 '19 at 20:29
  • that's a good question! I didn't pay overly special attention to that; maybe that would actually be worth asking ON – Marcus Müller Oct 05 '19 at 20:30
  • The tolerance may be given based on available pullup/down sink/source current limits. – MadHatter Oct 05 '19 at 20:35
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    I agree with Dampmaskin, tolerance is definitely a concern and I imagine they don't pay special attention to tolerances on pull-down resistors so it might be pretty poor tolerance. – Hearth Oct 05 '19 at 20:47