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I am building a simple circuit using PT7C4511 PLL Clock Multiplier. This chip has an OE pin that stops the output when LOW. By default (with no external signal driving it) it stays HIGH because of a built-in pull-up (270K).

What I want to do is to turn on and off the chip's output using a 2-pin header. However, I want it to be working when the header is shorted, and stop the output otherwise. To achieve this, I have connected 40K pull-down to OE pin, strong enough to overcome the internal pull-up, but not too strong, so that when OE is shorted to VCC, the pin can easily go HIGH again.

The design

I showed this to a friend with more experience in EE and while he agreed that it should work, he wasn't too excited about the design. He could not point to a specific issue, but the whole thing "smelled" to him.

Is he right? Why?

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    \$\begingroup\$ The answers seem to me quite right: you could consult the datasheet to see the minimum and maximum values of the internal pull-up (if indicated), and calculate the value of your external pull-down. \$\endgroup\$
    – linuxfan says Reinstate Monica
    Commented Feb 16, 2020 at 6:44
  • \$\begingroup\$ I would reduce R3 (maybe to 10K) to get closer to 0V but presumably you have checked 40K is guaranteed below Vil for that chip. Either way it will draw close to 10uA which would be disastrous for battery life in (say) a digital watch, but presumably tolerable for you. \$\endgroup\$
    – user16324
    Commented Feb 16, 2020 at 11:43
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    \$\begingroup\$ It would only smell bad if something causes it to overheat. \$\endgroup\$
    – Hot Licks
    Commented Feb 17, 2020 at 1:19
  • \$\begingroup\$ nothing wrong with voltage dividers so long as they dont let any smoke out \$\endgroup\$
    – old_timer
    Commented Feb 17, 2020 at 16:38

7 Answers 7

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Well, you can't really remove that 270K pull-up, so that just means you have to use a significantly smaller (stronger) pull-down. Also, as an on-chip resistor, the precise value of that pull-up is not going to be very well controlled and could vary by quite a bit. I would recommend going even smaller on the pull-down, perhaps 10k or even 4.7k or 1k.

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    \$\begingroup\$ Given that the datasheet only lists that 270k pullup in the typical column (not in the min or max), I would concur on the variability which may well be as high as 2:1 \$\endgroup\$
    – Peter Smith
    Commented Feb 16, 2020 at 14:36
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Yes, it smells. Use a transistor. This is the normal way to invert a signal. The OE input is a sourcing input. These are designed to be used with a switch which is either open or switched to common. It's not meant to have Vcc connected directly to the input! You have a switch (your header), which would be perfect, but you want the behaviour inverted - so invert it with a transistor. That's why it's called TTL. The resistor to ground also increases the noise sensitivity of the circuit needlessly.

enter image description here

Here when the header is open the base of the transistor is high and OE is pulled low. When the header shorts the transistor turns off and OE goes high via the internal pullup. I've shown 10k to the transistor base here, which is a bit greedy for power - a lot of values would work here, though. The higher you go the less power it draws, but the more sensitive to noise you become.

If you can spare the current and want a more noise resistant circuit you can also tie OE to Vcc with a parallel pullup (Rp) to the internal 270k. Absent a compelling reason not to, giving the input a lower pullup resistor here is probably a good idea.

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  • \$\begingroup\$ Ok, but why does it smell? What issues, even very small and/or theoretical could it create? \$\endgroup\$
    – Miloslaw Smyk
    Commented Feb 16, 2020 at 17:44
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    \$\begingroup\$ @MiloslawSmyk It's not idiomatic, you play teeter-totter with intermediate voltages, there's more noise sensitivity, and you're feeding Vcc straight into a sourcing input. A sourcing input is meant to be either open or pulled to ground. It's intended for a switch. Why wouldn't you do it this way? To save a transistor? If it's a spartan cheap-as-can-possibly-be-made challenge, sure, make it smell to shave a half-cent away, but otherwise I'd overwhelmingly prefer to just do it the normal, proper way. \$\endgroup\$
    – J...
    Commented Feb 16, 2020 at 19:47
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    \$\begingroup\$ @J... - I have to suggest that the onboard pullup on these chips is really only characterized for pullup when there is nothing attached to the package pin. When you connect up an external component, such as the NPN transistor that you show, then there is a need to drastically reduce the impedance of the pullup so that noise sensitivity is eliminated. Your circuit should show a 4.7K, 10K or even a 22K pullup resistor on the collector of the NPN transistor. \$\endgroup\$
    – Michael Karas
    Commented Feb 16, 2020 at 23:08
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    \$\begingroup\$ @MichaelKaras Any suitable transistor will have a high enough off impedance that 270k would function as a workable pullup. Especially for a low power solution I wouldn't add more load here unless the application really demanded it. You're right, though, I would probably add a more robust pullup for a serious application where noise immunity trumps power efficiency. I've added a note to the answer. \$\endgroup\$
    – J...
    Commented Feb 16, 2020 at 23:40
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    \$\begingroup\$ @MichaelKaras I get the point - I just don't think 270k is big enough to necessarily worry about in all cases. I don't disagree that a stronger pullup makes the circuit tougher, of course. I'm just not convinced that it is absolutely necessary. \$\endgroup\$
    – J...
    Commented Feb 17, 2020 at 15:24
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Your 40.2k pulldown is probably fine.

The datasheet, 'DC Electrical Characteristics' table on page 2 gives you all the info you need for this.
The VIL line tells you that 0.8V is the maximum value which the OE pin will recognize as being 'low'.
The R line tells you that the OE pin has a 270k Pullup.
You know you have a 3.3V supply, so with this info it's possible to calculate the maximum value pulldown resistor you can use and still have the input recognized as pulled low - and that value is 86.4k.
So since your 40.2k is less than half that you're well into the 'safe' range (you should expect about 0.43V).
The only other thing you might want to consider is putting a cap on that pin (since you're connecting it to a header and that might pick up some noise). I'd probably put a 100n there.

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    \$\begingroup\$ That assumes the pull up is actually 270k, but it could end up being significantly less as on - chip resistors can vary significantly between different components. \$\endgroup\$
    – alex.forencich
    Commented Feb 16, 2020 at 19:02
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I agree that this is not the best design. However I would not think that it smells like rotten cheese.

A much better approach is to lower the impedance all around and simply use a 10K or 12K ohm pullup on the pin and use the two pin jumper to GND to disable the output.

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  • \$\begingroup\$ Agreed, but only if we allow shorting to be the disabling action. I want the opposite. :) \$\endgroup\$
    – Miloslaw Smyk
    Commented Feb 16, 2020 at 5:44
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    \$\begingroup\$ @MiloslawSmyk - There a number of good reasons to avoid having direct connections to a voltage rail on an option header pin. This is part of the reason that what I suggested is the much better approach. Generally I do not see why it makes much difference which sense of a jumper option (plugged or open) is used but you have your desires and so be it. I could thus propose (a) a three pin option header so there is always an extra position to plug the shunt and (b) add a cheap NPN transistor and another resistor that inverts the sense of the option jumper to what you want. \$\endgroup\$
    – Michael Karas
    Commented Feb 16, 2020 at 9:59
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40K is not a very strong pull-down, it will be prone to noise and with a wire attached will have a poor fall time when the switch opens.

5K or less would be more appropriate.

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I don't see any problem with this providing the 40k resistor is sufficient to provide the low voltage for the input. Actually an elegant solution compared to the one requiring an additional transistor.

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If you don't mind having it the other way around (short to disable it), then an obvious solution would be to have just the jumper between OE and R3. When the jumper is not connected, the internal pull up will pull it high. When the jumper is connected, OE will be pulled to ground via R3. No other components required.

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