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I've bought a Davtron model 811B aircraft clock (14 Volts), but my power source when measured with an accurate multimeter, reads exactly 14,29 Volts. Is that a problem? I mean is that tension (14,29 V) "too strong" for a 14V avionic? Thanks a lot.

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    $\begingroup$ In my car that has a 12V battery, the voltage is 14.1V when the engine is running (and thus charging the battery). And when fully charged, the voltage is above 12V. I guess an aircraft shouldn't be much different. $\endgroup$
    – U. Windl
    Commented Oct 9, 2023 at 6:02
  • $\begingroup$ The nominally-12V line in a car might be as high as 14.5V when the battery is being topped off after a cranking-start, and as low as ~10V while the starter is actively turning the engine over (and a weak battery on a cold morning) This is why the console might not light up during cranking, and car audio often cuts out while cranking to avoid damage and popping sounds. $\endgroup$
    – Criggie
    Commented Oct 10, 2023 at 0:29

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No it is not, it's within engineering tolerances of the specification. Note also that to get meaningful measurements to 4-place accuracy requires manual calibration of the measurement device to a standard, if the measurement device was not self-calibrating. .

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  • $\begingroup$ Hi there! Thanks a lot for your prompt reply! I'm more confident to plug the chronograph in now, knowing the tolerance is ok. And another question, if I tried to light the clock up using "only" 12V instead of 14 V, would it damage the inner components? Thank you very much sir! $\endgroup$ Commented Oct 8, 2023 at 18:15
  • $\begingroup$ 12V would not damage the components, but the clock itself might not run at the right speed! -NN $\endgroup$ Commented Oct 8, 2023 at 19:55
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    $\begingroup$ @nielsnielsen, The idea that the speed of a clock—we're talking about a clock that's marketed as a navigation instrument for an aircraft—would depend on the DC supply voltage is... odd. $\endgroup$ Commented Oct 9, 2023 at 16:54
  • $\begingroup$ @SolomonSlow, the context is what would be the consequence if the drive voltage is out-of-spec for the device. This sort of thing is why the specs get written in the first place. It would be rash to assume a timekeeping device designed to run on 14 VDC would meet its accuracy specs at 12 VDC unless the manufacturer explicitly said so. $\endgroup$ Commented Oct 10, 2023 at 2:05
  • $\begingroup$ If I were writing requirements for a clock that was meant for use in an aircraft cockpit, I would propose that it should go dark instead of displaying incorrect time of day. It would be easy to make the display go dark if the supply voltage fell below some arbitrary threshold, and easy enough to design the timekeeping circuitry to be reliable at a still lower supply voltage. $\endgroup$ Commented Oct 10, 2023 at 2:15
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Note +0.29 V is only +2%, as already answered. This small overvoltage is likely not a problem:

  • I doubt the electrical generator voltage is 2% stable on an aircraft.

  • The accuracy doesn't depend on the exact voltage; there is a crystal-piloted oscillator.

But there is no voltage tolerance mentioned in the specs, and on the other hand it's very easy to drop the voltage in excess with a resistor.

The specs say you can use a source up to 30 V.

Model 811B operational input is 11V - 30V

Source: Davtron.

However this requires inserting a resistor in series.

enter image description here

The resistor drops the voltage in excess of 14 V. As the current is said to be 400 mA, for 28 V, voltage in excess is 14 V, and so the value would be $\small 14 / 0.4 = 35 Ω$. This is the value indicated in the documentation.

For 0.29 V in excess this requires $\small 0.29 / 0.4 = 0.725 Ω$. The standard value is $\small 0.75 Ω$ (with either 5% or 10% tolerance anyway). The minimum power acceptable by the resistor is ¼ W (maximum dissipated power is $\small 0.75 * 0.5^2 = 0.19 W$, more the time the fuse blows, but it's only a few ms).

enter image description here

With this resistor you have 14 V at the clock input. This will cost you 20 cents.

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    $\begingroup$ What makes you say that "the documentation is suspicious as with a resistor there is no limit actually"? If the documentation says that the upper limit is 30 V, then I certainly wouldn't connect it to 31 V or more even with a 10 kΩ resistor. A resistor doesn't drop any voltage until there's a current through it, and when you connect this device to power, there's no guarantee that it will begin drawing current before it gets damaged by the excessive voltage. (Correct me if I'm wrong. Maybe there is such a guarantee somewhere, but I would assume that there is not.) $\endgroup$ Commented Oct 9, 2023 at 3:04
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    $\begingroup$ As for the power rating, notice that for a 28 V system, the manual calls for using a resistor rated for 25 W even though the expected power dissipation is only about 9 W. That's probably on purpose! Resistors can get quite hot at their rated power, and an airplane's electrical system is a place where you particularly don't want anything to catch fire, so the recommendation to use a resistor with about 3 times the rated power makes sense. $\endgroup$ Commented Oct 9, 2023 at 3:11
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    $\begingroup$ The clock uses an incandescent display, and according to the manual that's all that's powered from the aircraft. Incandescent lights are tolerant to wide ranges of voltage and therefore are hooked up unregulated, though higher voltages reduces lifetime. The resistor is just a quick and dirty way of preserving bulb lifetime. $\endgroup$
    – user71659
    Commented Oct 9, 2023 at 4:05
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    $\begingroup$ @TannerSwett: That sounds about right; some failure mode could hypothetically pull the max current allowed by the fuse through the resistor, even if that's significantly more than the device draws in any normal state of operation. So yes, max power is fuse current rating times resister's resistance. In an enclosed space with poor ventilation, yeah, derate accordingly. Or just don't mess around with a resistor at all; if aircraft electrical systems are anything like automotive, 14.29 V is remarkably close to the actual rated voltage; it's likely fine with +-1 or 2 volts. $\endgroup$ Commented Oct 9, 2023 at 7:03
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    $\begingroup$ Thank you all for the answers. I've learned a lot.... very interesting this article about how diodes could protect the circuit from over voltage. Thanks a lot $\endgroup$ Commented Oct 9, 2023 at 19:25

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