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Question about a specific scenario:

In an (vehicle) environment where a "normal" voltage ranges between 24-29V I have a circuit that uses the NCV7805BTG 5V linear regulator to supply 5V for an ATTiny.

I want to protect this circuit from voltage spikes during (jump) starting or when there is heavy equipment being used, for example a winch, that may cause voltage fluctuations. I read that starting may cause voltage peaks up to 180V.

The NCV7800-datasheet is a bit vague (at least to me) about the absolute maximum voltage for this regulator. The way I read it, it's 40V.

I was thinking to use a TVS diode, but, as I understand it (and I'm new to this, so I may read it wrong), I would need a TVS with a working voltage < 29V and a clamping voltage of max. 40V. Correct? However, no such a TVS diode seem to exist. I recon the closest one is a 15KE30A, which starts breaking down at 28.5V and has a clamping of 41.5V.

What should I do? And how bad is it that if it starts breaking down slightly earlier (heat?/power loss)? Perhaps I should use Zeners instead (how?)?

Now, there is a chance that I'm overthinking this, so any field experience on the topic is welcome! Thanks!

-- reason I use the NCV7800 is that it's the only through-hole regulator I could find that can supply 5V and has an automotive rating. --

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  • \$\begingroup\$ A 7805 to step down from 24-29V to 5V is going to be dissipating a lot of heat if there's any significant current on it. Are you sure that's the way you want to do it? There is going to be upwards of 20V across that thing in normal operation. \$\endgroup\$
    – Hearth
    Commented May 7, 2017 at 13:22
  • \$\begingroup\$ ATTiny will draw less than 300mA.. more like 100-200mA i estimate \$\endgroup\$
    – svenema
    Commented May 7, 2017 at 13:25
  • \$\begingroup\$ 300mA*20V is six watts, which is quite a lot. \$\endgroup\$
    – Hearth
    Commented May 7, 2017 at 13:28
  • \$\begingroup\$ I hadn't realised it added up that quickly. See note below. Turns out the circuit takes about 0.2W idle, 2W peak. \$\endgroup\$
    – svenema
    Commented May 7, 2017 at 14:30
  • \$\begingroup\$ The circuit does? Or that's the power dissipated in your regulator? \$\endgroup\$
    – Hearth
    Commented May 7, 2017 at 14:31

1 Answer 1

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You're correct in your understanding.

However as long as the max breakdown voltage of the TVS is comfortably within your 40V upper limit, you'll probably be fine as the spikes will be largely suppressed by the TVS.

You want to make sure that the standoff voltage (VR or VSO) is above the voltage that you are planning on running at. If you will run at up to 29V normally, a 28.5V TVS is not ideal, a 30V one would be better.

You could additionally add something like a 10R resistor in series with the power line (with the TVS on the regulator side of the resistor). This would further limit the spike voltage by acting as a potential divider with the TVS.


However I would be more worried about your choice of regulator. A linear regulator converts all of the excess voltage to heat by dropping it across a transistor acting as a resistor. With a 25V input and 5V output, that means 80% of the power is wasted, and the regulator will dissipate 4 times more power than your load.

Even drawing 50mA which would be easily do-able with a small MCU would result in 1W dissipation in the regulator - it will get very warm. At you 100mA to 200mA estimate, you are looking at almost 4W of dissipation which will need some serious cooling - at the very least a large heat-sink and some decent air flow. With no heatsink, you are looking at a temperature rise of about 65*C/W (from datasheet) - so 4W would put you at a barmy 260*C above ambient.

A better option would be a switching regulator circuit to step down from 24-29V to something more reasonable like 6V or so, and then a linear regulator to bring that down to 5V. Using a linear regulator to go the last little bit helps reduce some of the switching noise from the DC-DC.

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  • \$\begingroup\$ an alternate simple solution to the heat problem would be to choose the resistor such that it already dissipates a good portion of the power (provided ofcourse you choose a resistor that can handle this). If current draw is lower, you will drop more over your regulator, but it's ok since the current is low. As it increases, the resistor takes over more and more of your power dissipation. \$\endgroup\$
    – Joren Vaes
    Commented May 7, 2017 at 13:53
  • \$\begingroup\$ @JorenVaes Still the same amount of heat has to be dissipated. All you are doing is moving it from the regulator to a resistor. \$\endgroup\$ Commented May 7, 2017 at 14:15
  • \$\begingroup\$ Ofcourse. I was thinking in line of temperature - spreading out over multiple dissipating devices will decrease needed heatsinking. (provided there is enough ambient flow) \$\endgroup\$
    – Joren Vaes
    Commented May 7, 2017 at 14:23
  • \$\begingroup\$ I hadn't really realised it added up that quickly. Just checked the circuit, it just takes 8mA when idle (which it will mostly do).. The peak current I saw it take was at 74mA. That is 0.2W idle to 2W peak. \$\endgroup\$
    – svenema
    Commented May 7, 2017 at 14:29
  • \$\begingroup\$ Sorry.. I had set my voltage incorrectly.. at 28V it takes 28mA when idle (which it will mostly do).. The peak current I saw it take was at 116mA. That is 0.8W idle to 3.2W peak. \$\endgroup\$
    – svenema
    Commented May 7, 2017 at 14:36

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