I am building a digital voltage gauge for my motorcycle for fun. I am buying a nice little display which has common ground, so that I can provide 5 and 12 volts from the same source. I was hoping to use a "7805" regulator to control the 5v, but do not want to provide the varying 12+- volts directly to the 7805. I would like to drop about 7 volts before I apply the power to the 7805. I looked at Using diodes to limit current to LEDs which seems to indicate that using diodes forward voltage drop is not a good idea for leds, and maybe using a zener diode is not either. If I just use a series resistor, I would have to guestimate and experiment to get in the ballpark (as I don't know the current draw), and, as I understand it, the voltage dropped over the resistor would vary with the applied voltage. Here are my questions:

  1. Is using a diode(s) to drop the voltage independent of current a bad idea here? (Is it always a bad idea?)
  2. Do I still get he same IIR loss for the voltage dropped whether it is a resistor or diode?
  3. Is there a conventional "best practices" approach I should be using?
  4. Lets go out into left field for a minute: other than heat-sinking the 7805, can I pot the whole mess with RTV silicon to vibration and moisture resist?

Always nice to know if the alternator is working OK. Not an electronics guy, just a tinkerer.


Thank you Nick and Thomas. I agree that a motorcycle is about the harshest environment one can choose-greater vibration, direct exposure to rainwater, probably greater variations in regulated voltage due to greater variations in engine (and so alternator) speeds. I had hoped to fuse and put the dropping resistor at the battery connection to provide the greatest protection for anticipated faults in the wiring run and/or the digital readout itself. Based on your comments, I will fuse at the battery connection, run fine (#28?) wire to a MIC2945 and let it do all the voltage dropping. I anticipate a mounting plate fastened directly to the 2945 heat sink, with the entire remainder of the assembly "glued" to the mounting plate with RTV silicon. Waterproofing is a concern, for which I will rely on the RTV silicon.


3 Answers 3


First, why would you do this? The 7805 will be perfectly fine with 12V. If you're passing a lot of current, use a heatsink. The diodes will only move power loss from the regulator to the diodes themselves.

  1. In this case it should be fine, but diodes do not drop a fixed voltage; the voltage drop varies depending on current in a non-linear fashion.

    So for example getting 5V from a 12V source using 10 diodes in series would be bad, and the voltage could range from as much as 7V to as little as 2V, depending on the current draw. The 7805 has a minimum voltage of 7V to output 5V at 1 amp, so as long as you can guarantee this 7V you should be okay.

    Note that the output of the 7805 will vary depending on input voltage by a few millivolts - it will be slightly higher at 12V than at 7V.

  2. The same loss happens no matter if you use resistors or diodes. The energy must go somewhere and it is converted into heat by either resistors or diodes.

  3. The best practice here is just to feed 12V into the 7805. Introducing more components into the circuit will make for more chance of failure.

  4. Why is a heatsink not practical? Here, I'm unsure if potting would work. Best to ask someone else.

You should be cautious when using the 7805 in an automotive circuit. The battery line is a harsh place. When you turn the headlamp(s) on your car/bike off, for example, the excess energy in the alternator has to go somewhere and it does - it appears as up to a 60V spike on the 12V line. And in some rare circumstances you can get a negative voltage on the battery line. Both of these situations will destroy a 7805, and the output could short, connecting the battery line to the 5V output, giving your load 12V instead of 5V.


1. Usually. 2. The energy will go somewhere

How much current are you drawing? A conservative junction-to-ambient Td for a TO-220 is 65°C/W, so much above 1 W or drawing 100 mA you will need some form of a heat sink. It doesn't have to be anything glorious, e.g. a screw, if you get really high you could screw the regulator directly to some plate of aluminum and have that be exposed in your enclosure.

Echoing Thomas's fear of vehicle DC power, I think motorcycles are worse than the typical car due to a much smaller battery and the more variable output from the dynamo. The standard operating voltage range (SAE J1455) is 9-16 V, faults of +24 V (double battery), -12 V (reverse battery), and some spooky transients.

3. Nevertheless, a stock 7805 should be just fine with a reverse blocking diode, and preferably a TVS to chop off any high (>30 V) transients.

Additionally, always remember to fuse vehicle electronics. Currents can be ridiculously high, and if you drag the bus down for very long, you might cause problems or even damage everything else like the ECU and ABS (not really standard on most motorcycles, but on cars). Even if you splice into a fused line, it can still be nice to only blow your fuse instead of one that will take other things with it.

4. I would think your environmental protection selection will depend a little more on the mechanical considerations of your other parts (gauges...) in addition to the regulator's power dissipation. RTV isn't the best heat conductor, so if you end up needing a good heatsink, you'll need to expose that to the air in some way. It was used fairly commonly at my previous company to protect the more vibration-sensitive parts (SMT electrolytics, large TH parts) in small blobs, but never to fully pot assemblies as it's fairly spendy (but for hobby projects, whatever :P).

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    \$\begingroup\$ The solution here is not to use a TVS, but a regulator designed to handle these spikes. MIC2954 is one example of such a regulator, rated to handle -20V to +60V without damage. \$\endgroup\$
    – Thomas O
    Nov 1, 2010 at 21:25
  • 2
    \$\begingroup\$ @Thomas O: I'm not sure how common they are in practice, but the inductive transients on standards I've seen go up to 300V, and there's also the mythic 100 V load dump that J1455 specifies (but that will never happen on anything but the largest trucks in the absolute worst conditions). \$\endgroup\$
    – Nick T
    Nov 1, 2010 at 21:29
  • \$\begingroup\$ I wouldn't expect any regulator to do that - the most I've seen is +75V and -30V, and those were automotive rated. (It was also rated to do 28V continuously for 2 minutes, "double battery jump start", nominally it takes in 12V.) \$\endgroup\$
    – Thomas O
    Nov 1, 2010 at 21:36
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    \$\begingroup\$ @ThomasO: Exactly :P, hence the TVS, which between it and the supply's impedance kill the transient. For some hobby thing I wouldn't bother with the TVS unless I had mysterious failures. \$\endgroup\$
    – Nick T
    Nov 1, 2010 at 21:42

The LM7805 is not sensitive to low current voltages exceeding its rated 35VDC input parameter; so if a 6V 5W zener is used in series with input VDC the initial minimum current requirement for the zener will not damage the 7805 at 41 volts and its output will still be 5.0 VDC as soon as any current is drawn by your device connected to it, the zener's operating voltage will reduce the 41V to 35 or lower should it ever go that high on your bike. If you think the 7805 is vulnerable, you may stack it with more than 1 - such as 2,3 etc all with the leads in parallel and effectively increase its rated output current proportionally. Also, there is a new regulator on Digikey.com, the LM317HV handles 60VDC in at 1.5A output which can be adjusted from 2.5-57V with a couple of resistors. Your project sounds like fun!

  • \$\begingroup\$ Putting a 6V zener in series between a 9-16v source and a LM7805 would frequently cause the 7805 to drop out of regulation (produce less than 5v output, and not be able to keep the output voltage stable). It typically requires at least 7.5V input to stay in regulation. Look up "regulator dropout voltage". \$\endgroup\$
    – Zeph
    Nov 21, 2016 at 18:12

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