I need to come up with a solution for a voltage regulator to be used in a vehicle, regulating ~12V from the car battery to 5V used by Atmel AVR microcontroller.

I've found this schematic on the Internet:

Car voltage regulator schematic

While I understand the most part of how this circuit works, I have a few questions about it:

  1. What's the purpose of R30 resistor on the input side?
  2. Why are there two capacitors on each side of linear regulator LM7805? This answer to another question might be the answer I'm looking for, but I'm not sure about it. If this answer is related to my question and the use of two capacitors is to reduce resistance and inductance, why are such different capacitor ratings used (0.1 µF and 470 µF)?
  3. Taking a single pair of capacitors, why is one of them polarized and the other one is not?
  4. Are there any drawbacks if using capacitors with larger capacitance instead of those displayed on the schematic?
  5. Are there any drawbacks if using capacitors with larger break-down voltage instead of those displayed on the schematic?

Thanks in advance.


2 Answers 2

  1. R30 limits the charging peak current to the capacitor somewhat, but at 1 ohm it still allows for 12 A, so of little use there. Also would limit the current through the zener if there are peaks above 20 V.

  2. The larger capacitors work less well at higher frequencies, and that's where the smaller ones take over.

  3. 470 µF in a non-polarized version would be expensive, but there would be nothing against it. All large capacitors are polarized.

  4. On the output it would give an extra load for the 7805 to charge it. On the input too for the battery, but that can deliver more than enough current.

  5. No, except that they're larger.

  6. Keep in mind that the input-output difference for the 7805 is about 5 V (12 V - 1 V for the diode and 1 V for R30 - 5 V out) and that at 1 A out the regulator will dissipate 5 W, so it will need considerable cooling (sizable heatsink) if you want to draw that much current.

  • \$\begingroup\$ Thanks for thorough answer, it's really helpful. Do you suggest using a resistor larger than 1 ohm? Also, there is a 0.5A fuse at the input, so the current will not go beyond that. Actually, I don't think it will go over 200 mA. \$\endgroup\$ Jul 3, 2012 at 12:29
  • \$\begingroup\$ If you go to 10 ohm, at 200 mA you have a 2 V drop, which relieves the 7805 a bit. You'll need at least a 1/2 W resistor, 1 W is better. \$\endgroup\$
    – stevenvh
    Jul 3, 2012 at 12:40
  • \$\begingroup\$ 5. Yes. Lower lifetimes for ecaps. \$\endgroup\$
    – Russell McMahon
    Jul 3, 2012 at 13:52
  • \$\begingroup\$ Resistor ratings are generally for continuous power dissipation. An examination of the data sheet for a typical 1W resistor indicates that it may have up to 100x the rated power applied at 1% duty cycle, provided the "on" time is no longer than 10ms. The circuit as illustrated, would initially put 144 watts through the resistor, which the spec does not allow for any duration (the power would exceed one watt for less than 1ms), but one could use either a 1W 2ohm resistor or a 2W 1ohm resistor without violating the specs thereof. \$\endgroup\$
    – supercat
    Jul 3, 2012 at 16:06
  • 1
    \$\begingroup\$ I would actually increase R30 to 6 to 8 Ω (increase power rating as appropriate) to knock the 1/2 A down 4-5 V; the resistor is an easy way to limit the power the linear regulator has to dissipate, and they aren't as picky about running hot. \$\endgroup\$
    – Nick T
    Jul 5, 2012 at 19:15

To expand on stevenvh's answer, for less-experienced designers:

  1. R30 is there to limit inrush current so the fuse doesn't blow. Otherwise when you first plug the circuit in, the input capacitors can draw lots of current until they get charged up, which can be enough to blow the fuse. As stevenvh says it may help protect the capacitors also, which can be important for some types of capacitors, especially tantalum, which can literally catch fire if mistreated too badly. The long-term current through the fuse will be comparable to the load current, which for a microcontroller would be small, so the inrush current at first power-on is sort of a special case and not a concern the rest of the time. If you want to measure inrush current try capturing it on a digital oscilloscope with a current probe. Also R30 can help limit current to the load under short-circuit conditions, notably in case someone has replaced a blown fuse with a piece of metal. 12A from the source can still fry every circuit in its path but at least it's low enough to probably not set your car on fire, which unlimited current from a car battery can be plenty high enough to do.

  2. The large-value capacitor is there for comparatively larger, slower-varying changes in load current (or input voltage, which can vary a lot in a car due to high battery currents during engine cranking). The small-value capacitor is there for smaller, faster-varying changes in load current. A real-life capacitor is not "ideal" and has its particular use cases, so putting different types of capacitors in parallel lets them shore up each other's weaknesses.

  3. Different types of capacitors can be polarized or not. For the values on that schematic you'd probably use aluminum electrolytic or tantalum for the large value and those would be polarized. For the small value you'd probably use ceramic and those would not be polarized, but if you chose some other type it might be. Polarization as marked is an assumption on the part of the schematic designer about what types of capacitors will be used. You don't describe a bill of materials but perhaps one is available at your source that lists the specific parts they had in mind.

  4. More capacitance is generally a good thing in voltage regulators, as variations in load current can be drawn from the capacitors rather than the input source (e.g. battery & wiring), which may have a high impedance and not be able to provide high currents on demand (while still being able to supply the average current). However it isn't necessary to overdo it, as the regulator's job is to actively use feedback to keep the output voltage close the nominal output. Digital chips aren't affected by a small amount of ripple or noise on the power input so long as it's within the specified amount (usually +/- 5%). Analog circuits like RF, audio, infrared etc. can be much more sensitive to power supply noise and need better regulation. Cost and size are important design factors, and minimizing them usually means including "just enough" capacitance. As stevenvh notes, more capacitance will take longer to charge up when first powered on (or more current to charge up in the same amount of time). How long it takes the regulator output to go from 0V to nominal when first powered can affect how the load circuit behaves if it's sensitive to that, notably if there are multiple voltage regulators (say 3.3V & 5.0V) and power sequencing is an issue.

  5. Higher-voltage capacitors may be more expensive, and someone could call you out on this during a design review for a real product. For a hobbyist it's just as well to overdesign things so long as size is not a concern, as a few pennies for parts more than makes up for the re-soldering effort if something gets fried.

  • \$\begingroup\$ Right on about fuses; they are meant only to prevent fire, not damage to a circuit. \$\endgroup\$
    – Nick T
    Jul 5, 2012 at 19:20
  • \$\begingroup\$ Matt, thank you so much for this detailed explanation. As you may have noticed, I am but a beginner when it comes to electronics; your guidelines will help me a lot on this, as well as on other projects I might chase in the future. @NickT, thanks for your thoughts as well, I will make sure to put a larger resistor there. \$\endgroup\$ Jul 5, 2012 at 23:08

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