Power supply filtering and reverse polarity protection

[dumbmonkey]

I think I need to start putting this sort of thing into my builds, just to be on the safe side.  Is there a "generic" circuit that most people use?  I've noticed from a couple schematics that it seems to just be a 1K resistor, 1N4001 diode, and 100uf cap coming off the 9v in.  Is that pretty much it?

[GibsonGM]

Yup, that's about it.  If the diode is in series, you'll have a .7v drop. Plus a little more from the resistor. For most purposes, that's not a big deal.  But it could be called overkill.

Personally, all I do is reverse-bias a 1N914 across my power supply input.  If the polarity is reversed, the diode will conduct and short out, (hopefully) saving the circuit.  If the power polarity is correct, it's as if the diode were not there (no voltage drop).
Filtering - I only use the typical cap-to-ground at Vref for filtering from a battery. A good regulated PSU should already be filtered, too, so I add no additional components unless required by a specific circuit....

[Schappy]

Ive seen cap values vary from 47uf to 100uf. Does it matter?

For the diode are you saying you put it across the power rail or is it from power rail to ground?

[R.G.]

An excerpt from http://geofex.com/Article_Folders/mosswitch/mosswitch.htm

Advanced Power Switching and Polarity Protection for Effects

^{Copyright 1999 R.G. Keen All rights reserved.}

If you're an effects builder or repairer, you know by now that it's a good idea to protect the circuitry in the effect from the consequences of a reversed battery.  While many older effects can withstand a momentary reversal, some more modern components can be damaged if you accidentally touch the battery contacts the wrong way during replacement. Sometimes this is subtle damage.
...
Ok, so reverse polarity is a Bad Thing. How do we prevent it? The standard wisdom is to use a diode. Not much cleverness needed here, and only two ways to do it - series and parallel. A diode hooked in series with the power supply will allow current to pass only in one direction. If you put the diode in the + lead, the diode takes all the voltage drop if the power leads are reversed. The circuit sees essentially no reverse voltage. The problem is that you pay for that protection in voltage. The diode's forward voltage drop is subtracted from every battery, so a fresh 9.5V battery becomes a middle-of-the-road 8.8V to the circuit that it's powering. And the battery "wears out" 0.6 to 0.7V early as well. Still, better than nothing, and you can get special Schottky diodes with 0.4V drops or maybe germanium diodes for 0.3V drops if you work at it.

A cleverer arrangement is to put the diode in parallel with the effect so it's reverse biased by the normal polarity. When the voltage is reversed, the diode conducts heavily and clamps the reverse voltage to no more than one diode drop. You don't even have to pay the diode forward drop in normal operation. Slick, huh?

As a side comment, it's a good idea for you to read all of geofex. There's a lot of answers in there to questions you haven't thought to ask yet.

@GibsonGM: You may want to reconsider that 1N914/1N4148. Those diodes are rated at 100ma of forward current and something like 1A of surge current. A fresh 9V battery connected reversed could burn one out. An AC power line adapter probably would, depending on the adapter. The 1N4000 series is rated at 1A continuous, and about 100A of surge current. These diodes will handle pretty much anything you throw at them in the effects world - except that they will burn out and die if you hook an AC power adapter to them. Only a series protector will help with an AC power adapter connected. The rest of that article and the followon "A Cheap -And Good- Power Protector" tell how to do that.

@dumbmonkey: The 1K value is awfully high for a series resistor. If you're going to do any serious work with effects, you need to learn Ohm's Law, which says that the voltage across a resistor is equal to the current through it times the resistance, or V = I*R.

A 1K resistor is the same as saying " one volt per milliampere of current". So the 1K resistor drops the voltage your circuit sees by one volt per millampere of current drain. That's fine if your circuit only draws micro amperes, but for instance, an indicator LED draws from two to 20 ma on average. That's a loss of two volts on up. A 1K series resistor is a huge thing to put in series with the power.

[GibsonGM]

Good call, R.G., thanks.  My way, you'd only get the "oh whoops, don't do that again!" and the diode burns out (or shorts, ouch), leaving you exposed next time.  With the 1N4xxx, the oops can be repeated ;o)

[R.G.]

Good call, R.G., thanks.  My way, you'd only get the "oh whoops, don't do that again!" and the diode burns out (or shorts, ouch), leaving you exposed next time.  With the 1N4xxx, the oops can be repeated ;o)

My style of debugging, when find that something doesn't work, it to try it again a few times.  It's made me fond of the idea that doing something wrong once shouldn't let the smoke out.

[GibsonGM]

Yes, I've noticed your site shows that you've 'looked into' keeping the blue smoke for yourself, LOL. 
My style often tends to "ok, this is what a reverse-biased diode will do; let's do it", without really following up with a failure analysis to realize that it may create a short, or simply cease to be part of the equation after it fails.  Duh!  ;o)