Protecting a power supply

[Bill_F]

I would like to use a 9v DC walwart instead of a battery for testing circuits. Is there any way to protect them, in case of shorts or reverse polarity. Reason I'm asking is I've already fried one and don't feel like making it a habit of replacing these things.

Thanks,
Bill

[smoguzbenjamin]

You're gonna use AC on a ckt that was designed for DC? Am I getting this?
I think it might be a good idea to buy a 9V DC wallwart They're cheap enough, only about 5 euros and they work just fine!

[Peter Snowberg]

There are a number of ways luckily.

One of the simplest is to use a small light bulb that's rated at 12V and about three times the current your effect is using. That's just a general figure.... a 12V 100ma lamp should work for most uses. When your effect is working correctly, the lamp will not get enough current to light up. If you have a short, the lamp will light which raises the resistance dramatically and limits the current. It works like a little circuit breaker.

Another way is to use a automatic resetting fuse. These things generally look like a ceramic cap dipped in colored epoxy. Try the mouser catalog at one source.

An even better way is to use a regulated power supply which is built around a regulator that can take being short-circuited. One of my favorite regulators is the LT1129 adjustable regulator. I love chips from Linear Technology.

When it comes to protecting the actual effect circuit, if you run the power through a schottky diode (like a 1N5818) first, that will stop any power from flowing if it's reversed. That will take .3 volts from your circuit, but it's a small price to pay. If you use a small lamp in series with your supply (9V 200ma) and a reversed schottky diode across the power inputs (with the band pointing towards the + lead), that's perhaps the best solution. Under normal operation the lamp just acts like a tiny resistor. If you reverse the power connections, you see it in a hurry and the circuit is protected by the low forward voltage of the schottky diode.

Best of luck to you... and remember.... electronics run on blue smoke.... let the smoke out of the parts and they stop working.

Take care,
-Peter

[Bill_F]

I think it might be a good idea to buy a 9V DC wallwart

Sorry, that's what I meant. I've changed the question above to reflect that.

Thanks

[Bill_F]

There are a number of ways luckily.

One of the simplest is to use a small light bulb that's rated at 12V and about three times the current your effect is using. That's just a general figure.... a 12V 100ma lamp should work for most uses. When your effect is working correctly, the lamp will not get enough current to light up. If you have a short, the lamp will light which raises the resistance dramatically and limits the current. It works like a little circuit breaker.

Another way is to use a automatic resetting fuse. These things generally look like a ceramic cap dipped in colored epoxy. Try the mouser catalog at one source.

An even better way is to use a regulated power supply which is built around a regulator that can take being short-circuited. One of my favorite regulators is the LT1129 adjustable regulator. I love chips from Linear Technology.

When it comes to protecting the actual effect circuit, if you run the power through a schottky diode (like a 1N5818) first, that will stop any power from flowing if it's reversed. That will take .3 volts from your circuit, but it's a small price to pay. If you use a small lamp in series with your supply (9V 200ma) and a reversed schottky diode across the power inputs (with the band pointing towards the + lead), that's perhaps the best solution. Under normal operation the lamp just acts like a tiny resistor. If you reverse the power connections, you see it in a hurry and the circuit is protected by the low forward voltage of the schottky diode.

Best of luck to you... and remember.... electronics run on blue smoke.... let the smoke out of the parts and they stop working.

Take care,
-Peter

Thanks Peter for taking the time to write all that.
I'm pretty new at this and pretty clueless so I might be way off. Could I build a simple circuit that is outside the box for testing purposes that has both the lamp and the diode in it? This could sit inbetween the walwart and circuit and they could both plug into this.

Like this:

[smoguzbenjamin]

Should work fine, Bill.

[Peter Snowberg]

Your schematic is perfect, but keep in mind that the real utility from the diode in that circuit comes from knowing it's always connected in a known way so if you have a plug (or other disconnection point) after the diode, you don't really get any protection from it. Diodes are fairly small and inexpensive so I would suggest just adding one to your board where the power enters (many commercial effects include this feature) and putting the lamp in-line, maybe even at your power supply.  

Take care,
-Peter