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It says "... a 9V AC/DC wall adapter with a standard 2.1 mm barrel plug"!?
If I have 9V-800mA, is that a standard wallwart?
In the spec it actually says 9V, 500 mA plug-in. Those 300 mA must be really significant if it smoked the 1N4148 that was across the regulator.
How do you make 9V, 500 mA power supply?
Thanks.

My mistake despite the fact that I know mA are only used for what the source needs. The wallwart output is 9V"AC".
Sorry.

are you sure the polarity was not backwards ?
any device is only going to use just what it needs.. you cannot force more current into a device.the device will use what it needs, and the rest is just power to spare.

The datasheeit I checked says a 1n4148 is only a 1/2 watt part so there is planty of power (9V *300mA is 1W) to toast it (though a forward biased diode would attempt to clamp the 9V to less than 500mV. Contradictory to that it is rated to 450mA. So that supply probably had more current than it was rated for.

Wall warts are designed cheaply and voltage regulation is the primary concern, they are typically not calibrated current limited devices. Consider the current rating as a minimum availability of current for the load (where voltage is still regulated) not a max value limit. Transformer based wall warts (old style) are secondary winding limited while switching wall warts will likely be more true to their spec. However even they will have some margin above the rated current so that voltage stays regulated at the max spec. Likewise, a given load may be speced above what it really needs for the sake of having a margin of error to amke up for cheap sources

You can force more current in most loads, it just requires more voltage. (consider ohms law)

My mistake despite the fact that I know mA are only used for what the source needs. The wallwart output is 9V"AC".
Sorry.

You mean the Load not the source. Just make sure the supply is greater than or equal to the supply current requirement of the load. Using that 800mA source will power a 500mA load just fine without too much extra juice to get out of control in the event of a failure.

That diode, sounds like it was a reverse connection protection device. You can probably replace it with a 1N1004 or other 1W or greater device if the leads fit in the same holes. In power circuits, diode specs usually aren't that picky.

The wall-wart output is 9VAC. When I measured it, used AC setting on my VOM, I got some like 11.47VAC or so. The load requires 9V and 500mA and that is 9VDC. It provides power for the PIC33F uC. Like you say, it was a reverse protection device and that AC going negative just fried the diode.
I need to get 9VDC wall-wart.
Thanks.

After using a variable power supply to make adjustments and the project is all trimmed then I try to find a walwart.
The old lm317 and 15V 500mA walwart and pot is a minamal standard needed. I think its very handy and worthwhile in project design.

most of the 9V I use are 300mA closer to a 9V battery.
I have found the output voltage of the walwarts to significantly vary from what is printed on them.
some are half rectified some use crumby ecapacitors some just too noisy.
I prefer AC walwart so I can improve the rectifier and filter at the project board. (depending on the application)

So, what size of AC wall-wart do you use to get the specs like 9VDC and 300 to 500 mA?
Thanks.

THeres a long and a short answer.

The long answer is >9VAC and at least 500mA. Any regulating curcuit you use, needs a bit more voltage than you want on the output. Exactly how much more depends on the type of rectifier and regulator, and how much energy you want to waste as heat. You need to design it backward. Start with your output requirements, choose a regulator circuit and select a rectifier and smoothing filter (caps) appropriate for it. Once you knowthese, you will know howmuch losses they incur and what input voltage and current to supply it.

A short answer: use a 12V, 1A transformer and you'll have plenty of headroom but probably need a heatsink. Remember the difference between your input and output voltage needs to be dropped across the regulator and dumped as heat.

Also be careful with peak to peak and RMS measurements of AC sources. Don't mix them up.

Just one other thing to add to Hacklesup good suggestion. Any transformer feeding a capacitive input (as opposed to resistive or inductive choke style) with full wave rectification should be capable of supplying 1.5 times the end circuits load requirements in amps. This is due to the fact that transformers are rated in RMS and the resultant DC power extracted is actually 1.4 times that by virtue of charging and keeping that smoothing cap (s) DC at peak AC voltage. Example- 500 ma load requirement; 750 ma transformer rating.