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I like to scavenge some old (or new) stuff and reuse parts. In the picture above, you can see what I recently found inside a Philips disc/cassete/radio player and I think it can be a nice power supply and with a little tweek, just can be boxed.
Got 10 ACV in the transformer output and 13.45 DCV in the rectfication board output, so we got voltage enough, but I didn't know how many current I get.
Judging by eye, seems like 1A transformer, but I want to be sure.
How can I get a realistic measure? Google info about seems a little confuse to me, so slap me in the face (any math will be welcome) and teach me a little, please!

Anybody?

Any idea?

You could use a bit of maths in the form of Ohm's Law,but you would first need to convert the AC from the mains,into DC....


Another way would be to see if there was any info written on the plastic casing,the transformer came from,about power consumption,that might give you an idea of how much current the power transformer can supply,but I reckon that it is a safe enough guess that the power transformer looks like it should be able to supply at least 1A......



And judging by the thickness of those yellow coloured wires coming from the power transformer's secondary winding,yep 1A looks right....

Hey, DrNomis thanks for your comment!

I suppose that I can put a resistor in series w/ my DMM and read the measure, but what wattage you suggest for that resistor?
The case was very dammaged to read anything, but the device uses 6 x 1.5v type "C" batteries...does it help?
I look for the #part Philips info but can't found nothing useful.

Let me know what you think

Cheers!

The amount of current your circuit will put out will be directly related to the amount of current your power source can supply.

The amount of current an alkaline battery can deliver is roughly proportional to its physical size. This is a result of decreasing internal resistance as the internal surface area of the cell increases.

A general rule of thumb is that an AA alkaline battery can deliver 700 mA without any significant heating. Larger cells, such as C and D cells, can deliver more current.

Applications requiring high currents of several amperes, such as high powered flashlights and portable stereos, will require D-sized cells to handle the increased load.

Long story short... I bet you can get at least 1A out of it.

Grendahl wrote:Long story short... I bet you can get at least 1A out of it.

Thanks, Grendahl.
In fact, I compare it with one 1A transformer I got, there's no big difference (maybe 1 or 2 mm)
I just wondering how to measure output current with some more accuracy, just to learn something more about.

can't you just put your DMM in series with the circuit?

i'm going to move this to the FAQ section.

That'll tell you how much current it is delivering, but not how much it's capable of delivering.

This same topic floats through my head on a regular basis.

If you try to pull more current than the transformer can handle, will the transformer:
a) become damaged and fail ?
b) simply not deliver anymore current than its capable of ?

I've got a dead valve amp here which i've attributed to EL34s being put in place on 6L6s, and the extra heater current being pulled by the EL34s killed the power transformer (1.5A each compared to 900mA each, from memory). Whether this is an accurate assumption or not, I'm not sure.

Any thoughts?

RnFR wrote:can't you just put your DMM in series with the circuit?

Yes (kinda 975mA or so) but...

salocin wrote:That'll tell you how much current it is delivering, but not how much it's capable of delivering.

So... putting my DMM and it's all? End of the story?
Well...when I make my search I found a lot of people in the net arguing each others about the correct way to get it done, ergo, my question to the forum.
Sorry guys for my "Noob Of The Year" attitude but i'll do it again, for sure