AC adaptor polarity

Load the supply to 300mA and measure the output voltage.... If fact, load it a little more and see what you get.

I would think that a regulated supply would be better. You can build one quite easily should you wish.

If you know that the original power supply was un-regulated there is no reason to get one now. An unregulated adapter with the same data will work fine.

You can measure the voltage of a battery, you say.

When you do that, can you see a polarity indicator on the display? Can you see a difference if you shift the probe leads, does it show a minus sign if you put the probes on backward?

If you can see this, you know how to find the plus contact on a battery, and that is marked on the battery too, so you can check that your DMM works as it should. You will know after trying this that the red probe lead is the one you should use on the plus side. If you shift the leads you get negative voltage values.

Now you should be confident enough to measure the voltage that comes from the new adaptor, and find out what polarity it has.

In practical terms you may have to find a piece of wire you can insert into the contact, because the probe will not get into the contact.

Anyway, you fix the mechanical contact with both the inner and outer metal in the contact, and measure the voltage and polarity.

If it needs changing you can cut the cable and connect it again, with the other polarity.

When you measure the voltage of an unregulated adaptor without a load you will get a significantly higher voltage, like 16-19 Volt, and this is nothing to worry about as long as both the voltage and the current is the same as on the original adapter.

If this chess computer is very important to you you could, of course get a regulated supply, and test it before connecting it. That would remove this possible source of error: If the adapter is made for much higher current it will give higher voltage when loaded too light. If also the power supply circuits of the computer are really badly designed this could cause some problems. Very unlikely but not completely impossible.

You're going to need a power resistor to load the supply. You have E=I*R so rearrange the equation for R since that is what you want to find. R=E/I You are going to be playing with R so you don't have to be exact. R=14V/.350A R is about 40 ohms, lower values will increase the current. The power dissipated by the resistor will be P=I*E P=.350A*14V P=4.9W use a 10W resistor and don't burn yourself!

You can build a regulated 9 volt supply most easily by obtaining an LM7809 three terminal regulator. A couple of small capacitors wouldn't hurt either. Page 6 from this pdf shows you how to connect the regulator:

You will need to bolt the regulator onto a heat sink which you can scavenge from a broken power supply or simply cut up an aluminum can.

You test the current by placing the meter leads in series with the load resistor. Don't forget to plug the test leads into the correct socket and set the meter to its highest current range. The power supply + output connects to the load resistor, the other side of the load resistor connects to the meter - lead. The meter

• lead then returns to the supply -. + to - to + to - in series....for current. To measure the output voltage, the meter, reset for volts is placed across the supply output + to + an - to - in parallel.

Another option is to get an unregulated supply with voltage switch, so it can be set to positions marked 3V 6V 7.2V 9V 12V. With such a supply you can safely start at a low voltage, connect the computer with correct polarity, step up one step at a time, until the machine works as it should. I have used that several times. Then I fix the switch with tape and use it for that device.

These stepped but unregulated adapters are much cheaper than regulated supplies.

Another idea is to build a regulated power supply yourself. You need an npn power transistor which can handle 300mA and a few watts power.

Connect the collector to the output of an unregulated supply, the emitter becomes your new regulated output, the base we lock at 9.6 Volt, because the emitter will always be 0.6 Volt lower than the base, so we get 9V out if the base is kept at 9.6V.

We can use a resistor and a 9.6V zener diode to fix the base at the right voltage.

Hi

I have purchased a Chess computer over the internet that requires 9v 300ma, centre positive.

It is coming without an ac adaptor.

I have a couple of old adaptors that are 9v 300ma but there are no polarity markings.

A know only a little about electronics, although I only use my multimeter to test my rechargeable batteries usually.

Apart from checking for smoke..., how can I determine the polarity of the adaptor.

Also, am I better off spending a little more and just buying an adaptor that is regulated? The chess computer is extremely expensive to replace and I don't want to risk it.

Thanks

Darren

Thanks Tom.

That all made sense and I tried it, finding that: I have one adapter that is centre negative (not suitable for this application) and one adapter that is centre positive (suitable for this application).

I did find however that this 9v adaptor is putting out 13.3v.

I'm not sure if the circuitry of the chess computer will have some for of protection to allow for that fluctuation.

Do you think I should get a regulated adaptor? (When the chess computer arrives.)

or

Wait until the thing arrives and there may be some information about voltage tolerances?

or

Not be concerned about this extra voltage.

(The adaptor's 50hz too if that info. is of any use.)

Thanks again.

Sincerely

Darren

multimeter

the

respect

Thanks Lord Garth

That's a little beyond me at the moment I'm afraid.

How would I load it? With a light bulb? Resistor?

I'll have to check the math etc. as it's been a long time. E=I*R (amps = volts * resistance?)

Then attach the contacts of the multimeter as before?

Sorry I'm not that educated on the issue.

I'll get there eventually.

Thanks again.

of

Thanks Roger.

Cutting and swapping the cable would fix the polarity issue.

Yes the mm has a polarity indicator and I've cleared that up now.

And now it becomes apparent to me that the voltage would be higher when the adaptor's not under load.

Perhaps a regulated adapter would be safest as the chess unit is 600 Aust. dollars to replace or 300 to import. (Big profit margin.)

I will try and find out the spec's of the original adaptor perhaps too. (Reg/unreg.)

Thanks group. You've been extremely helpful.

Darren.

Darrren, that should be: E=I*R ( volts = amps * resistance)

...jerry

Oops that's right Jerry. Thanks.

Thankyou for all the advice gang. I feel confident I have all I need.

I shall now put your advice into practice.

Darren