Need Help Improving A 240v->9v Transformer

[Neta]

Hey guys,
I recently got my hands on a multimeter and decided to check my 9v power supply. Turns out it gives 13.69v!
But there's an upside - the voltage is constant. So I was thinking maybe I can form some kind of a voltage divider to reduce to voltage to 9v?

Tell me what you think!

Thanks guys

[antonis]

Propably you've full wave rectification so the 13.69V are quite alright...
(secondary voltage X 1.44 minus some diode voltage drop...)

If you've checked the voltage stability with "heavy" load, you can create a voltage divider but the actual resistor values depend on Rload...


P.S.
Better choice is a fixed voltage regulator..

[R.G.]

I would be more definite about this. It is impractical to use a resistor voltage divider to power any significant load. The resistors needed will eat at least ten times the power of the load they can run and get any kind of semi-stable output.

Definitely - if you want 9V out, use a three terminal 9V regulator like the 7809.

[Neta]

Because power is resistance times current and more load is more current? Did I get this right?
And how is 13.69v alright? I thought the manufacturers were supposed to calculate the values so that even with the full wave rectifier (which I think usually drops 1.4v), the output is still 9v.

Now the things I read in "electronics for guitarists" is starting to come back to me. Seems a regulator is the obvious choice..

Thanks guys!

[GibsonGM]

Power is I^2 * R, but you were close.  Yes, more "load" implies more current, and all things being equal will mean a voltage drop. 

Yes, it is related, in the sense that if you use a resistive voltage divider you must have about 10x the current in the divider than what you wish to draw from it, to prevent sag - which is of course related to that relationship.

The manufacturers of transformers like this rate them AT A CERTAIN CURRENT DRAW.  You may find that 13.69 is very close to 9V at "300mA" or whatever was on the housing.   That is as close as you're going to get for this kind of circuit.  When you're getting started, this can be confusing (it was for me), until you really think about that "9V @ 300mA" or whatever they rated it at.    What you have is an UNREGULATED transformer/rectifier rather than a real REGULATED power supply...

Like they said...get yourself an LM7809 and you'll have a very nice 9V supply for hobby use (pedals...).  The 'extra' volts you see with no load will be used to run the regulator.   This is a good start.  Down the road, you get to try to figure out the voltage/current ratings of unknown transformers, and come up with ideas for how to use them

[Neta]

That makes a lot of sense! The power supply was for a 500mA electric device, I just took it and added more jacks to it.

So "load" usually means greater current consumption, and in our discussion more pedals connected, and sag is when there's not enough current for all the devices?
And what's the difference in operation terms between lack of voltage and lack of current?

Of course, thanks for the help

[GibsonGM]

"Load" simply means "something acting to use power at the output of a power supply".   It can be a LOAD RESISTOR, which is also the bleeder for the filter caps in some power supplies.   Or it could just be the device you're trying to power.   In THIS case, they expect a load that will eat 500mA, and so designed it to drop the voltage to about 9V at that level of current consumption.

When you connect something that draws about 500mA, that 13.7V will drop to NEAR 9V.  Within tolerances.  This is an unregulated power supply, remember. It is not SUPPOSED to be exact - the thing they designed it to run with didn't need that level of precision.
So the voltage DROPS DOWN from that high to where it's designed to be....that is the LOADED VOLTAGE.   To make the output STAY at 9V, consistently, you need to rectify the output and add a voltage regulator.

What you measured before (13.7v) is the UNLOADED VOLTAGE.  Totally normal for what this was made to do.

Now say you draw 500mA, and read 9V.    You then connect something that draws 600mA, and you read 8.2V.  That is voltage SAG.  The supply cannot keep up with the demand made on it, even though that demand is out of the device's specifications.  YOU will say it is sagging; the manufacturer will say you're not using their transformer for what it was meant to power.

Lack of voltage means you're suffering EXCESSIVE CURRENT DRAW.  "Lack of current" is almost same thing...we would say the transformer lacks the AMPACITY to supply the desired current.    As current flows thru a resistance, it causes a voltage drop - naturally.  When that drop becomes excessive, THEN it is a problem.  Not until. 


Remember - ANY transformer will sag at some point - they cannot supply ALL the power demands we might make on them - just what they were designed for     IMO, the best way to utilize them for what we do is to connect a voltage regulator (LM7809).

[Neta]

Thanks for clearing everything out!! Now I fully understand.
All of a sudden my previous replies seem quite dumb...

Thank you again!!