Ohms law question

[9 volts]

Hey there I have an ohms law question. I have an 18 v transformer that I need to drop to about 800mA. Looking at a variation on ohms law:
current = voltage/ resistance
Eg 18/200 =.9 (900mA)
Does this need to take into account the current of the transformer in order to be correct or will a 200r wire wound resistor always give 900 mA?
Thanks

[feddozz]

If I understand what you said, an answer cannot be given to your question.

You need to know the impedance of your load because, the resistance you need to use in ohms law is the total resistance you want to apply.

I.e. suppose you need to supply your .8A at 18V using your 200Ohm resistor above to a 1M resistor. You will have guessed your total reistance is no more 200ohm but 1.2M ohm.

Now apply ohm law and you current value is now 10 times smaller!

[GibsonGM]

Hi 9,

Well, you're on the right track, but with Ohm, we have to really pound in the basic principles to make them stick. It's not easy, and takes time!   It happens to us all.

You have voltage, which is "pressure"  - think of a water tank.    You have water flowing from the tank down a pipe into a....bathtub or something.    The flow of water in the pipe is current.     If you have a valve at the bathtub, a faucet - that is resistance.

Ohm's law means that the current (I) flowing in a wire equals the voltage (E) divided by the resistance (R) in the circuit.    I = E/R    This re-arranges in many ways for convenience, more later.

You don't "drop current", altho you can drop voltage.  A circuit DRAWS current...so, a light bulb connected to a battery will DRAW whatever current it wants to...it is a physical property of the light bulb.   The power source simply either can, or cannot deliver that current...which is really 'how many electrons per time period' are being asked for, and given.    That's called the "ampacity" of the power supply.


OK, to your original...if you wanted to make 800mA flow in a circuit CONNECTED to a transformer, then you get .8A = 18/X
X = about 22 ohms, for 818mA.     This current is also flowing in the transformer secondary, correct.

What are you trying to do, or figure out, 9?   This is what I mean about 'fundamentals'...hard to wrap your head around at first, but don't let that deter you!!

After you know what resistor gets you the desired current (22 ohms, 800mA) you have to figure out the POWER thru it.   Power in watts = Voltage X Current.

So,  18 X .8 = 14.4 watts.   Most stompbox resistors are rated at 1/4 watt...so, your resistor will light on fire if you do this in the real world       Power calculations like this are one of the most important things you can do before putting something like this together, ha ha!    It's why so many ppl burn out their potentiometers. 

The current will be "about correct", since in the real world components aren't exact, and here, you're deal with AC, not DC, so you're using RMS values.   Close enough to make sure the thing works without lighting up.   At times we must also take into account PRIMARY current, since nothing is free...you can overtax the primary as well and burn things up.

[9 volts]

Thanks I'm trying to build a power supply for my alesis multimix6 fx. It takes two lots of 18v ac.
The only transformer I could find is 30va so I hoped to get the current down using resistors on each of the 18 v transformer secondary. Thanks again

[FiveseveN]

Why do you want to "get the current down"?
You stick both your phone charger and your vacuum cleaner into the same hole in the wall. One uses about 5 W, the other 1000 W. How does the mains outlet know how much power to let into either one? It doesn't, see GibsonGM's reply. A circuit will only draw as much current as can flow through it.
The original PSU for that mixer is 19 W. Your 30 W transformer will do just fine.

[9 volts]

Super! Thanks. I will re read Gm's post.
Thank again. I haven't posted here for a while and it's great to get such a quick response.

[GibsonGM]

It's ok, 9, it is easy to get lost when new, and sometimes even us old guys still do!!  I get something wrong about once a week, ha ha.


Now, what I WOULD be a little thoughtful about (and I know PRR or someone will see this, and chime in)...is that a specific transformer is probably going to say "18VAC @ 800mA"....if you only are drawing 200mA out of it, the voltage will be HIGHER, by a good amount, often, as this is not what is called a 'regulated power supply'.    What is really happening is that the load you put on the trafo makes the voltage "SAG"....to where the manufacturer has designed it to be, if your load matches their data sheet.    If your load is less, you may get too high a voltage...

Do you have a DMM?  Measure the output of the transformer if you can, "unloaded".  Don't be surprised to see 28VAC or something!     Can you tell us what is "written on the back of the mixer, by the power plug-in?    It sometimes says what it draws....rather than "getting the current down", we may find you need to get the VOLTAGE down...

If you have not hooked up a transformer to line current yet, disregard this and come back for safety info if needed, ok?  Wall power can kill you, we don't want that!!!

[antonis]

Wall power can kill you, we don't want that!!!

I completely agree with the first sentence but I have my doubts about the second one.. 

[GibsonGM]

Wall power can kill you, we don't want that!!!

I completely agree with the first sentence but I have my doubts about the second one.. 

You mean the very first and second sentences of my entire reply, right Antonis?  The quote got chopped     

We want 9 to learn how to mess with line power, if he doesn't know already, lol...wire nuts, tape, set it up, clear your body, plug in, read meter, unplug....that stuff. 

[9 volts]

Ha, I'm good with line current. (built several tube amps and synths- though several years ago...). Though I guess most accidents occurs when one gets alittle too comfortable/slack etc.
The mixer has two regulators inside 7915 and 7815. So I guess they should take care on the voltage issue. My initial concern arose based on some roland equipment....the sh101 and tr606 both will apparently cook if connected to the bigger current than required. (from memory this was due to have no regulator inside). I figured a small mixer with FX may have had a similar power setup.
ps the mixer doesn't have any info written near the power input. Just a three prong input.
Thanks again
I'll have more time this weekend to test it out.

[antonis]

the sh101 and tr606 both will apparently cook if connected to the bigger current than required. (from memory this was due to have no regulator inside).

Maybe you mean bigger VOLTAGE or you should study more about Voltage & Current sources..
(which are absolutely theoretical concepts but let you "embed" terms like Voltage, Current, Load e.t.c and their interaction..)

You can't connect to "bigger" current anything because you can't "force" extra current to it (other that the required from it's demands..)
(as long as the "force" source remains unchanged..)

Sir Mike told you about the water inflow analog of current but I think that you've loosely skip it... 

[GibsonGM]

This is what I was getting at, 9!  It is very easy to read a quick thing about voltage and current, think you understand it, and then move on.  And then make the same mistake over and over, ha ha!  I have done so many many times...we all have.   It is ok, keep reviewing the concepts and they will make sense, and it will not take a long time, either. 

The MIXER draws the current that it wants - IT is in control, so to speak.  Like a sponge taking on water. 

The power supply sits there with a pressure on its output, if you will.   It is allowing the device to draw the current that it wants, it is NOT pushing it into the device.

If the power supply doesn't have enough ampacity - ability to SOURCE current - the voltage sags, maybe enough to make the device inoperable.   In fact, an unregulated supply WILL sag, naturally, and if designed correctly it will end up where it needs to be from this sag.    So, "unloaded" (no device attached), you'd measure a higher voltage.

If a power supply is replaced, as in your case, and is unregulated (not your case, it appears), and has more ampacity than the original one (a higher unloaded voltage...), it may not sag to where it should be for proper operation, which may damage the device.    Meaning...the draw of the device may not be enough to pull it down. 

Here is the takeaway for your problem as I see it:

Your 3-terminal devices (the 78XX, 79XX regulators) WILL take "any" DC power and regulated it, within their device specification limits.    BUT - if the amount of DC is too much - if you ask them to drop too much extra voltage - the device may get too hot to operate unless it's properly heat sinked.   If your replacement trafo results in this condition, this can happen.

I am assuming that inside the mixer is a rectifier and a few caps, since you say you the mixer calls for an AC supply.  When you apply your 'new AC voltage', I'd suggest measuring the DC going to the regulators if you can FIND it (probably microscopic, LOL!).  Best case is that it's a few volts over what the regulator puts out, but not much more, as again - the more it has to drop, the hotter it will get. 

[antonis]

BUT - if the amount of DC is too much - if you ask them to drop too much extra voltage - the device may get too hot to operate unless it's properly heat sinked.

Pardon me Sir but your verbalization is rather confusing... 

It could be interpreted from our friend like something " If we ask the regulator to drop much voltage then it draws more current" or vice versa..

Maybe you should insist about the "fixed" current drawing (from the load) and refer to the product of current x voltage drop which you could name it something like "Power Loss" or similar..

[9 volts]

Super! (I'm always up for learning!) it does make sense. (Beginning too!). So if I do a voltage measurement on the input voltage leg of the 7815 and 7915 I will get a measurement of the converted dc post rectifier pre regulator. Then check that the voltage is within the spec of the 78xx.
You're right a diode bridge and capacitors. The power board pcb thankfully isn't surface mounted. If it was I wouldn't have got this far...might take photo over the weekend.

Just to clear up the Roland Sh issue, regulated would be needed if there isn't a regulator in the circuit as unregulated would have too much voltage. So the mA is not the issue, it's the voltage. (Correct?).
Once again thanks, I know this is all a little off the topic.
Seems like more problems would arise from not enough current, than too much. It's a grey area for me. I measure voltages, resistance etc but the mA part of the dmm.....well I guess it's not just there for decoration! Time to study a little more. Keep you posted.
Ps there is a large heat sink is in there.
My understanding is that the power board produces 15- 15+ and 5+ (For the digital part).  But also has to have the 48v for the phantom power option. Hence going for dual 18 v. I've pulled bipolar 15 off 12v transformers (voltage increase after diode bridge). So I guess original 18v adapter is needed for this cover all the demands.

[GibsonGM]

BUT - if the amount of DC is too much - if you ask them to drop too much extra voltage - the device may get too hot to operate unless it's properly heat sinked.

Pardon me Sir but your verbalization is rather confusing... 

It could be interpreted from our friend like something " If we ask the regulator to drop much voltage then it draws more current" or vice versa..

Maybe you should insist about the "fixed" current drawing (from the load) and refer to the product of current x voltage drop which you could name it something like "Power Loss" or similar..

Nah, it's something that one has to get over, Antonis...a hurdle of sorts.  Feel free to chime in! 

If you ask a regulator to drop an input DC voltage to a regulated voltage (which is what it is meant to do!), it will generate heat.  It is trying to 'dump' the excess voltage.  That is why yours has the metal heat sink.  You need a couple of volts over your regulated voltage to operate the regulator (so, for 12V regulator, you want 13.5, 14 volts input).     If you make the input voltage much greater than the output voltage, however, all that extra drop must be accounted for.  The  way we account for it is by using a heat sink to dissipate the heat that is generated.   This is much like how automobile brakes get hot from using them on a slope.  The greater the slope, the hotter your brakes will be.  This is 'power dissipation'.  Actually, it's measured in Watts, and is the 'second half' of Ohm's law (goes with it). 


Yes, 9V, you would want to know what is coming out of the diode bridge, going TO the regulators!  Exactly.

Whether 'regulated' or not is the issue is hard to say from here. If the circuit HAS a regulator, then regulation is important...I would hazard a guess that the Roland does in fact use regulated power.    In something else, like a stompbox distortion, they are designed AROUND a nominal voltage like 9V.   As long as any caps, ICs etc. inside are rated higher than the voltage you apply, you are fine (altho you will begin to experience greater headroom, and less distortion!).   

The Roland and Alesis are going to be more complex devices that likely depend on a well-regulated power supply for noise immunity and 'consistency' of operation. Plus, it's far easier to get those voltages from easy to use IC regulators than resistors and stuff

Read up more on the relationship between current and voltage....it's really very interesting once you get past the geek part, ha ha!   Do some experiments, even, with a small safe power supply and resistors.     Draw more current, voltage will drop.    Decrease demand for current, voltage will rise (with UNregulated supplies, that is).  A voltage DROP across a resistor indicates higher current flow IN the resistor!   Basic electronics, and boring until it sinks in - then it becomes useful and interesting      

What YOU are working on now is something that designers put together, knowing the draw and the voltage they want.  So, the original power supply is going to be best, IMHO....much easier for YOU, honestly!   $39.95 and the problem is solved, or you have to dig in and do a lot of work to reinvent what they have already done for you!