A Few More Questions About AC/DC Power Supplies...

There's no such thing as constant current, yet there is a constant and also variable load. Even voltage drop a little under load.

Load(resistance) dictates circuit current. Elementary electricity - Ohms Law V= IR.

Residential Voltage is at 110Vac in the US, some other countries 220Vac -constant Voltage Supply, Yes?

It's not a beasty power supply you got there. Only that your resistor has a lower power capacity, it burnt.

No, you are wrong. There certainly are constant current sources. Every operational amplifier has at least one in them. Most LED lighting supplies are constant current sources. It is also very common for a battery charger to operate initially as a constant current source.

No offence intended but electrical engineering is a field you do not know enough about.

I apologize, I missed to consider other application and focused about load (like house appliances, motors, refrigerators etc.) but would it apply to the OP scenario? Yes?

There's is a voltage regulator and current regulator too, I admit. But, the fact that they existed, it is conversely that, the root cause is voltage and current fluctuation, Yes?

Oh, and I never claim to be an expert of that field, it just came to be that I knew a little about it. I apologize if it offended anybody.

OP's got an extra bonus.

As resistors heat up, their resistance increases, and thus the current decreases.
In this case, 8 volts into a 330 ohm resistor should result in 8/330 ~~ 0.024 A, dissipating 8²/330 ~~ 0.194 W. What wattage rating is your resistor?

It's either an increase in the "load resistance", or a desperate attempt by the power supply to limit its output current to forestall its destruction from the abuse it's getting.

Please have another look at his pictures!

The said transformer is indeed red and surely the transformer from hell! :-)

For those who want to know more about a NIMH quick charge supply, Linear technology has a nice application note.

I forgot to answer your question about humming. Typically this is an indication that the transformer inside is saturating. As James Doohan's most memorable character, Montgomery Scott, would often say "I cannot give you any more!"

This is not always a sign of trouble. Some circuits are designed for the transformer to be saturated in part of the operating cycle. In other circuits a buzzing transformer is the sound heard just before something releases the inherent smoke hidden inside every piece of electronics. Once the inherent smoke gets released the part no longer works as planned.

As usual redfred nailed it: You have a constant current supply. Welcome to the smoke theory!

Unless you are trying to design circuits at the component level, you might want to invest in some type of electronics experimenters kit. Look for one with a power supply, breadboard, various components, specific projects, and a manual. Personally I like the Arduino's (you will also pick up some programming), but there are a lot to choose from online.

Oh, and it wouldn't hurt to brush-up on a multimeter tutorial. Forgetting to change leads when going from current to voltage readings is a popular way to fry things.

I'll embellish on what others have said.

You would benefit from reading some internet tutorials on the basics of electronics.

You will need to understand algebra and one of the most fundamental formulas in electronics is Ohm's Law:

I = E/R where:

I = current in Amps
E = Volts
R = Resistance in Ohms

Another formula of interest is:

P = E • I where

P = power in Watts.

Understanding those formulas will give you a basic understanding of electricity and why the resistors heated up and burned out.

Also, a more subtle effect is that heat will change the performance of electrical devices, particularly resistors, which tend to drop in resistance slightly as temperature goes up.

Conversely, a typical failure mode for a resistor is to increase in resistance when overloaded beyond their power ratting in Watts, eventually burning out and becoming infinite in resistance. That is what happened to your resistors.

I = 8 VDC / 330 Ω = 0.024 A (or 24 mA).

P = 8 VDC • .024 A = 0.2 Watts

Both of these numbers suggest that what you did was not exactly as you described. If the resistors you used were 1/8 Watt they probably would burn up, but if they were 1/4 Watt resistors they would survive.

However, the meter should not read 0.84 A, so you did something very wrong - likely hooking the ammeter up in parallel as Fred suggested. Bad boy!

The other question to ask yourself is if the resistors were really 330 Ω and not something else like 33 Ω, or even .33 Ω.

Rearranging the Ohm's Law equation to solve for resistance to draw .8 Amps:

R = E / I = 8 VDC / .8 A = 10 Ω

Your resistors or the improperly executed connection made by the meter created a load resistance of 10 Ω or less.

Extra credit reading:

Unregulated Power Supply Design

I don't believe I made a "parralel circuit" (Look at the diagram...did I?) but I often strip wires extra long and do not insulate them to give me places to more easily measure. Please look at the picture and tell me this: QUESTION A) If this circuit was running, and I put my multimeter leads on each YELLOW STAR.... would that have shorted out my multimeter? Assuming that the area by both stars were not insulated, of course. (Because I've definitely done that more than once...)
Thank you for confirmation this is a constant current power supply. But, how can one tell? I mean when I plugged the adapter in, I did a multimeter test between the negative and positve sides to test for current: the current DID drop while I measured it...but it did so EXTREMELY slowly... would a non fixed curent adapter drop quicker? Does it depend on the multimeter? ....so many variables....
QUESTION B) If it was *TRULY* a fixed current adapter, shouldn't I have seen it hold steady at one amperage? or perhaps drop .01, then rise again/right itself? Because I did not see this- I only saw an extremely slow descent...
On another note, it is official....I've officially fried all three of my multimeters. To be fair, ONE arrived Dead on arrival (A gift from my uncle: an antique from the 70's...) ANOTHER cost me ten bucks at walmart (worked fairly well until I messed up with it more times than I can remember...) and THE LAST ONE I got for free from Harbor Freight with a 2 dollar purchase. So, for awhile I think I will just draw diagrams and ask you guys whether or not they are correct. I will concede I am not yet ready to be "MacGuyver." I'll start off drawing circuits and ordering the right parts, lol...
Thanks everyone for your help once again. This is rapidly accelerating my learning, and I hope it's not long before *I* am helping some others out. Lyn: I plan on reading that link upon finishing this.... Thanks everyone.

Unplug the + lead in your breadboard, connect it to your meter.

Plug the neg meter lead into the place where the + lead was.

The load has to run through the meter.

The meter has to be able to handle the load.

Such a good response anonymous. And thanks everyone for making my learning process so much easier.
As usual, I will have to put this in my own words to clarify, and please tell me if I "screw it up" when doing this.... "When measuring Amperage, both the negative AND positive leads should be placed after the resistors and as close to the "positive"/ground terminal as possible." (Because we want to sense the ending amperage in a circuit.)
...did I get the jist right? Or did I screw it up somewhere along the way? Let me try another...
When measuring Voltage: ....l don't even know to be honest, lol. Electonictheory.com and electronics for dummies both had me believe voltage was constant throughout the circuit, so I never thought before about hwere I placed them.... (I am not yet advanced enough to begin my experimentation with voltage dividers. THat was step 2... However, I've found a lot of steps between 1 and 2...) Let me try and understand this. Are you telling me that if the Topmost diagram was a SERIES circuit instead of a parrelel (IN other words: Imagine it as the bottom diagram rather than the top...) I would place one lead BEFORE the resistor and one lead AFTER? If i wanted to measure voltage?
(My brain feels like it wants to explode, lol. The books I read made electronics seem so easy, but there have been a lot of details they've obviously neglected.... thanks everyone for your patience. Has been much appreciated.)

"When measuring Amperage, both the negative AND positive leads should be placed after the resistors and as close to the "positive"/ground terminal as possible." (Because we want to sense the ending amperage in a circuit.)

No. There are some key concepts that you are not grasping. People in this thread and your last thread have written about Voltage as PRESSURE, and current as FLOW. Those are 2 entirely different things.

When you say: "When measuring Amperage, both the negative AND positive leads should be placed after the resistors ". You are basically asking us whether the front bumper of a car is moving at a faster speed than the back bumper! It doesn't matter where you measure the speed, it is the same.

I/we can give you analogies, but only if you give US a frame of reference, in terms of what you understand. Do you understand hydraulics (flow of fluid through a pipe)? Do you understand mechanics (in terms of horse power, speed/velocity)? The relationship of those to electronics is not perfect, but some analogies can be drawn that might help.

Sadly, I am involved in a startup of some equipment this next week, so I will not be able to help a lot. But if you give us a clue to your background understanding of how things work, I am sure I, and others, will try to help you understand. But you have GOT to give us something to work with.

What is your background understanding of STEM?

First of all welcome to CR4.

Please forgive me I'm not normally rude but I can't resist: perhaps you should have used Christopher Lloyd's other alter ego for your avatar until you get a little better at this:-

Keep trying to understand what Anonymous Hero is telling you, especially by looking at his pictures above.

And, always start with Ohms law. Volts = Amps x Resistance.

You are not being rude at all. XD. It is a very good suggestion. I am several decades away from developing the quantum flux capacitor. Also...may I commend you for the wonderful pictures of Fester? God bless Christopher Lloyd.

Lyn: Within the first 15 seconds of reading your link I stumbled upon something that made me think....
" Before designing any power supply the load requirements must be known. It is always a good idea to take the worst case scenario when making this decision. For example if your circuit is designed to draw 1 amp at 12 volts, assume that component tolerances are 20% and design to meet these requirements with at least 20-50% reserve current, in this example I would design a power supply which could safely deliver 12 volts at 1.5 amps without overheating. "
In the example it talks about giving "reserve power." but IMMEDIATELY when I would measure the amperage of this ac/dc connector in the circuit it would start me off at around .82 amps current- even though it was designed to put out 1 amp and would do so if I just measured current from one side of the adapter to the other. For every amperage drop below .82 it would FIGHT ME. 1 amp to .82 amps... Seems awfully close to that 20 percent component tolerance. And awfully close to the idea that this device was trying to keep me within a 20% component tolerance...
A good read. Will finish it now.

Please (re)read RedFred's post #3, you are not using your multimeter correctly. An ammeter must be inserted in series with a load, not across it (in parallel). An ammeter is a very low resistance device so as to introduce as little voltage drop as possible. When you attach it to the yellow stars you are putting what is essentially a dead short across your power supply, and since it is a current source it will attempt to feed current into it up to its rating, and beyond if there is no regulation in it.

The humming you are hearing most likely means you have reached its rating and it is working as hard as it can, a situation known as saturation.

PS- You may not have fried all those multimeters, many of them have an internal fuse that opens when you pass too much current through the ammeter portion, especially if you place the leads across a power source as you have.

Oh, rest assured.... I've screwed up MULTIPLE TIMES with each meter, lol. Enough times that if they fried, I don't blame them. So I guess the silver lining is the most expensive of them cost 10 bucks....
I will chalk it up to a learning experience. P.S: I know they are fried because I can try to measure the voltage on a battery and get nothing. if I try to measure the amperage, I get the same result or a number which varies rapidly.... They are fried, and I'm the one to blame. I will go back to theory for a bit, increase my understanding, then get a better/More foolproof multimeter when I'm ready to come back to experimentation.

Look at AH drawings in #17 again. Where he says "load" read "resistor". Then the line coming from your power supply to the A meter is a cable/conductor/lead/wire then from the other side of the A meter is another wire that goes to the resistor. Then from the resistor back to the power supply is another wire. NOW the important bit. On your meter is different connection for measuring Amps. Usually marked AC/DC 10A +. PUT YOUR PLUS (red) LEAD IN HERE.

When using a multimeter go through these steps each time you test.

1. set the range/type of test. (in this case set to amps, plug lead into correct hole)

2. test the meter. ( in this case put leads briefly on each end of an ordinary carbon battery. )

3. be wary of shorting out leads.

Look again at the drawings you should work out that they are both connecting the load (resistor) in series. To measure amps you change the meter and connect it into the series circuit. to now go and measure the volts you unplug the red test lead and put it into the hole marked V,Ω +. then put those leads on either side of the load (resistor ). Your meter is now in parallel. In the drawing the two lines from either side of the M are the test leads from your meter.

If you don't change your meters' leads to measure A you will blow it up.

Jim

Step one has the meter set to amps.

Step two has you placing both leads across a battery (power source).

Wa that what you intended to write?

Yes, but specifically an ordinary dry cell carbon battery. As this is a dead short a very hot finger could result. This just tests that the set up is testing amps. A NiCd battery will get very hot or go bang and so given this guys level of understanding perhaps I should have left it out altogether. I am just repeating what I was taught so long ago.

On many occasions upon 'testing the meter' I have found nothing. Pull the leads, reinsert and test again, OK. Cheap meter, little use, Murphy or wrong setting (AC vs DC)etc. So it is something I would still recommend to do. To check Amps I use a D cell dry carbon type battery.

I did also put those specific steps in brackets and say "in this case" because in this case we are trying to get the OP to understand that he has to pull the lead and plug it into a different hole to test A. You tried twice with what I thought was perfectly lucid and simple to follow directions but to no avail. I had a go; so lets see if he gets it now. Like most OPs he's probably gone and left us to discuss the topic 'till it dies. Please correct me if you think I am off track in any way. I, for one, value your input.

Jim :)

you might like to try Docircuits an on line electrical circuit simulation program. When you make a mistake, the virtual components don't burn up. After you simulate a circuit and test it, then you can build it on your breadboard.

By the way, try to resurrect that analog multimeter, They are great for learning, as you can watch changes in voltage and current when your study gets into capacitors and inductors.

I'm looking into this right now...thank you for your suggestion. I was looking for a valuable resource like this. Before DoCircuits I was looking at circuit lab. I also dabbled (one youtube tutorial...) With EAGLE, but it didn't seem to have a good option to "test" my schematics....seems an invaluable resource for finding components that fit, however.

First, replace your power source! That package is a specialized power source that is expecting a particular kind of a load. It will change its operation between current and voltage source depending on the load it sees and how long it sees that load. This will quickly confuse a student since no status lights exist on the supply that will tell you what is happening when. You want a small size constant voltage power supply for your testing.

A 9V alkaline battery would be a good choice. This will produce voltages high enough that your meter will easily read what is happening but will be unlikely to damage components, wires, test instruments or student as you make learning errors. The drawback of using a 9V battery is the output voltage will eventually drop when you pull too much of the power out of the battery.

Second, when you use your multimeter the test lead polarity is nothing but a polarity orientation of the meter. In other words, if you test the voltage of your 9V battery and put the plus lead of the meter to the plus terminal of the battery you will read about +9V, depending on the age of the battery. If you put the negative lead of the meter to the plus terminal of the battery the meter will read -9V. The voltage coming out of the battery has not changed, only the orientation with the meter. [I assume you can figure out where the other lead of the meter should go in both scenarios.] When you use your multimeter as an ammeter the convention is a positive current reading means that current is flowing into the positive end of the meter and out of the negative end.