@Juicyblunts Please stick to one thread on the same topic otherwise life on AAC will quickly become very confusing.

Moderator action: merged new thread into existing thread.

 

Okay so the 7805 is unnecessary. I suspected this. The circuit will blow the transistor. This is great information. But can anybody tell me why or how to wire it differently? As for proper procedure for measuring current, I can see where I might have lead some to believe I simply touched the leads to the battery and expected a reading. To clarify: after failing to obtain a current measurement within the drawn circuit, I used a separate breadboard with the battery leads plugged in, and a couple random resistors. I understand you can't probe the two terminals of a battery and expect a reading, just poor wording on my part. I will check my multimeter internals today for a blown fuse nonetheless. Any advice though on how to make this circuit function properly? When I try without the 7805 the pot gets really warm, is 12V too much for the 10K pot, or vice versa? 10K is the smallest value I have.

 

the pot gets really warm

Sounds like something is wired wrongly. If you have the 1k (confirm it really is 1k) base resistor in place and the pot correctly wired then the maximum current through any section of the pot track will be <12mA and the maximum power dissipation in the pot will be ~40mW. Unless the pot is tiny it should get slightly warm but not excessively so.

 

A 10k pot will not get warm or hot if connected across 12V in a certain manner.
However, you will destroy a pot of any value if you connect it incorrectly across a 12V source.

You need to pay attention to a number of technical details, related to electricity and electronics, each one of which will cause you a lot of grief.

Take one thing at a time.

Lesson #1 - How to correctly use an ammeter
Fix your ammeter. Learn the correct way to use an ammeter. We will teach you how to do that. Focus on this one thing and you will learn the first thing about the magic and secrets of electricity and electronics.

 

Thanks Alec_t! I am going to go over everything again and again until I figure this out. It's great to have things like iCircuit when planning these circuits but it's unfortunate that it rarely agrees with reality. I just can't figure out why it has worked for me several times before and now suddenly does not work at all. Strange but with all the help I'm sure it will be sorted out soon. The potentiometer is a pretty standard size I think, the wires are soldered to the leads (clean solder job I've checked to ensure no shorts) so it's a decent size. I was tired last night after spending all day building the frame and finally getting around to setting up the electrical components so I could very well have missed the simplest of thing. Also: update on multimeter problem. It was not a blown fuse, rather this thing was missing the fuse completely!!! I've had this thing for a few years but never really used it after purchasing a good high quality meter a couple years ago. I keep it in my toolbox at work though so I used this old one since I had it handy. Anyways, thanks guys for encouraging me to look into the fuse!!!

 

MrChips, thanks for the advice. I'm in my 3rd year apprenticing for heavy duty mechanics so I've learned hands on in class how to use a multimeter for testing current, voltage and resistance. I understand how to use it and mostly how it works (at least how it works in theory). I also learned most of what I do know about electricity and electrical circuits on this very site. The trouble is that I have spent more time working on and learning about mechanical systems in the last year than I have electrical so I might be a little rusty in some places. I do tinker frequently in my free time but usually with much lower voltages, for example I also am currently working on an 8 Bit breadboard computer highly influenced by Ben Eater's breadboard computer series on YouTube though I've implemented several of my own changes, the point being that though voltage is important in such a project, it doesn't have to be monitored too much. This is the first project outside of automotive/ heavy duty that I have worked with voltages over ~5V, and is my first project experimenting with transistors and potentiometers, so I am not ashamed to admit that I am outside of my comfort zone here, hence why I have asked again for help on this.

 

The reason the fuse is missing is the same reason the pot is getting hot.

Let's talk about the fuse thingy for the moment. You might learn something new.

 

I'm interested. . .

 

I'm interested. . .

The multimeter still works on the voltage measurement function. There is no protection fuse. Why is that?
It all comes back to the all important Ohm's Law which states that the current in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance.

In simple mathematical terms, we say

I = V/R

The resistance of a voltmeter is very high, typically 10MΩ. Hence typically, a voltmeter can be subjected to 1000V and experience no damage.
In such a situation, the voltmeter is seeing 1000V/10MΩ = 100μA

The manufacturer of a testmeter wisely installed a protection fuse in the ammeter function of the testmeter. Typically, this is a 300mA fuse. Why did they go to the trouble of protecting the ammeter. Because they know that someone is going to do something stupid and blow the meter. The manufacturer would love you to have to buy a new meter. Of course, replacing a fuse is a lot cheaper than replacing a DMM.

The resistance of an ammeter is very low. Let us say the resistance is 30Ω. With a 300mA protection fuse, V = I x R = 0.3 x 30 = 9V.
10V across the ammeter will blow the fuse.

Why is the resistance of the voltmeter very high,10MΩ, while the resistance of the ammeter is very low, say 30Ω?
That is because when the voltmeter is trying to measure voltage across two points, you want the voltmeter to take as little current as possible. That is, you do not want to disturb the circuit you are trying to measure. An ideal volt meter should have infinite resistance and should look like nothing is connected to the circuit.

When the ammeter is inserted in-circuit to measure current, you want the voltage across the ammeter to be as little as possible. Again, you do not want to disturb the functioning of the circuit. An ideal ammeter would have 0Ω resistance.

The bottom line:
To measure current in a circuit, you must break the circuit and insert the leads of the ammeter where you made the break. What is protecting the meter is the total resistance of that part of the circuit. If the circuit resistance is too low you will blow the fuse.

(It is very likely someone blew the fuse in the past and didn't bother to replace the fuse because they did not have any at hand. Always keep a stock of meter fuses. Been there, done that. We all make mistakes too.)

If you are comfortable with this we can move on to why your pot is getting hot, and that is for the same reason, Ohm's Law.

 

The multimeter still works on the voltage measurement function. There is no protection fuse. Why is that?
It all comes back to the all important Ohm's Law which states that the current in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance.

In simple mathematical terms, we say

I = V/R

The resistance of a voltmeter is very high, typically 10MΩ. Hence typically, a voltmeter can be subjected to 1000V and experience no damage.
In such a situation, the voltmeter is seeing 1000V/10MΩ = 100μA

The manufacturer of a testmeter wisely installed a protection fuse in the ammeter function of the testmeter. Typically, this is a 300mA fuse. Why did they go to the trouble of protecting the ammeter. Because they know that someone is going to do something stupid and blow the meter. The manufacturer would love you to have to buy a new meter. Of course, replacing a fuse is a lot cheaper than replacing a DMM.

The resistance of an ammeter is very low. Let us say the resistance is 30Ω. With a 300mA protection fuse, V = I x R = 0.3 x 30 = 9V.
10V across the ammeter will blow the fuse.

Why is the resistance of the voltmeter very high,10MΩ, while the resistance of the ammeter is very low, say 30Ω?
That is because when the voltmeter is trying to measure voltage across two points, you want the voltmeter to take as little current as possible. That is, you do not want to disturb the circuit you are trying to measure. An ideal volt meter should have infinite resistance and should look like nothing is connected to the circuit.

When the ammeter is inserted in-circuit to measure current, you want the voltage across the ammeter to be as little as possible. Again, you do not want to disturb the functioning of the circuit. An ideal ammeter would have 0Ω resistance.

The bottom line:
To measure current in a circuit, you must break the circuit and insert the leads of the ammeter where you made the break. What is protecting the meter is the total resistance of that part of the circuit. If the circuit resistance is too low you will blow the fuse.

(It is very likely someone blew the fuse in the past and didn't bother to replace the fuse because they did not have any at hand. Always keep a stock of meter fuses. Been there, done that. We all make mistakes too.)

If you are comfortable with this we can move on to why your pot is getting hot, and that is for the same reason, Ohm's Law.

That's great info thanks! It's also the same reason I used 2 resistors when testing the battery isolated from my EL circuit, as 1/2 the voltage should be dropped across one resistor, though I'll be honest I didn't take into consideration calculating total resistance, I just wanted some kind of reading, even a blown fuse would have been preferrable to the 0.00 (had I only known the fuse was missing!!!). So what is with the pot getting hot?

 

That's great info thanks! It's also the same reason I used 2 resistors when testing the battery isolated from my EL circuit, as 1/2 the voltage should be dropped across one resistor, though I'll be honest I didn't take into consideration calculating total resistance, I just wanted some kind of reading, even a blown fuse would have been preferrable to the 0.00 (had I only known the fuse was missing!!!). So what is with the pot getting hot?

Potentiometers are known by various names, pot for short, variable resistor, voltage divider, rheostat, volume control, taper, fader, pad, trimmer, slider, etc.
Essentially, it is a fixed resistor with a wiper or slider that slides along the body of the resistor.
Diagrammatically, it looks like:

A pot is a three terminal device. The total resistance is the resistance from end to end, A to B as shown in the diagram above.
For example, if the pot is specified as a 10kΩ pot, then the total resistance from A to B is 10kΩ.

If you connect a 12V source across the A and B terminals, the current through the pot is I = V/R = 12V/10kΩ = 1.2mA.
There are three ways to calculate the power dissipated by the pot.

P = I x V
P = V x V / R
P = I x I x R

All three formulas will arrive at the same answer.
In this example, P = I x V = 1.2mA x 12V = 14.4mW
The pot would not even get warm.

Some folks may attempt to connect the pot as a variable resistor by using terminals A and W alone, or B and W alone.
This works fine so long as you do not turn the pot to its minimum setting.

Suppose you dial the pot from its maximum resistance of 10kΩ down to 10Ω. With a 12V source, you have increased the current from 1.2mA to 1.2A!!!

Potentially, that is equivalent to 14W. At that point you will smoke the pot, burn out the resistance track at that end of the pot, in effect destroying the pot.

Bottom line
Limit the maximum current the pot will ever have to conduct for all positions of the dial.

Are you ready for the next lesson on how to create a current regulator for your EL lighting?

 

Potentiometers are known by various names, pot for short, variable resistor, voltage divider, rheostat, volume control, taper, fader, pad, trimmer, slider, etc.
Essentially, it is a fixed resistor with a wiper or slider that slides along the body of the resistor.
Diagrammatically, it looks like:

A pot is a three terminal device. The total resistance is the resistance from end to end, A to B as shown in the diagram above.
For example, if the pot is specified as a 10kΩ pot, then the total resistance from A to B is 10kΩ.

If you connect a 12V source across the A and B terminals, the current through the pot is I = V/R = 12V/10kΩ = 1.2mA.
There are three ways to calculate the power dissipated by the pot.

P = I x V
P = V x V / R
P = I x I x R

All three formulas will arrive at the same answer.
In this example, P = I x V = 1.2mA x 12V = 14.4mW
The pot would not even get warm.

Some folks may attempt to connect the pot as a variable resistor by using terminals A and W alone, or B and W alone.
This works fine so long as you do not turn the pot to its minimum setting.

Suppose you dial the pot from its maximum resistance of 10kΩ down to 10Ω. With a 12V source, you have increased the current from 1.2mA to 1.2A!!!

Potentially, that is equivalent to 14W. At that point you will smoke the pot, burn out the resistance track at that end of the pot, in effect destroying the pot.

Bottom line
Limit the maximum current the pot will ever have to conduct for all positions of the dial.

Are you ready for the next lesson on how to create a current regulator for your EL lighting?

Yes take me to school on this lol. I got the lights to work, but now the pot is still getting hot so I'll wait to see what more you can teach me about this before I destroy anything.

 

So you got rid of the LM7805 and rewired the circuit and EL driver.
Show us a circuit diagram of your new setup.
Provide us with links to the EL light and the EL driver.

 

Yes I got rid of the 7805 but that's the only difference. The pot is now directly connected to 12V rather than being connected through the 7805. I am considering doubling the pot resistance and trying that. As for a link to the exact EL wire and Driver I purchased, it was an eBay purchase from China, about 6 months ago so I can't access the item page, but found a similar one. Not much for info though http://www.ebay.ca/itm/1-2-3-4-5-20...ry-USB-/302226995542?var=&hash=item465e21f956

 

According to the diagram posted, you are likely to blow the transistor for the reasons we have already discussed. There in no resistance in the loop from the 12V battery to the transistor. Once the current through the base to the emitter is high enough, the transistor presents a short circuit across the 12V battery, high current will flow and burn out the transistor.

The EL wire is powered by a driver module. Where is the driver module connected with respect to the circuit you have shown?

 

The driver module is between the 12V battery and the transistor, that's what giv s the resistance to prevent a short circuit. I have a better schematic here hopefully this will clarify, what I mean. The EL driver is a 12V driver and has a red and a black wire. I have the red wire connected to 12V, the black wire to the transistor collector, so it is essentially held high at 12V until the transistor is switched and allows connection to ground through the transistor (hence why the black wire is connected there).

 

If your circuit is wired correctly the pot should not get hot. The highest current flowing from the top terminal of the pot to the wiper is 12mA.

How much current does the driver take at full 12V?

 

The only way I can see your pot getting hot is if you have mis-wired it like this :-

Double-check the wiring. The centre pin of the three on a pot is usually the wiper, but you can't guarantee it.