Like the idea but I only have one meter (unless it's implied that the measurements are taken independently from the same meter).

Here ya go!
http://www.ebay.com/itm/7-Function-Digital-Multimeter-/181687957100?pt=LH_DefaultDomain_0&hash=item2a4d72366c

 

$3 ? That's outrageous!! I have at least a half dozen of these and haven't paid for one yet. Just got another free one yesterday.

But yeah, I guess $3 is pretty good if you don't have a HF near your travels. Can't drive very far on $3 anymore.

 

Just got back from the local electronics shop. I got a cheap MM for $10. It was just not worth the time, effort, and gas money to save $6 to drive to HF 15 miles from my house.

I also picked up three 1-foot lengths of buss wire (tin-coated copper) of varying gauge to add to the conductor mix and a 1 foot length of aluminum shielded coax (which will act as our stranded aluminum after I strip it off).

I then hopped over to HD and picked up small spools of solid 18 gauge aluminum, copper, and galvanized steel in the picture wire hanging section, each for under $4.

I then verified at home that my PAD-234A still works after 22+ years and can provide us with a 5V or 12V DC power source (my meter actually measured 4.89V across tje ground terminal for the 5V) so I won't need a battery setup. It's actually this exact one that another forum member posted right here on this site.

So we might be able to perform that suggested resistivity experiment after all. I have a bunch of resistors (amongst all of my ICs/LEDs/etc from my digital design classes) and now I just need to know which one (or which acceptable range) would be best suited for our purposes so that the measurements we take will clearly show how the resistivity/conductivity of each different wire (or different gauge of same wire) under test differs from one other.

All of your continued support is greatly appreciated. I will report back tonight or tomorrow after I have a chance to tinker around with all of these ideas, tools, & matrrials with my son.

 

Just got back from the local electronics shop. I got a cheap MM for $10. It was just not worth the time, effort, and gas money to save $6 to drive to HF 15 miles from my house.

I also picked up three 1-foot lengths of buss wire (tin-coated copper) of varying gauge to add to the conductor mix and a 1 foot length of aluminum shielded coax (which will act as our stranded aluminum after I strip it off).

I then hopped over to HD and picked up small spools of solid 18 gauge aluminum, copper, and galvanized steel in the picture wire hanging section, each for under $4.

I then verified at home that my PAD-234A still works after 22+ years and can provide us with a 5V or 12V DC power source (my meter actually measured 4.89V across tje ground terminal for the 5V) so I won't need a battery setup. It's actually this exact one that another forum member posted right here on this site.

So we might be able to perform that suggested resistivity experiment after all. I have a bunch of resistors (amongst all of my ICs/LEDs/etc from my digital design classes) and now I just need to know which one (or which acceptable range) would be best suited for our purposes so that the measurements we take will clearly show how the resistivity/conductivity of each different wire (or different gauge of same wire) under test differs from one other.

All of your continued support is greatly appreciated. I will report back tonight or tomorrow after I have a chance to tinker around with all of these ideas, tools, & matrrials with my son.

Just be aware that you will have to increase the value and wattage rating of the ballast resistor if you go above 1.5 volts. It's that fellow Georgi Ohm again. If you increase the voltage, you increase the current. If you increase the current you increase the power dissipation. (P = I x V)

 

Just got back from the local electronics shop. I got a cheap MM for $10. It was just not worth the time, effort, and gas money to save $6 to drive to HF 15 miles from my house.

I also picked up three 1-foot lengths of buss wire (tin-coated copper) of varying gauge to add to the conductor mix and a 1 foot length of aluminum shielded coax (which will act as our stranded aluminum after I strip it off).

I then hopped over to HD and picked up small spools of solid 18 gauge aluminum, copper, and galvanized steel in the picture wire hanging section, each for under $4.

I then verified at home that my PAD-234A still works after 22+ years and can provide us with a 5V or 12V DC power source (my meter actually measured 4.89V across tje ground terminal for the 5V) so I won't need a battery setup. It's actually this exact one that another forum member posted right here on this site.

So we might be able to perform that suggested resistivity experiment after all. I have a bunch of resistors (amongst all of my ICs/LEDs/etc from my digital design classes) and now I just need to know which one (or which acceptable range) would be best suited for our purposes so that the measurements we take will clearly show how the resistivity/conductivity of each different wire (or different gauge of same wire) under test differs from one other.

All of your continued support is greatly appreciated. I will report back tonight or tomorrow after I have a chance to tinker around with all of these ideas, tools, & matrrials with my son.

One more tip that will help get you started. The actual resistance of your sample wire is going to be very low relative to the ballast resistor. You might consider measuring exactly 1 meter length of your sample wire and wrapping it around a cardboard tube. Don't let the windings touch each other!

 

Just be aware that you will have to increase the value and wattage rating of the ballast resistor if you go above 1.5 volts. It's that fellow Georgi Ohm again. If you increase the voltage, you increase the current. If you increase the current you increase the power dissipation. (P = I x V)

What should the maximum current in the circuit be to keep things safe?

 

What should the maximum current in the circuit be to keep things safe?

You should keep it under 200 mA for a couple of reasons:
1. Your MM reads up to 200 mA then jumps to a 10 amp scale. The 10 amp scale does not have the resolution required for this experiment. Therefor you need the 0 to 200 mA scale so you can read the current to 3 significant digits.
2. As the current increases, the ballast has to dissipate the power in the form of heat. From your research you will note that the resistivity of materials is also a function of temperature. The standard resistivities are typically specified at 20 deg C. If the ballast resistor gets too hot, its resistance will change as will the voltage drop across it. So you will be chasing a moving target.

I have been referring to the resistor in the circuit as a ballast resistor. By now it should be obvious that the actual function of this resistor is a current limiter. Good luck!

 

So a 100 ohm resistor that would limit the current to 50 mA should be adequate right?

 

So a 100 ohm resistor that would limit the current to 50 mA should be adequate right?

This would indicate you are using a 5 volt power supply so the answer is Yes, but..........
There's always a but isn't there?

Consider your circuit. It consists of a battery and essentially two resistors - the current limiter and your sample wire. The sum of the voltage drops across the two resistors equals the battery voltage. Now, the larger we make the resistance, the larger the voltage drop across the current limiter. If we make the current limiting resistor too large, then the voltage drop across your sample will be virtually unmeasurable with the equipment you are working with. So, the trick is to make the current limiting resistor as small as possible while still keeping the current under 200 mA. Does this make sense?
It may take a little manipulation od Ohm's Law to come up with the optimum value. This is a golden opportunity to get your son involved in this phase.

 

Well, I have some encouraging initial findings.

I'll start with the basics....

I connected the 5V terminal of my PAD-234A to a 100 Ohm resistor on a breadboard (that is jumpered to the 5V terminal). The other end of the resistor was connected to the ground terminal via gator clip.

Using my "better" MM (Commercial Electric HDM350) I measured 4.88V from the 5V terminal across the ground terminal. Not perfect but it will do.
Across the 100 Ohm resistor I measured a 4.79V drop and the current measured 47 mA

I then swapped out the HDM350 and connected the $10 Velleman DVM810 MM in its place. It was only to display a value of 50mA (close enough, as long as I'm under 200 mA).
But I needed that MM to be in the circuit as the ammeter so I can more accurately (and I use that term loosely) measure the voltage drop with the HDM350 across each of the wires I would be testing.

I then started to connect 12" lengths of varying test wires (according to the schematic supplied by bertz) in between the resistor and the ammeter and measured the voltage drops across them.
The measurements were all in millivolts however they were all slightly different.

There were as follows:

*16AWG tin-coated copper buss wire 0.3 mV
*18AWG tin-coated copper buss wire 0.5 mV
18AWG solid copper 0.7 mV
*20AWG tin-coated copper buss wire 1.2 mV
18AWG aluminum 1.5 mV
18AWG galvanized steel 2.3 mV

*The buss wire gauges are actually my estimates. The owner of the shop cut them for me off the spool rather quickly and charged me next to nothing for them so I didn't complain.

Some conclusions I'm going to take a stab at:
1) At 18AWG, tin-coated copper appears to be slightly more conductive because of the smaller voltage drop. Is that suppose to be the case?
2) The zinc in the steel wire I assume increases the wire's resistance. Normally steel is more conductive.
3) The conductivity of the materials in decreasing order is copper, aluminum, and steel

At this point, what can I do to my test setup/method to make the voltage drops a bit more dramatic (without going overboard with cost). A previous post recommended wrapping a considering amount of wire around cardboard instead of using short 1 foot sections. I'm open to additional suggestions. I didn't take a stab yet at bertz's suggestion of calculating actual resistivity vs. standard restivity because the voltage readings were so small.

 

Using my "better" MM (Commercial Electric HDM350) I measured 4.88V from the 5V terminal across the ground terminal. Not perfect but it will do.
Across the 100 Ohm resistor I measured a 4.79V drop and the current measured 47 mA

This would indicate that the voltage drop across your conductors is 0.09 volts. What size conductors are you using? This is not a show stopper, just an observation.

1) At 18AWG, tin-coated copper appears to be slightly more conductive because of the smaller voltage drop. Is that suppose to be the case?

I don't understand. The 18 AWG has a voltage drop of 0.5 mV and the 16 AWG has a voltage drop of 0.3 mV. This is exactly what you would expect. Look up the formula for specific resistance and you will see it is a function of cross section area. 16AWG has a greater cross section than 18AWG.

2) The zinc in the steel wire I assume increases the wire's resistance. Normally steel is more conductive.

The zinc is plated on and has a negligible effect (cross-sectional area again). Steel is 99.8% iron alloyed with a bit of carbon. So we are comparing iron to zinc. Zinc is actually more conductive than iron. (1.624 x 10^7 Siemens/m vs 1.04 x 10^7 Siemens/m)

At this point, what can I do to my test setup/method to make the voltage drops a bit more dramatic (without going overboard with cost). A previous post recommended wrapping a considering amount of wire around cardboard instead of using short 1 foot sections. I'm open to additional suggestions. I didn't take a stab yet at bertz's suggestion of calculating actual resistivity vs. standard restivity because the voltage readings were so small.

Try droping the value of your current limiting resistor in half and increasing the length of your test pieces. Also, in order to compare apples to apples, all your test pieces should be the same wire gauge and length.

 

What size conductors are you using? This is not a show stopper, just an observation.

Well the way my PAD234 is setup up I have a very short (probably 2" max of I believe 22 AWG) lead coming from the screw terminal of the 5V point and terminating in the top row of my bread board. All of the top rows of my bread boards are then jumpered to that 5V connection. Perhaps I should disconnect the jumpers to the other bread boards? One end of my resistor connects into the top row of the breadboard for the 5V connection and the other terminates in any random spot on my bread board (just so it's not dangling). I then have a gator clip that connects that side of the resistor to the GND terminal. I then measured the voltage drop across the resistor by touching the leads on both side of the resistor.

I don't understand. The 18 AWG has a voltage drop of 0.5 mV and the 16 AWG has a voltage drop of 0.3 mV. This is exactly what you would expect. Look up the formula for specific resistance and you will see it is a function of cross section area. 16AWG has a greater cross section than 18AWG.

My mistake, I should have been more specific. My statement was referring to the comparison between the 18 AWG copper and 18AWG tin-coated copper. Just pointing out the fact that the tin-coated copper has a smaller (0.5mV) drop compared to a standard copper wire (0.7mV drop) of same gauge.

The zinc is plated on and has a negligible effect (cross-sectional area again). Steel is 99.8% iron alloyed with a bit of carbon. So we are comparing iron to zinc. Zinc is actually more conductive than iron. (1.624 x 10^7 Siemens/m vs 1.04 x 10^7 Siemens/m)

Okay, so what we've measured for steel is pretty much right on.

Try droping the value of your current limiting resistor in half and increasing the length of your test pieces. Also, in order to compare apples to apples, all your test pieces should be the same wire gauge and length.

I don't have any 50 ohm resistors but I assume I can just use two 100s in parallel. This should get me something closer to 100mA. Or even three in parallel to get me 150mA. How long would you recommend I make the lengths so it's still practical.

 

Well the way my PAD234 is setup up I have a very short (probably 2" max of I believe 22 AWG) lead coming from the screw terminal of the 5V point and terminating in the top row of my bread board. All of the top rows of my bread boards are then jumpered to that 5V connection. Perhaps I should disconnect the jumpers to the other bread boards? One end of my resistor connects into the top row of the breadboard for the 5V connection and the other terminates in any random spot on my bread board (just so it's not dangling). I then have a gator clip that connects that side of the resistor to the GND terminal. I then measured the voltage drop across the resistor by touching the leads on both side of the resistor.

A picture is worth a thousand words

I don't have any 50 ohm resistors but I assume I can just use two 100s in parallel. This should get me something closer to 100mA. Or even three in parallel to get me 150mA. How long would you recommend I make the lengths so it's still practical.

You're on the right track with the resistors. All your samples should be the same wire gauge & length. I would use 1 meter lengths wound on a cardboard tube. 1 meter is a standard unit (e.g. Siemens/m). By the way, most science fair protocols require units to be in the metric system. Here's a handy reference for wire gauge dimensions. Good luck!

http://www.powerstream.com/Wire_Size.htm

 

Great, thanks. Will report back as soon as I have additional info.

 

After some extensive debate, my son and I have come to the conclusion that at his age he should instead pursue an electrical circuit project which will result in more predictable experimental results. After tossing around a few ideas we have narrowed it down to a study of Ohm's law. Specifically targeting the question....how do circuit designers limit current flow to prevent damage to sensitive components.

After covering the basics of Ohm's law, what current is, what voltage drop means, etc....he will then start out by calculating & measuring current and voltage drops using some basic configurations of resistors in a circuit (e.g. one resistor, two identical ones in series, two identical ones in parallel, two different ones in series/parallel, networks of resistors) to verify that V=IR holds true. We will be using a small breadboard, two multimeters, various configurations of AA/AAA/9V batteries in holders that have breadboard leads, and a lot of different resistors ranging from 8 ohms to 1000 ohms (I have resistors that cover the 10K+ range but we won't be using those).

Once he covers the basics as described above, he will then attempt to calculate a resistance value that will provide optimal illumination of an LED to emulate the thought process involved in (elementary) circuit design. In one case the resistance will be too high and the LED will be dimly lit (or not at all). In another case it will exceed the max current rating and show that it permanently malfunctions. Then he will use Ohm' law and the specs of the LED to determine the optimal current (and corresponding resistance value).

Does this project idea sound like a better approach?

The complete project won't be due until May or even June so we have plenty of time to add to it to make it more sophisticated. Can anyone offer suggestions on what we could add as additional experiments?

He won't really be able to demonstrate much at the science fair (due to the amount of kids participating). All he will have to present is a tri-fold poster board (like everyone else) with photos of meter readings/drawings/graphs/etc to prove his calculations. We could try asking permission for him to bring in his tablet on which we could create a short demo video (that can play on an endless loop) but we'll cross that bridge if we have both the time to do this and the approval of the teacher.

In order to prepare myself to take on this endeavor with him I have a couple of questions I'd like to ask...

1 I have a small ziploc bag full of LEDs in various sizes, colors, and shapes from my digital design courses from 2o+ years ago. I have no idea what the specs for each of these LEDs are. Is there an easy way to determine this? My memory is really fuzzy on this after solely developing software for the past 20 years.

2) If an LED's current is exceeded, with the LED go up in smoke or just cease to function. I ask because I've never actually attempted to do this.

Thanks again, and thanks for all the input on our previous project idea.

 

1 I have a small ziploc bag full of LEDs in various sizes, colors, and shapes from my digital design courses from 2o+ years ago. I have no idea what the specs for each of these LEDs are. Is there an easy way to determine this? My memory is really fuzzy on this after solely developing software for the past 20 years.

I would assume they are all 20mA LEDs, and target to run them at 10-15mA for a test. They should be plenty bright at that level. If they seem dim, reduce the resistor and take the current up a little at a time.

2) If an LED's current is exceeded, with the LED go up in smoke or just cease to function. I ask because I've never actually attempted to do this.

At the extreme, yes, they can pop like a fuse. But often it's more about shortening their life and making them fade too quickly. So you might get away with running a 20mA at 30mA. It'll seem really nice and bright. For a while. And then it will grow dim and eventually it may quit altogether.

 

I would assume they are all 20mA LEDs, and target to run them at 10-15mA for a test. They should be plenty bright at that level. If they seem dim, reduce the resistor and take the current up a little at a time.

Will do. We'll use these LEDs as our guinea pigs.

At the extreme, yes, they can pop like a fuse.

That's what we're hoping for. The LED would remain intact I would imagine.

But often it's more about shortening their life and making them fade too quickly. So you might get away with running a 20mA at 30mA. It'll seem really nice and bright. For a while. And then it will grow dim and eventually it may quit altogether.

So after the initial phase perhaps we should buy just a single type of LED (a batch of them in fact) that has a specific rating so we can see what happens when the rating is gradually exceeded. In one case we'll show it burn and fade out, in another case an instantaneous burn out.

 

That's what we're hoping for. The LED would remain intact I would imagine.

No! I don't know just what it takes, but you could pop one and release the magic smoke. Putting one directly on the poles of a car battery might do it. Plugging into the wall outlet might produce a blinding flash in addition to producing shrapnel before the breaker blows.

 

No! I don't know just what it takes, but you could pop one and release the magic smoke. Putting one directly on the poles of a car battery might do it. Plugging into the wall outlet might produce a blinding flash in addition to producing shrapnel before the breaker blows.

Okay...... I guess that's what you meant by extreme. We'll just want to show it fizzling out. A safe controlled blowout for lack of a better term.

On the topic of LED ratings, what does the voltage rating imply (forgive me, I'm trying re-learn some principles of electronic components).

The following is listed in the specs for a 5mm red LED at Radio Shack:

• Typical Voltage is 2.25, with a maximum voltage of 2.6V

Suppose I have a 6V power supply connected to a 300 ohm resistor. The current flowing in the circuit is 20mA. Most of the voltage drop is across the resistor, a negligible amount is across the leads between the resistor and battery terminals.

Now I add the LED into the circuit. Is the following explanation correct?

Since the supplied voltage (6V) is greater than the voltage range of the LED (2.25V to 2.6V) the LED lights up and as a consequence the voltage drop of the 300 ohm resistor is reduced by the LED's voltage drop (from ~6V to around ~3.4V to ~3.75V). By Ohm's law this results in a total circuit current that is in the 11mA to 12.5mA range. Does this sound right?

When the supplied voltage (e.g. 1.5V) is less than the LED's voltage range and the LED does not go on what happens then?

Or do I have it completely wrong?

 

On the topic of LED ratings, what does the voltage rating imply (forgive me, I'm trying re-learn some principles of electronic components).

All LED's have a forward voltage rating, typically 2.4 volts for red, but it can vary a little. If you google LED's you can find a chart or table that will give you the forward voltage rating of various colors.

Consider, the attached schematic. The supply voltage is 6 volts and the Vf is 2.4 volts. That means the LED is dropping 2.4 volts. Now most common LEDs like to run at 20 mA or less. Lets say we want this one to run at 15 mA. This means that we will have to drop (6 - 2.4)volts across the current limiting resistor. Now that we know current (.015 amps) and voltage (3.6 volts) we can plug these into ohms law and we get R = E/I or 240 ohms. If you know any two you can figure the other.

The power dissipation is 3.6 x .015 = 0.05 watts so a 1/4 watt resistor will do just fine.

Just for giggles, hook up an LED directly to your battery and see what happens. You wont get an explosion, but at least you know.

Also be careful to observe polarity on the LEDs. They have a flat spot and one of the leads is shorter than the other. This is the cathode and always gets connected to the negative terminal.

I started my grandson on exactly this type of project and now he can recite Ohm's Law chapter & verse. Good luck!