Actually I just realized that this won't work with 50Ω and 100Ω resistors as my current limiting resistors.

For the 50Ω resistor the power dissipated in it will be (.15)(.15)(50) = 1.125 watt which is far greater than the 1/4 watt it is rated for.

For the 100Ω resistor, the power dissipated would be (.082)(.082)(100) = .67 watts (which still exceeds its rating by almost a 1/2 watt).

I would need something in the vicinity of a 500Ω resistor so the current would be ~18mA and the power dissipated across the 500 Ω resistor would be (.018)(.018)(500) = .16 watts

I think what it boils down to is that for this to work the power rating of the load resistor NEEDS to be less than the power rating of the current limiting resistor by at least half.

Or does it? Any suggestions on what I should use for my load resistor and what I should use for my current limiting resistor?

Since they are in series, they will have the same current. The power dissipated in a resistor (or anything else) is

P = I·V

But using Ohm's Law it can be written as either

P = (V/R)·V = V²/R

or

P = I·(I·R) = I²R

Since they have the same current, the latter form is the one that makes the most sense to use. So the ratio of the powers in the two resistors is

\(
\frac{P_1}{P_2} = \frac{I_1^2R_1}{I_2^2R_2} = \frac{R_1}{R_2}
\)

Since I1 = I2.

Thus, if you use a 100Ω resistor and a 10Ω resistor, the 100Ω resistor will be dissipating 10x the power as the 10Ω resistor.

Another demo you might consider is show that with a 100Ω 1/8 W resistor and 12V you get about the expected 1.2 A (the meter on the 10A range should have a sufficiently low burden resistance to yield a good measurement) and that the resistor either burns up or gets extremely hot since it is dissipating 10x the power it is rated for.

Then use Ohm's Law (and a tiny bit of math) to show that putting 10 resistors in series to make a string and then 10 strings in parallel will yield an effective resistance of 100Ω. Then demonstrate this by showing that it still draws 1.2 A from your supply, but now the resistors don't even get warm since each is dissipating about 1/10 of the power it is rated for.

 

Unless I've done something wrong in my spreadsheet calculations, I can't figure how I can get the 1 Ω 1/8 watt load resistor R2 to dissipate an excessive amount of current without frying the current limiting resistor R1. Unless I misunderstood the suggestion and my thinking is backwards?

 

Ok, in myspreadsheet, I have recommended several values, but they may have to be tweaked.... All current limiting resistors should be 5W or greater but you may be able to get away with 1W resistors.

To "blow" it up, use no current limiting resistor.
Try a 4.7hm resistor to make it warm; but it may blow up as well.
A 44 ohm resistor should not blow it up, but it should get warm.
To run it safely, use a 100 ohm resistor.

6vdc    Ohms    Watts
        0       3.000
        5       0.896
        44      0.130
        100     0.059

 

Ok, in myspreadsheet, I have recommended several values, but they may have to be tweaked.... All current limiting resistors should be 5W or greater but you may be able to get away with 1W resistors.

To "blow" it up, use no current limiting resistor.
Try a 4.7hm resistor to make it warm; but it may blow up as well.
A 44 ohm resistor should not blow it up, but it should get warm.
To run it safely, use a 100 ohm resistor.

6vdc    Ohms    Watts
        0       3.000
        5       0.896
        44      0.130
        100     0.059

5W minimum for current limiting resistors.....now it's starting to make a little more sense.
So I should use a 1/8 watt 1 Ω resistor as the load and 4.7Ω, 44Ω, and 100Ω resistors as the current limiting resistors with ratings of 5W or greater. Does that sum it up?

 

Just got back from stocking up at the local mom and pop electronic shop.

Here's what I picked up:

10 watt ceramic rectangular wirewound resistors (in packs of 2):
5.1Ω, 47Ω and 100Ω.
I also happened to have a 10Ω 10W resistor in my supply at home already.

1/8 watt flameproof resistors (in packs of 6):
3.9Ω and 6.8Ω....these are the smallest values they carried

On the way home I passed by and stopped in at the local Radio Shack and picked up the 500 pack of 1/8 watt carbon-film resistors which among the assortment contained the following:
1Ω (qty 10)
2.2Ω (qty 5)
10Ω (qty 10)
15Ω (qty 5)

So I guess we're set to start experimenting. Any words of advice before we begin....especially safety? Our workbench is in the cellar.

 

In general, fumes from smoking electronic components contain some nasty things. If you have reasonable ventilation the smoke from a single component isn't much to worry about. But if you are in a small cellar without reasonable ventilation it's not the best thing to be breathing in. So either set up a fan to clear the smoke out the window (with a door open to allow fresh air to be drawn in) or leave and give the smoke time to dissipate before continuing.

Also, make sure that nothing very flammable is too close to the components you expect to get hot -- a couple feet of clearance is usually adequate. If you have a fire extinguisher, keep it near your workbench. If you don't have a fire extinguisher, get one and keep it near your workbench. You'll probably never need it, but it's far better to have it and not need it than the other way around.

 

In general, fumes from smoking electronic components contain some nasty things. If you have reasonable ventilation the smoke from a single component isn't much to worry about. But if you are in a small cellar without reasonable ventilation it's not the best thing to be breathing in. So either set up a fan to clear the smoke out the window (with a door open to allow fresh air to be drawn in) or leave and give the smoke time to dissipate before continuing.

Also, make sure that nothing very flammable is too close to the components you expect to get hot -- a couple feet of clearance is usually adequate. If you have a fire extinguisher, keep it near your workbench. If you don't have a fire extinguisher, get one and keep it near your workbench. You'll probably never need it, but it's far better to have it and not need it than the other way around.

Time to move the workbench into the garage! It's actually a portable fold-up architect's desk that was given away to me.

 

Time to move the workbench into the garage! It's actually a portable fold-up architect's desk that was given away to me.

Don't go panicking too quickly, but a well-ventilated area (with an easy escape route) is best if you can do it. But lots of electronics has been done in basements and attics.

 

Don't go panicking too quickly, but a well-ventilated area (with an easy escape route) is best if you can do it. But lots of electronics has been done in basements and attics.

Okay, I'll see how things work out with the first one in the garage.

 

Look for the new thread I am starting on this topic momentarily.