Is this a fuse or a resistor?
I have been having difficulties with a PSU construction for a formerly battery driven (unknown amperage, 18 v ) electric drill and first needed help in identifying a part; picture included:
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I am familiar with fuses clear like colorex shows and smaller opaque ones.
In the final decision, I vote a power resistor.
Always learning is my stupid goal because when I attain the most knowledge and experience I will die !
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Can you test it properly, in situ? If not, desolder one leg and meter it.
It should read somewhere between 0 and a small ohm value (like 1 to 10 ohms). That's also consistent with heat--a low value resistor passes lots of current, so it gets hot. Consequently it's large and beefy so it won't fail. It's also a consistent value for a resistor meant to limit current in a motor.
I'm not sure why it's there--but I'm not very knowledgeable about motors personally. Maybe it's there to limit current and prevent damage to the batteries (or the motor?) if the motor is switched on when it's being mechanically prevented from rotating...
A fuse? It wouldn't show much resistance either if it were whole. A fuse shouldn't be very hot on the exterior, unlike a power resistor. It would be infinitely high resistance if it were blown, but it couldn't get hot either.
A very high value resistor would pass very little current, and be closer to an open circuit.
Gmoon, I am not trying to be argumentative, I am trying to understand, so please bear with me; you wrote: That's also consistent with heat--a low value resistor passes lots of current, so it gets hot.
Wouldn't a higher ohm resistor get even hotter? As it would be converting more "resistance" to heat, I would think. If I am wrong, please help me understand (I DO understand that most "power" resistors are very low in ohm ratings, I have quite a selection of ceramic and giant wire wound ones).
Sorry, gh, I really don't know why your meter reads like that. You're sure there's no other components; i.e., another current path when you test it, right?
Not a dumb question; not argumentative. Two ways to look at the resistance / heat thing--one is (sorta) logical, the other is mathematical.
Logically, if greater resistance == more heat, then an open connection (air!) is VERY high resistance, and would be the hottest of all. That's clearly not the case. You need current to flow to generate heat.
But wait...a very efficient conductor (thick wire) with low resistance (and lots of current flow) doesn't generate much of heat, either. So why does a resistance value that's seemingly midway between the two generate heat?
OK, turning to math. You really need to look at more than one Ohm's Law equation to make sense of it...
First, remember that current is the the same for all series components--if a motor draws 0.5 amps, then a resistor in series also draws the same amount of current, 0.5 amps.
So:
P = I2 X R (watts = amperes squared X ohms)
If a motor draws .5 amps, and the connecting wire is 0.005 ohms:
Power = 0.00125 watts
Not much heat...
If a motor draws .5 amps, with a series resistor of 5 ohms:
Power = 1.25 watts
Lots more heat!
So let's get ridiculous:
If a motor draws .5 amps, with a series resistance of 100000 ohms:
Power = 25000 watts!
HUGE AMOUNT OF HEAT! LITERALLY Unbelievable!
What's wrong with that last scenario?
Volts = watts / amps
25000 / 0.5 = 50000 volts
You'd need 50000 volts to "push" 0.5 amps through a 100000 ohm resistor. That's the missing link here... We're forgetting the Ohm's Law interaction between current, resistance and voltage. You push enough current (by upping the voltage)--then yeah, a larger resistance will convert all that to heat. IF you could find a power resistor with a value of 100K (hint: you MIGHT find something like that at a power station).
Yeah--with enough voltage we can even overcome an open connection and generate massive amounts of heat (think lightning!).
So by raising the series resistance, the voltage needs to increase to maintain the current draw.
Obviously, there's a practical "sweet spot" for motors & series resistors, based on their current draw, their inductive load (in ohms) and the supply voltage (and the size / wattage rating of the resistor). If I were to guess the load resistance of the drill, something like 10-20 ohms would seem reasonable. I call that a "low impedance" device!
So any series resistance will be smaller that than that (10% ? 15% ?), or the voltage that would be needed to "push" the current would quickly rise to an impractical level...
Thank you SO much for taking the time to explain it in greater detail for me. Suddenly I feel like a noob :-)
For most practical circuits, the voltage and the load are fixed. So increasing the series resistance increases waste heat. But only up to a point. Beyond that point the series resistance limits the current, and the heat as well.
Hey, I've just picked things up over the years as a DIY hobbyist, too--and for most of those years, didn't use the math either.
I learn best with practical examples--and better by far when it's an immediate problem I'm trying to solve.
You can still cull some cool info from your drill project, just with the tools on hand:
-- Measure the current draw of the motor (ammeter in series), both turning freely and also under a mechanical load.
-- If the power resistor is in series, there will be a voltage drop (over both). Take a voltage reading across the motor itself. Try that when the motor is working hard, too.
From those two facts (volt, amps) you'll be able to find the real "load resistance" of the motor (in ohms) using Ohms Law. Finding the load via math is more accurate than measuring resistance directly--the motor is an inductor, not a resistor. Also, it will probably change it's load when it's "doing work," and at different speeds, too.
Ohms = volts / amps
-- Measure the voltage over the resistor, too.
From the examples in my (loooong) post above, it should be plain that two devices in series DO draw identical current, but DON'T consume the same amount of power. You can work that out, too.
You might not NEED to know this for your project, but understanding stuff like this opened up a new world for me...
I do "need" to understand a lot of this stuff, for two reasons: #1: I am always curious to learn more about things I am interested in (thus all the reading I have been doing on theoretical particle physics, and genetics), and #2: for use in helping me design things (I have gotten really good at building things from a schematic, but have yet to be able to "cobble together" an original project). So, thanks again.
I will remove one of the legs of the resistor if I want to measure it, in case that is causing me the "mis-readings". :-)
Mathematically I have no chance of understanding advanced physics ;-). But the high school-ish level stuff is still great. Also, I had a class in college where the prof showed us how Einstein used basic equations to arrive at his conclusions regarding time dilation. Fantastic.
I had great satisfaction converting my little "radio tube" amp filaments from AC to DC. Due to the voltage increase that comes with rectification, I needed a new series resistor value. I outlined it here on my ible; it's a very similar exercise to our above discussion. All thanks to Ohms Laws.
Just set yourself a practical problem--you'll solve it, I'm certain.
(I do have an old textbook on "introductory circuit analysis." Never took a class, but it's a good reference.)
It must be there for a reason--can you trace/draw a schematic of the whole shebang? We don't really know if it's part of the charging, control, or has another purpose...
Putting the power wires back the way they were, I used another "power" resistor" that was a little higher (by about 10 ohms or so) in it's place, but I guess it wasn't heavy duty enough as it burnt a hole in the side of it almost instantly on swithing on the drill's motor.
It would be interesting to see a schematic. I'm not super conversant with motor stuff, and would like to learn more.
This motor is obviously not polarity restricted, as there is a mechanical switch on the drill to reverse it's direction.
If so, I wonder if the power resistor is there to prevent the battery from overheating. A drained battery would draw considerable current, and could be damaged without current limiting.
On theoretical particle physics, I found Quintessence by Krauss to be very enlightening without the need for much of the math involved. I am about to start The Elegant Universe by Greene, to get a little more from another point of view. None of this will actually do me much good, career-wise, but it is one of my many interests.
I'll have to see if I can find a decent, and not too expensive book along the lines of what you mention (intro circuit analysis) to get me "up to speed". :-)
I think it may be a heavy-duty-resistor, which has been fine-tuned by cutting that groove (conductive on the outside, they're usually covered in paint)
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There are semiconductor ( I think called polyfuses ) that are both a resistor
and a fuse and reset after switched off and also can be damaged.
I agree with Wesley666, I would check the location of where the part is in reference to where the power comes in...
If you intend on reusing the board I would (once it is identified for sure as a fuse) replace it and add in a fuse holder,not only would it make it easier if it blows, but I personally think it would make it a bit safer, as I personally wouldn't trust those solder connections....
Here's a pic of the fuse in an old desktop PSU:
I am not a technician, so don't trust me.