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:
44
answers
|
Answer it!
|
sort by:
active |
newest |
oldest
The very clean laser cut not all the way around a ceramic base strongly suggest a resistive element. I do not recognize the edge of the laser cut weather metal or resistive material it could be obscured by the laser burn.
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 !
A
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 !
A
I believe you Gmoon and lemonie are right. It shows open in the MM (or very High resistance) and yet heats up when I apply power to it (18 vdc, 4 A).
Nov 14, 2011. 6:12 AMgmoon
says:
It could be either. It does look a lot like a wire-wound power resistor (but somewhat like a fuse, too).
Can you test it properly, in situ? If not, desolder one leg and meter it.
Can you test it properly, in situ? If not, desolder one leg and meter it.
tests as open (or very high resistance) but does get very hot when a bit of amperage is applied to it.
A low resistance should show "closed" or "short" when I use the ohm meter on it, shouldn't it?
Nov 15, 2011. 6:18 PMgmoon
says:
I dunno about the meter--sometimes the digital ones are a little funky. Digital meters also have accuracy problems reading low resistance--even on the lowest setting (which you should be using, maybe that's part of the problem). That's due to the accuracy factor--if the meter accuracy is 2%, but the scale is 0-200 ohms, that's a potential error of 4 ohms (2% of 200)--a huge amount when you're checking a 1 ohm resistor.
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.
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.
Put it this way, when I touch the two "meter probes" together, it reads "0", if I have them separated it reads "infinite". If I touch them to my dry skin within millimeters of each other, it reads a high resistance, and if I measure a 1 ohm resistor, it reads 0 to two ohms.
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).
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).
Nov 16, 2011. 10:48 AMgmoon
says:
Danger! Long reply ahead...
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...
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...
Ok, that is VERY helpful (in all the years I have fooled around in electronics, I never really learned to math involved to my embarassment; except for some very simple things like adjusting the frequency of a 555 output etc.
Thank you SO much for taking the time to explain it in greater detail for me. Suddenly I feel like a noob :-)
Thank you SO much for taking the time to explain it in greater detail for me. Suddenly I feel like a noob :-)
Nov 17, 2011. 6:07 AMgmoon
says:
I probably should have summarized the "lecture" more simply:
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...
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...
Measuring the the motor's current draw would require me to disassemble the mechanism more then what I really want to, I think. That circuit board with the resistor in question on it, drives 2 LED's, one for power on, and one for "charge" (when I had a battery for it).
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". :-)
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". :-)
Nov 18, 2011. 4:58 AMgmoon
says:
Yes, you may need to lift one end of the resistor to get an accurate reading...
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.)
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.)
Ok, yes it is a 22 - 22.5 ohm resistor. Removing one leg did not extinguish the red CHARGE light, but changed the green power light to yellow (I am assuming it took the diode out of the circuit with it, and it is a bidirectional like I see on many monitors). I still don't know why it gets so blazingly hot though. If I "put it back in the case" it would literally melt the case after just a few SECONDS of drill use.
Nov 21, 2011. 6:06 AMgmoon
says:
OK, resistor confirmed.
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...
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...
I can do that when I get back to it. Just an Update though: I thought I might have had the input wires reversed.....studidly, I switched them and a small spark let me know that this was a mistake. Now the resister IS open :-)
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.
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.
Nov 22, 2011. 5:08 AMgmoon
says:
Hope you didn't fry anything else when you reversed the polarity...
It would be interesting to see a schematic. I'm not super conversant with motor stuff, and would like to learn more.
It would be interesting to see a schematic. I'm not super conversant with motor stuff, and would like to learn more.
It doesn't appear that way (unless it is on that little board). I bypassed it since it only seemed to be there to show that the battery was #1: functional, #2: in correctly, or #3: the drill was "running" (LED #2). The drill motor still runs...and for the short time I operated it, it didn't overheat. But I will get you that circuit in a bit.
This motor is obviously not polarity restricted, as there is a mechanical switch on the drill to reverse it's direction.
This motor is obviously not polarity restricted, as there is a mechanical switch on the drill to reverse it's direction.
Nov 22, 2011. 7:40 AMgmoon
says:
Is the PCB actually the battery charger for the drill? Somehow I assumed it was part of the drill itself. But the LEDs would contradict that.
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.
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.
The "battery pack" needed to be removed and plugged into the "charging station", and yes, I guess I did take this little board from the charger......hmmmm (sorry, I still hadn't had time before work to draw out the traces of the board....it isn't complicated so when I have 2 minutes I will get that done.....when I remember *sigh*)
Nov 22, 2011. 8:53 PMgmoon
says:
Just a thought--a battery-charging circuit might not be the best power supply for actually running the motor... :-)
I am in agreement now that I have fried a second resistor. It would be nice to have the LED light when I have it plugged in however, but that shouldn't be too hard to accomplish without the other "junk" :-)
Nov 23, 2011. 8:11 AMgmoon
says:
I bet you have all the hardware you need to (re)design a viable power supply...assuming the transformer, etc. can handle the current draw of the motor.
I had thought that the "circuit" was totally unattached to anything, but upon re-examination found it still moored to the motor, not to mention the two LED's I referred to. Live and learn :-)
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". :-)
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". :-)
Apply the multimeter to it.
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)
L
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)
L
It appears as an "open" circuit on the meter, yet when I apply the power from my psu, it gets VERY hot, VERY fast (nearly as hot as my soldering iron as near as I can tell from the burn it gave me).
If it had a low value, wouldn't my Ohm meter show it as either "closed" or short? It displayed "infinate" on the highest setting (unless you mean low WATTAGE, then I agree)
How does it get hot if it isn't a conductor?(I don't get that)
L
L
Well it apparently conducts "some current", but the resistance is "either so high the meter can't read it" or my meter had gone coo-coo :-)
Still would be interesting to see the solder side.
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.
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.
Hmm, I can get a picture of it, once I get home again (tomorrow) but IIRC it only has solder points (and traces) on the other side. If this is a fuse, it gets VERY hot without actually blowing (since the drill continues to work regardless of how hot and how long I run it). I am sticking with "resistor" at this point and now am thinking how best to drive the circuit without overheating the thing. I mean, sure I could BUY a PSU for the drill but that's not only no fun, it requires money :-)
Hmm looks similar to those cylindrical house fuses...once you peel the labeling sticker off.... The end caps definitively in my opinion give it away... I have yet to see any resistor with end caps like that... I think that it might have originally been a fuse that was designed to be put in a fuse holder on the board, but the manufacture decided to cut cost by just soldering it directly to the board...
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....
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....
I believe Iceng, Gmoon and lemonie are right. It shows open in the MM (or very High resistance) and yet heats up when I apply power to it (18 vdc, 4 A).
Nov 13, 2011. 9:11 PMWesley666
says:
I would put my money on Fuse. The metal end caps look more like a fuse, as in Colorex's picture, you can see that fuse has metal caps on each end. As well, I have seen fuses that have a brown or white piece of paper material glued around them, as well as fuses that instead of being a hollow tube with a thin piece of metal are solid all the way through and are generally brown (Like the one pictured) or white on the outside. How close is it to where the power comes into the board? If there is a place for one of those adapters/wallwarts or an AC plug very near it, then it is probably a protection fuse.
this puppy was dark like this before I applied power for the first time.
tests as open (or very high resistance) but does get very hot when a bit of amperage is applied to it.
Definitely a fuse. Hard to say what value it is, though. Do you have a multimeter to test it with?
tests as open (or very high resistance) but does get very hot when a bit of amperage is applied to it.
Apparently a burnt fuse. That's exactly how they look in old computer PSU's. Of course they should be clear glass, but this one looks burnt.
Here's a pic of the fuse in an old desktop PSU:
I am not a technician, so don't trust me.
Here's a pic of the fuse in an old desktop PSU:
I am not a technician, so don't trust me.
Well, it is still "heating up" when I apply power to the drill's motor, which is what bugs me.
 |
