EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: haizaar on July 19, 2019, 02:13:17 pm
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Good day again guys,
My BM235 DMM prints "MAX 10A HBC FUSED" in a ammeter port. I'm probably missing obvious, but does it mean I can't measure more than 10 amps even on 3.3 voltage?
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Yes. If you need to measure higher current in a low-voltage circuit, you need an appropriate 4-terminal “shunt” resistor that typically produces 100 mV across the voltage terminals with a specified high current through the current terminals. These are specialty items. “Ammeter shunts” used with analog panel meters usually produced 50 mV at full-scale current, for use with 1 mA meters with 50 ohm resistance.
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The fuse limits the maximum current independent from the voltage in the circuit (unless way to high) .
The fuse can blow at more than 10 A, though it may take quite some time if only a little more. Still it's not a good idea, as the proper HRC (HBC) fuses are relatively expensive (e.g. $5-$15).
Anyway it is difficult to measure a high current directly with the DMM in a low voltage system, as there can be quite some drop:
some 50 mV for the shunt, some 500 mV or so for the fuse, some 200 mV possible for the leads and contacts. This drop can upset a low voltage system.
A separate external shunt or clamp on meter may be the better choice in such a case. The external shunt may very well get away without a fuse if the system is known to be safe without it.
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As discussed in many threads, the UT210E looks like a good clamp meter for such measurements or search on ebay for "current shunt" as mentioned from Kleinstein, both of choices.
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Thank you everyone for elaborate responses. It's very helpful.
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The fuse limits the maximum current
This may sound like nitpicking, but since this is the beginner forum, and phrases like this easily get out of context and get stuck with us, I want to clarify that fuses do not limit current*, nor is that their intent. This is an important distinction because beginners tend to think fuses can be used to protect parts that are sensitive to overcurrent, such as transistors.
From the design viewpoint, it "limits" the current you can design with (and I think this is what you meant), but it doesn't actually limit the physical current.
*) In reality, they do limit current due to their internal resistance, but orders of magnitude higher than their nominal ratings, and varying with voltage, and this is often poorly specified; a 10A fuse may limit the current below 1000A at voltages below 100V, for example.
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The fuse limits the maximum current
This may sound like nitpicking, but since this is the beginner forum, and phrases like this easily get out of context and get stuck with us, I want to clarify that fuses do not limit current*, nor is that their intent. This is an important distinction because beginners tend to think fuses can be used to protect parts that are sensitive to overcurrent, such as transistors.
From the design viewpoint, it "limits" the current you can design with (and I think this is what you meant), but it doesn't actually limit the physical current.
*) In reality, they do limit current due to their internal resistance, but orders of magnitude higher than their nominal ratings, and varying with voltage, and this is often poorly specified; a 10A fuse may limit the current below 1000A at voltages below 100V, for example.
Just making sure - the standard 20A fuse in my car, will it burn out on over 20A? Or are you saying that it can still allow, say, 1000A for 0.001 secs and the to burn?
In general, if I think about use as a melting wire, then my intuition tells me that melting point is related to power (heat), meaning that if a fuse can hold 1A on 1000V, then it should hold 1000A on 1V as well. Does this make sense?
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Just making sure - the standard 20A fuse in my car, will it burn out on over 20A? Or are you saying that it can still allow, say, 1000A for 0.001 secs and the to burn?
In general, if I think about use as a melting wire, then my intuition tells me that melting point is related to power (heat), meaning that if a fuse can hold 1A on 1000V, then it should hold 1000A on 1V as well. Does this make sense?
Car fuses tend to have limited datasheets, cheap ones may be anything, but let's look at this:
http://www.farnell.com/datasheets/1633611.pdf?_ga=2.254648512.998935643.1563954505-942271309.1553682867 (http://www.farnell.com/datasheets/1633611.pdf?_ga=2.254648512.998935643.1563954505-942271309.1553682867)
For the 20A fuse:
110%*20A=22A - 360 000 seconds (100 hours) before it ever opens
200%*20A=40A - opens between 150ms and 5 seconds
600%*20A=120A - opens in less than 150ms
Now the existence of these numbers already makes it obvious that the fuse does not limit the current. Whatever current flows, will flow happily, and only eventually is the circuit disabled.
150ms for 6x overload is a massively long time! Semiconductors, for example, die in hundreds of microseconds. The purpose of fuse is to prevent wire insulation melting and burning. Wires tend to have enough thermal mass so that an overload for 5 seconds is not a problem.
Now, to the actual current limiting.
The cold resistance of the fuse is 3.38mOhm. If you look at a suddenly increasing load (short, for example), the initial current happens when the fuse element is still cold, so use the cold resistance. Assuming cables, fuse holders, batteries etc. have total internal resistance of 3mOhm, then the current is limited to I=U/R, for example,
at 10V: 10V/(3.38mOhm+3mOhm) = 1567A, or
at 30V: 30V/(3.38mOhm+3mOhm) = 4702A
With these numbers, the cables, batteries and whatnot are already "limiting current", and the fuse is contributing to this limiting. It is highly likely that in a typical car 20A circuit, there are way more resistance than the assumed 3mOhm, and hence the current is limited by everything else, not the fuse. Worst case, there is so much resistance that the current during a short is limited to a value lower than the blow rating of the fuse!
Now what does the "interrupt rating" of 1000A@32VDC mean? It means that it cannot safely interrupt fault currents over that. But we just calculated that at 30V, the current's going to be 4700A. So this fuse is unsuitable for that particular example - you need to increase the circuit resistance elsewhere to provide current limiting for the fuse! Or pick a fuse with a higher interrupt rating - or use it at a lower voltage, which both decreases the current, and allows a higher "interrupt rating" (although unspecified for lower voltages in this datasheet, so you need to assume).
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Siwastaja, thanks for elaborive answer. This is very interesting.
Looking at Bm235 specs I see:
Overload Protection:
uA & mA: 0.4A/1000V DC/AC rms, IR 10kA, F fuse or better
A: 11A/1000V DC/AC rms, IR 20kA, F fuse or better
Now I understand "IR 10kA" meaning - it means that above 10kA the fuse will explode, or create an arc, etc., i.e. will not terminate the current safely, right?
Still my question remains. I'll rephrase it: taking the spec you linked to as an example:
* At 200% of 20A it will open in between 150ms to 5s. Even on 1V? I.e. if I connect it to lab power supply that will output 1V and unleash 40A current, will it open?
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The IR rating is the current the fuse is supposed to reliably interrupt. At higher currents the fuse may fail to interrupt the current and possibly explode up / or just fuse internally. There is still a chance it could brake the circuit (especially with AC and low inductance), but beyond specs.
At only 1 V one may not be able to get more than twice the rated current through the fuse, especially with some other voltage loss from the cables. If the higher current actually flows (e.g. not much other resistance) the fuse will (should) blow - the fuse will only see the possibly higher voltage after it has blown, before it does not even "know".
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Now I understand "IR 10kA" meaning - it means that above 10kA the fuse will explode, or create an arc, etc., i.e. will not terminate the current safely, right?
Your understanding is not quite correct, the 10 kA means the fuse is rated to interupt a fault current of 10 kA.
Still my question remains. I'll rephrase it: taking the spec you linked to as an example:
* At 200% of 20A it will open in between 150ms to 5s. Even on 1V? I.e. if I connect it to lab power supply that will output 1V and unleash 40A current, will it open?
Voltage has nothing to do with how fast a fuse will open except that it will generate more current through that fuse. As the fuse is designed to open when the fuse element gets hot enough to burn out this is calculated for a specific current flowing through it. The fuse will have a voltage drop across it as the fuse wire in the cartridge does have some resistance, but this resistance changes as the fuse heats up and in most cases is irrelevent to the designer who has specified the fuse.
The voltage rating on a fuse says how much voltage can safely be applied across it when the fuse has opened.
So to answer the question if you do put 40 A through your 20 A it will open.