Calculating Current Draw For Input Circuit

Hello all, we have a circuit here from an old system that keeps blowing fuses every now and then. We have no idea what the right size fuse should be as there is no documentation for it and the fuses have kept getting changed over the years. I would like to take the connected components and based on their rating calculate a proper fuse size. I have a circuit feeding 15 Flex I/O 1794-IA16's and all the sensors connected to them. I assumed i would take the sensor specs which most have a spec of max 100mA with some having a max of 300mA. According to the 1794 sheet each input point has a max amperage draw of 14.81mA. So each point for the 100mA sensors could pull a total of 114.81mA, and if I had 16 sensors connected to one of my 15 flex cards and the sensors were active it would total 1837mA. Multiply that by 15 cards and I get 27,555mA. Is that correct in how i figure the fuse size needed for this?? It seems quite a bit high and i don't think its right but i don't know how to figure this. Especially since 82 of the 106 sensors are currently on right now and the circuit is only drawing 1.16 amps....I greatly appreciate any help or to be pointed in the direction of some documentation on how to do this.

Ignore the 100mA and 300mA. That's only the maximum load the sensors can drive.
In your case, they're driving about 15mA, so well under the load spec.
.1481 x 15 = Total Amp load is the calculation.

You have 1.16 Amps with 86 on. 1.16/86. That's 14.14 mA per sensor.

Look at the two triangles here. I memorize them by calling them the EAR and the PIE. EAR is Voltage-Amps-Resistance and PIE is Power-Amps-Voltage.
When I learned it, the E was Electromotive Force (voltage). Not sure why I is Amps.

Thanks keithkyll,
I was leaning towards that but i didn't want to guess. So the sensors only drive as much current as is required for the input point to activate? In this case according to the flex I/O spec each points max current is 14.8mA. That seems to make the most sense to me. I appreciate the help very much!

The 100mA that you are assuming to be drawn by one sensor will be most likely the max continuous current that the sensors output can handle. Most of the times this is written on the sensor itself.
To assure myself that this is right I looked for a manual for a PE that we are using.
In the actual manual (the one that comes with the box) the current drawn by this device isn't even given. I had to look for the technical datasheet to find it.
The sensor is able to continuously output < 100mA, but the current consumption is max. 30mA.

Manual:
https://www.sick.com/media/dox/1/61...27_3_de_en_fr_pt_it_es_ja_zh_ru_IM0014161.PDF

Datasheet:
https://www.sick.com/media/pdf/1/11/411/dataSheet_WL27-3P2430_1027769_en.pdf

In total you are now looking at a max current of 30mA + 14.81mA (the current max. drawn by one FlexIO input point) per Sensor, for some inductive sensors I found a current consumption of only 10mA and the typical current for one FlexIO input might be lower. That would make a total consumption of 3.68A for 106 sensors in a 50/50 mix. If they aren't fed by another circuit you would have to add the current that the FlexIO network adapter draws.

keithkyll said:

When I learned it, the E was Electromotive Force (voltage). Not sure why I is Amps.

Click to expand...

Been using the same since a teenager and never thought about why I is amps. Found this in Wiki:

The conventional symbol for current is I, which originates from the French phrase intensité de courant, meaning current intensity. Current intensity is often referred to simply as current.

Jerry

Thanks MaReKA,
I can't find the current consumption for the sensors I have. I have AB 875CP 2 wire AC capacitive sensors. The spec sheet lists the load current as less than or equal to 300mA but there is no Current consumption value. However, in the same book a few pages down under the DC version of the sensor, it lists both the load current AND the current consumption.....why list it for one sensor and not the other...especialy the same family and same book...only difference is that one is AC and the other DC.

Thanks JerryH.
MaReKA. I would agree that we need to add both loads, but only for DC sensors. If the output was sinking, that wouldn't affect the power needed by the sensor electronics.
In this case, the sensors are AC. The 1794-IA16 is a 120V input module. That means the sensors are most likely 2 wire, and in series with the input. Sensor electronics load doesn't add to the current draw, it adds to the voltage drop. It "steals" a few volts from the circuit for itself.
The OP made a current measurement with 86 sensors on. That verifies assumptions above.
What's missing is fuse type, amps, and reason for blowing. Is it load, or an intermittent short somewhere?

keithkyll
Super helpful information, all of it.
Is that why the spec sheet doesn't list the current consumption? Because there basically isn't any?

keithkyll said:

...In this case, the sensors are AC. The 1794-IA16 is a 120V input module...

Click to expand...

My fault, I haven't used AC sensors in years and automatically assumed it would be DC.
But if you ever run into a 1794-IB16...

To make the best of it and help me understanding AC inputs a little better I read into the manual and in a footnote it is said that 2-wire sensors can have a leakage current of <2.5mA and a Ion of at least 5mA. Does that mean that the current consumption of a 2-wire sensor is below that figure? If it would exceed 5mA would it turn on the input?

ldrewes said:

keithkyll
Super helpful information, all of it.
Is that why the spec sheet doesn't list the current consumption? Because there basically isn't any?

Click to expand...

Yes.

MaReKA said:

My fault, I haven't used AC sensors in years and automatically assumed it would be DC.
But if you ever run into a 1794-IB16...

To make the best of it and help me understanding AC inputs a little better I read into the manual and in a footnote it is said that 2-wire sensors can have a leakage current of <2.5mA and a Ion of at least 5mA. Does that mean that the current consumption of a 2-wire sensor is below that figure? If it would exceed 5mA would it turn on the input?

Click to expand...

It draws a couple of mA (<2.5mA) when off. That can be a problem with supersensitive input modules. You might see a slight glow on the input module LED. If it triggers it, you need to add a load in parallel with the input module. Unpleasant. For the 1794-IA16, the minimum current is 5.49mA, so 2.5mA won't be a problem.
For the 5mA figure, the input module must draw at least 5mA, or the sensor won't work. The 1794-IA16 draws about 15mA, so we're good.

It gets easier. The spec sheet for the 1794-IA16 has a footnote: "AC inputs compatible with proximity switches with leakage ratings of Ileak < 2.5 mA and Ion minimum = 5 mA".

Careful when sizing fuses....You could have a 1000 devices on a single fuse, but you better not size it larger than the current ampacity of your wire. If it's 16 awg, then fuse shouldn't be larger than 7.5 amps. Almost always, fuses are sized according to the feeder it's protecting, not all the individual devices...By the time you add up all the devices, you're not protecting any of them with a fuse. Size the fuse to the wire.

Information is greatly appreciated. Great information! Always love this forum.

robertmee said:

Careful when sizing fuses....You could have a 1000 devices on a single fuse, but you better not size it larger than the current ampacity of your wire. If it's 16 awg, then fuse shouldn't be larger than 7.5 amps. Almost always, fuses are sized according to the feeder it's protecting, not all the individual devices...By the time you add up all the devices, you're not protecting any of them with a fuse. Size the fuse to the wire.

Click to expand...

Yup. Don't fret about the devices, size the fuse based on the feeder ampacity.

A current meter will go a long way towards giving you your answer.

If you do measure the current, and come up with a number that is nowhere near your fuse size, say an amp or two on a circuit protected at 8 amps, look for devices that are shorting out.

GE used to make a switch that occasionally would cross its double pole contacts. If you had a neutral or common connected to the switch to power the indicator light, you would have a short and the control fuse would blow.

On systems with hundreds of limit switches this could be a bit of fun to find the offending switch. The general procedure was to go to the terminal strip where all of the switches got their power, and split the circuit in two. Measure your resistance to ground, and if the fault cleared, then the fault is in the half that you cut off. If it didn't then it is in the other half.

Keep splitting the "halves" until you are down to just one wire (which may feed multiple switches). Leave this one loose, and re-connect and power things up. Find the part of the machine that isn't doing much, and you will find your bad limit switch in this group.

For the old GE switches, a sharp rap with our GE radio (they were indestructible) would jar the contacts loose and it was unlikely that the switch would fail again. And since a new switch had just as good a chance of failing as an old one, we learned to not bother with replacing them.

Occasionally (extremely rarely), you would have a switch that would look okay to a meter set on resistance, but would short out if power was applied to the circuit. To troubleshoot this, wire up a pilot light across the (blown) fuse's terminals and as long as you have a short, the light will be on. Follow the previous isolation procedure to identify the bad circuit.

ldrewes said:

Is that why the spec sheet doesn't list the current consumption? Because there basically isn't any?

Click to expand...

It won't list the current consumption, because that's entirely dependent on what else is in the circuit.

If you wire your sensor directly into a PLC input, well, a PLC input has a very high resistance, so ohms law will tell you that your current will be very low.

On the other hand, wire your sensor into a relay - a relay may have a lower resistance than a PLC input, so the current consumption will be higher.

Wire your sensor into a dirty great contactor coil, which has lower resistance again, and your current will go even higher.

The "maximum current" is basically the highest current the sensor can handle. If the maximum current is 300mA, you can't use it to drive a load that draws anything more than that.

Seeing as most of the time these days a sensor is directly wired into a PLC input, the actual current will be extremely low - but nobody can tell you *exactly* what it will be without knowing exactly how it's wired, and what else is in the circuit.