What is a SHUNT? (Used to measure Current) + How to make a DIY version

so what exactly is a shunt well a shunt is used to measure current for example the amp meter on the front of my power supply uses a shunt to measure the output current from the power supply so in a nutshell a shunt is just used for measuring current but what does a shunt look like exactly well for example this is an axial style shunt and this is a high current shunt rated to handle up to 50 amps and this rather ordinary looking wire jumper on the circuit board is also a shunned so shunts come in a wide range of varieties so what exactly defines as a shunt then well in a nutshell a shunt is nothing more than a conductor which has some value of resistance typically in the Milione range basically any conductor such as an ordinary resistor this wire rod or a length of copper wire could be made into an improvised shunt provided it has some resistance and we know the value of said resistance a typical multimeter is useful for measuring the resistance of components down to say 1 ohm however they are unsuitable for measuring the incredibly low resistance of the shunt for example you can see my meter reads all zeros while measuring the resistance of both shunts now although the resistance is incredibly low it is there and that's what makes a shunt useful so basically a shunt is a low value resistor okay great but how does the shunt help us measure current well that's a great question and weirdly to measure current with a shunt we actually measure voltage let me explain have you ever found yourself situated in a paddock full of cows and need to order circuit boards yeah me neither but if I did I would use this video sponsor jlc PCB 5 circuit boards cost as little as two dollars they offer fast production time and with a multitude of design options you're only limited by your imagination ordering is as simple as going to jlc PCB comm uploading your Gerber files and choosing your design preferences you can also choose any color solder mask at no additional cost and if you're new to designing circuit boards then check out my co-ed circuit board series to get you started look look Daisy free circuit boards so using my whiteboard I'll draw a shunt with a value of 1 ohm next I'll add a 1 amp load over here and lastly we have a power supply the power supply voltage is irrelevant for the purposes of calculating current it could be 10 volts or 100 volts or any other random number it really doesn't matter we don't use the supply voltage in any way to calculate the current okay so I've drawn this diagram what's the shunt connected to the positive rail but the exact same setup could be used on a negative or ground rail it really doesn't matter now remember a shunt is basically a resistor and when current passes through a resistor there is a voltage drop across the resistor so if we attach our multimeter leads on each side of the shunt we will be able to measure the voltage drop across the shunt when current is flowing with a 1 amp load the multimeter will read a voltage drop across the shunt of 1 volt this voltage drop is directly proportional to the amount of current passing through the shunt so at 1 amp of current we have a 1 volt drop now if we increase the load to 5 amps the voltage drop will also increase to 5 volts in this example I have deliberately used round numbers to keep the math simple but what if we don't have a precision 1 ohm shunt maybe I'll shunt has a random value so let's give our shunt a new value zero point eight six seven ohms looks pretty random to me and our multimeter is reading one point one nine volt drop across the shunt now in this new scenario we have a load but we don't know how much current it's drawing however we do know the value of the shunt and the voltage drop across it from these two numbers we can calculate the current draw from the load so how do we do that well get ready for two hours of hardcore math no I'm just kidding someone made a calculator to do all the work for us if you watched my last video on Ohm's law you probably saw this one coming if you aren't familiar with Ohm's law then go watch that video then come back to this video ok so into my shunt resistance which was zero point eight six seven ohms and the voltage drop across the shunt which is one one nine volts press calculates and now we know the current draw is one point three seven amps and that's how a shunt is used to measure current so now you know how to calculate the current from any value of shunt using this method pretty cool eh okay time to put theory into practice with a demonstration I will use this ten milli ohm shunt and this 12 volt bulb will act as my load the shunt is connected between the power supply and load I'll then attach my multimeter leads across the shunt to measure the voltage drop just as I showed in my whiteboard drawing I'll blank out the ammeter on my power supply so there's no cheating here alright time to fire it up okay so my multimeter is reading 42 millivolt drop across the shunt it's time okay let's go back to Ohm's law calculators to work out the current I'll enter my shunts value of 10 Mille ohms and the voltage drop was 42 millivolts press calculate and that works out to be 4 point 2 amps of current now if I uncover the ammeter on my power supply it reads 4.3 amps now you might be wondering why my calculation was 100 milliamps less than the amp meter displace that's down to a couple of factors for instance the shunt I'm using has a tolerance range of plus or minus 1% and even though my Broman meter is a fantastic meter measuring down in the sub millivolt range with pinpoint precision is not what this meter is intended to be used for so we'll have to forgive this 2.3 percent error margin okay so now you know what a shunt is and also how to use one so let's move on to making our own DIY shunt and a good place to start would be to use a common resistor with a low value like this 10 ohm resistor so after replacing the commercial shunt with my 10 ohm resistor it's time to test oh well that didn't last long if we freeze the video we can see the voltage drop was over 11 volts across our 10 ohm shunt that equates to over 13 watts of power to dissipate more than 24 times what the resistor is rated to handle hmm ok well how about we use this beefy hundred watt 8 ohm power resistor I'll replace the puny burnt resistor with my powers of stir and retest hmm something's not right here the bulb is barely glowing and the current draw should be more than four amps and to top it off the resistor is getting really hot so what's going on here well the issue is the eight ohm resistor has too much resistance to allow the circuit to function as normal there is a ten point volt drop across the resistor if we do the math that is more than 14 watts of power been wasted as heat that leaves less than 1.5 volts to power the bulb so that's why it's barely glowing if I switched back to the 10 milli ohm shunt I used earlier not only does the bulb run as normal but the shunt isn't hot at all in fact I can hold my fingers around it in the setup only point 17 watts is converted to heat compared to my 8 ohm resistor at a whopping 14 watts and exactly the same setup so l8 empowers us to shunt isn't practical to use we need something with much lower resistance something around 10 million start I have a spool of 1.25 millimetre enameled copper wire I could use to make a shunt I will use this handy wire resistor calculator on command e-comm I need to know the resistance per meter so enter and 1 meter for length and the wire diameter which is 1.25 millimeters and that calculates to be 13 Malone's per meter so in theory I'd only need around seven hundred and seventy millimeter lengths of wire to achieve a 10 milli ohm resistance so I'll cut a link slightly longer than I need to allow a margin for fine tuning I placed a piece of masking tape slightly lower than the theoretical length I need and proceeded to sand off the enamel insulation up to the masking tape the reason will become clear in a moment the question is how do we precisely measure the correct length of wire to achieve a value of 10 Malone's well I can't use my multimeter because it lacks the resolution to accurately measure 10 Malone's but it can measure millivolts with a reasonable degree of accuracy so here is the method first I need to run a precise amount of current through the wire so I'll set up my power supply to limit the current to one amp now I could connect the wire across the output of my power supply but the only problem with that as the wire is basically a dead short and the voltage is almost zero this means controlling the current with a reasonable degree of accuracy is difficult to get around this issue I'll connect my 8 ohm power resistor and series with my copper wire then reconnect the power supply now the voltage is around 8 volts and the current is fairly stable I'll set my multimeter to read volts and connect the negative probe to one end of my copper wire now I can slide the other probe along the copper wire until my meter reads 10 millivolts which is right here I'll mark that spot with a marker and then cut the wire at the mark no in theory this length of copper wire should have a resistance of 10 Malone's and if you're wondering how I calculated that well of course it was Ohm's law to prevent shorts where the enamel has been removed I used a piece of heat shrink to insulate the wire so now we have a calibrated length of wire we can use as a shunt however it's a bit bulky to use in its current form so I wrap the wire around a pipe to make it smaller and used a couple of cable ties to retain the shape now the only danger in doing this as we have inadvertently made an inductor which depending on your application can be very undesirable so I straighten the wire out and found the midpoint of the wire I then bent the wire back on itself like this then I can once again wrap the wire around a pipe like this to make a coil now because we folded the wire back on itself half the wire is wrapped in a clockwise direction while the other half is counterclockwise this creates two equal and opposite magnetic fields that cancel each other out resulting in no inductance pretty cool eh so how does our DIY 10 mil ohm shunt compared to its commercial counterpart well let's go back to our test setup what's the light bulb just like before the shunters connected in series was the load on the positive rail and my multimeter will be measuring the voltage drop across our DIY shunt okay let's fire it up so according to the meet of the current passing through our DIY shunt is 4.7 amps which is around 400 milliamps higher than reality that works out to be an error margin of 8.8% considering that we didn't use any special or expensive equipment to make our DIY shunt I think that is pretty reasonable however if we take into account that commercial shunts are more compact offer a higher degree of precision and in most cases are relatively cheap I think I'll stick to buying my shunts so if you found this video useful please give it a like it would be much appreciated helps out mercifully and if you have any questions or suggestions for future videos leave them down in the comment section below thank you very much to my supporters on patreon you guys Rock you helped make content like this video possible and I will see you all in the next video bye for now

Channel: Schematix

Views: 138,284

Rating: 4.9369869 out of 5

Keywords: shunt, current, amps, ohms, law, how, to, measure, make, diy, ammeter, multimeter, DMM, multi, meter

Id: _3GPaFuqrxs

Length: 11min 44sec (704 seconds)

Published: Wed Jan 15 2020