1
\$\begingroup\$

Please excuse my awful math/s. Can anyone please help me calculate/convert the mV reading from a 50A shunt? For example if i was reading 1.5mv how do i convert please. Please keep it simple as math is (obviously) not my strong point! Thank you


Slightly embarrassed at the state of my maths but thank you all for comments. However going with (50/75) x the Amp reading isnt agreeing with my two ammeters!

The ammeters are reading 2.06 and 2.12, and the DMM is reading 12.0mV - so - (50/75)x12 =8 oops. Any ideas folks?


It is a series circuit 12v battery, fuse holder with cheap ammeter plugged here, switch, bulb, earthed through chinese shunt with second ammeter reading here along with DMM. Appreciate shunt and cheap ammeters may not be 100% but im not getting close even. How might ammeters not be reading total circuit current pls?.


Schematic (description below) enter image description here

The two ovals at (a) represent a fuse holder (fuse removed). into this is inserted an ammeter (c) which accepts the removed fuse (b). (d)represents the shunt with the other ammeter attached and this is where the voltmeter reading are taken off either side of shunt. Excuse poor schematic drawing/symbols etc.

schematic

simulate this circuit – Schematic created using CircuitLab

Schematic added by Transistor. OP to edit / correct.


Update - PROBLEM SOLVED Brilliant - I was taking the mV reading from the wrong place, on the large, outer terminals when I should have been reading from the the smaller terminals. Here I get 3.5mv which then gives me a close amperage figure (50/75*3.5=2.4).

Lesson learnt - thanks to all - admin please feel free to edit this to make it more useful to others.

\$\endgroup\$
  • \$\begingroup\$ Are the ammeters reading the total current in the circuit? A diagram may help.. Do the ammeters have a scaling factor? \$\endgroup\$ – Solar Mike Jan 29 '18 at 21:29
  • \$\begingroup\$ Appended a description of circuit to post if anyone has any other ideas. \$\endgroup\$ – Nick Hudson Jan 30 '18 at 17:56
  • \$\begingroup\$ So where is the circuit? I think we all want to find out what is happening here... \$\endgroup\$ – Solar Mike Jan 30 '18 at 18:09
  • \$\begingroup\$ I described it above ... I'll upload a schematic later if poss (grandchild in the house). \$\endgroup\$ – Nick Hudson Jan 30 '18 at 18:26
  • 1
    \$\begingroup\$ So, have I understood correctly that at "d" you have the shunt which has an ammeter connected with it - if so, then these should be in series ie cable in : shunt : ammeter : cable out and the voltmeter should be across the shunt only, if they are in parallel this will cause an incorrect (low) reading on the voltmeter due to resistances in parallel. \$\endgroup\$ – Solar Mike Jan 30 '18 at 19:01
2
\$\begingroup\$

The calculation of measured A = Vm * 50A/75mV is correct (Vm measured in mV).

You need to connect the meter directly to the shunt and you need to use the "inner" terminals on the shunt (usually they are smaller screws) to go directly to the meter (and nowhere else) and the outer terminals on the shunt carry the high current.

It should look like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The resistors marked "?" represents wiring resistance and internal resistance of the shunt.

If you were to (say) connect the meter to ground rather than directly to the shunt then the resistance of the connections would add to the shunt resistance and you would get a significant error most likely.

enter image description here

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Brilliant - this is the answer. On the smaller terminal I get 3.5mv which then gives me a close amperage figure (50/75*3.5=2.4). \$\endgroup\$ – Nick Hudson Jan 30 '18 at 19:44
  • \$\begingroup\$ @NickHudson So, I gave you the correct calculation, but the issue was the circuit construction - well done for sorting it. \$\endgroup\$ – Solar Mike Jan 30 '18 at 19:54
  • \$\begingroup\$ Yes. the calc was the only one it could be!! The help here is really useful and very informative. I'm learning!! ta \$\endgroup\$ – Nick Hudson Jan 30 '18 at 20:19
2
\$\begingroup\$

If 75mv is the full scale reading for 50A, then :

Edit 2 : So

(50A / 75mV) * 1.5mV = 1A

Where 50A is max shunt current, and 75mV is full scale reading and 1.5mV is the reading obtained by the OP. Note, one can see that the mV, as units, will cancel leaving the result to be A.

Edit : the shunt has already been calibrated, you may find small cuts in it to "match the resistance to give 75mV at 50A.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ "... so (50 / 75) * 1.5 should give you the current.". Um, that will just give you '1 A' no matter what the mV reading is. \$\endgroup\$ – Transistor Jan 29 '18 at 20:56
  • \$\begingroup\$ @Transistor So (50 / 75) * 22mV = 14.67A, not the 1A you suggest... \$\endgroup\$ – Solar Mike Jan 29 '18 at 20:59
  • \$\begingroup\$ @Transistor If I follow your calculation in your comment below, why does it give the same as my calculation? \$\endgroup\$ – Solar Mike Jan 29 '18 at 21:01
  • \$\begingroup\$ I know you know how to calculate the true reading but you have no variable in your equation. How did 1.5 become 22 mV? \$\endgroup\$ – Transistor Jan 29 '18 at 21:01
  • \$\begingroup\$ Ah, I see the 1.5 mV in the OP. It's not clear in your answer. \$\endgroup\$ – Transistor Jan 29 '18 at 21:02
0
\$\begingroup\$

You need to provide the ohm value of the shunt. Without knowing that this cannot be determined from the provided information.

Edit: Yeah, I missed the "75 mV" part in the title. With that its just a matter of multiplying mV reading by 50/75.

My Bad.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ This is not an answer to the question, when you have enough rep you can leave comments. \$\endgroup\$ – Voltage Spike Jan 29 '18 at 20:39
  • \$\begingroup\$ @Norm: There's enough info in the question to calculate the resistance, if required. Nick has supplied the full-scale current and voltage and the resistance can be calculated from these. In any case, it's not required. \$ I = mV \times \frac {50}{75} = mV \times \frac {2}{3} \$ and the answer will be in amps. The formula doesn't mention resistance. Welcome to EE.SE. \$\endgroup\$ – Transistor Jan 29 '18 at 20:52
  • \$\begingroup\$ This does answer the question. But wrongly. If Op knows current through the shunt and the voltage across it, he knows the resistance. R = V / I or 0.012 / 2.09 = 0.00057 ohms \$\endgroup\$ – Passerby Jan 29 '18 at 21:32
  • \$\begingroup\$ Your answer is nearly fixed. The title says 75 mV not mA. It's tough in here! Don't be discouraged but when writing an answer it's a really good idea to check that you have actually answered the question and proofread before hitting submit. There are too many guys here who know what they're talking about! \$\endgroup\$ – Transistor Jan 29 '18 at 22:49
  • \$\begingroup\$ +1 since content doesn't belong in the title, and I missed it originally for the same reason. This was the OP's fault, not yours. \$\endgroup\$ – Olin Lathrop Jan 30 '18 at 20:51
0
\$\begingroup\$

Look up something called Ohm's Law. It says that the voltage across a resistor is the current thru the resistor times the resistance.

You know the voltage across the resistor. Go find what resistance your shunt is. Then divide the voltage you see by that resistance to get the current.

For example, if this is a 1 mΩ shunt, then with 1.5 mV across it, you have (1.5 mV)/(1 mΩ) = 1.5 A.

Added

I now see that you say this shunt causes 75 mV across it with 50 A thru it. That means it is really a (75 mV)/(50 A) = 1.5 mΩ resistor. To get the current, plug in this resistance and the measured voltage into Ohm's law. For example, if the reading is 1.5 mV, then the current is (1.5 mV)/(1.5 mΩ) = 1.0 A.

In summary, divide the voltage reading by 1.5 mΩ, and you get the current.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ So, if you have 75mV then 75 / 1mΩ is surely 75A ? But the OP says it is a 50A shunt... \$\endgroup\$ – Solar Mike Jan 29 '18 at 20:51
  • \$\begingroup\$ @Sol: First, the 1 mOhm value was a example to show how to do the math. Note "For example, if ..." in the last sentence. Second, the OP hadn't said what the resistance nor maximum voltage of the shunt was when I wrote this answer. \$\endgroup\$ – Olin Lathrop Jan 30 '18 at 11:45
  • 1
    \$\begingroup\$ The Op had given sufficient information in the original post - can't understand how you missed it... \$\endgroup\$ – Solar Mike Jan 30 '18 at 19:50
  • \$\begingroup\$ @Sol: The information was originally hidden in the title only. In any case, I have updated the answer to his specific case. I don't see what your problem with this is. \$\endgroup\$ – Olin Lathrop Jan 30 '18 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.