# Requesting help with a 0-300mA on shunt circuit

#### Dimis

Joined Oct 24, 2017
15
Dear all as i am a new member please forgive any mistakes.
I am asking help for the following.
I have a 0-300mA qauge with damaged internal resistors.
I have remove them but are burned so no possible to measure values any more.
In order to check the rotating coil i have measure the coil resistance alone, that is 2,28Ohm.
Then i have apply 1,37 dcv from an AA batery and in series i had connect a 100ohm pot, the full scale deflection (300mA) comes when the pot is set at 71 ohm, pot measured after dc voltage removed.
Second measurment i take is again with the circuit operative and FSD adjusted to 300mA as before,
the voltage across gauge terminals was 0,035V.
I would like to connect this mA in my power suply which operates from 400 dcv up to 680 dcv with max load 250 mA.
Can someone please calculate the two resistors i have to connect (shunt and series) and give me an information regarding the wattage also?
Dimis

#### AlbertHall

Joined Jun 4, 2014
10,417
From your measurements I make the movement FSD current 18.7mA. Given the movement FSD voltage of 35mV, that makes the shunt resistor for 300mA 0.124Ω. The dissipation in the shunt resistor would be 11mW.

#### Dimis

Joined Oct 24, 2017
15
First you mention one resistor 0,124 Ohm/11mW that is the shunt or parallel as far i understand, but how about the other one existing previously in series.
Moreover is this calculation taking into account the power supply parameters i have give above?
SAFETY FIRST as usual, please let me know whenever you can
Brgds
Dimis

#### AlbertHall

Joined Jun 4, 2014
10,417
One problem is that the accuracy of the reading will be based on your measurements so you might want to include a pot to adjust the final full scale current. See the end of this article for how to do this and how to calculate the values. This method will also allow you to use a more standard value of shunt resistor.

#### Alec_t

Joined Sep 17, 2013
11,665
I trust the meter insulation, between either terminal and any part of the meter which can be touched, is able to withstand >680V?

#### Dimis

Joined Oct 24, 2017
15
Gendleman, thank for your kind replies,
AlbertHall, i was there before trying to understand how i can calculate both resistors but i am not sure, thats for i was asking help with this post.
Alec_T i believe that after calculations the circuit voltage will not affect the mA as only a tiny portion will go through, and of course within operational limits, if is that you mean, otherwise please explain.
So i will wait for your posts once again for both shunt and series resistors required in order to be safe within above power supply (400-680vdc @ 250 mA maximum).
Shunt: ______ohms @ _____ watt
Series: ______ ohms @ _____watt
Brgds
Dimis

#### Dimis

Joined Oct 24, 2017
15
Maybe it worth to mention that the two resistors was previously within the qauge was not more than 1/2 watt, if that helps.
I remain
Dimis

#### Alec_t

Joined Sep 17, 2013
11,665
the circuit voltage will not affect the mA as only a tiny portion will go through
That is neither here nor there. There will be only a small voltage difference between the two terminals, so if either of the terminals is at 680V then so will the other one be. If you then touch the meter at any point you need to be sure that leakage current via the meter casing can't zap you.

#### Dimis

Joined Oct 24, 2017
15
AlbertHall on tip from the article
In this example the 50 microamp meter movement with a 2700 ohm internal resistance will be used.
Is this the moving coil resistance or a resistor positioned within the qauge like the two i remove from the old one?
Brgds
Dimis

#### Dimis

Joined Oct 24, 2017
15
Alec_T, thanks but i am still fighting to get that, will not be so dificult but i am tottaly confused,
Any new approach?
Brgds
Dimis

#### AlbertHall

Joined Jun 4, 2014
10,417
AlbertHall on tip from the article
In this example the 50 microamp meter movement with a 2700 ohm internal resistance will be used.
Is this the moving coil resistance or a resistor positioned within the qauge like the two i remove from the old one?
Brgds
Dimis
The 2.7k is the resistance of the movement.

#### Dimis

Joined Oct 24, 2017
15
Dear all hello again,
Thank you for your kind efforts to help me and i am really appreciate your efforts,
but although i read again and again the article i cant make my mind on that.
To my amateur calculations i think i need something around 1,2 Ohms shunt and about 470 Ohms in series with the qauge coil.
I got the reason for the shunt in our circuit but the other resistor in series i dont understand, is a resistor which plays the same role like the pot i use to test the voltmeter with the batery after dismantling it lets say for final qauge adjustment to the printed value on bezel? And if yes do i realy require this resistor?
Brgds
Dimis

#### AlbertHall

Joined Jun 4, 2014
10,417
If you want the meter to accurately read current then you will probably need to be able to adjust it to read the correct value.

#### JoeJester

Joined Apr 26, 2005
4,369
Did the thread starter zero the meter before that test. 18.7 mA is a wierd number. 20 mA is more likely unless the zero was way off. I recommend the tape do another test after ensuring the meter reads zero before the supply any current.

#### Dimis

Joined Oct 24, 2017
15
Thanks AlbertHall and JoeJester,
I have manage to identify the previous resistors these was 0,82 Ohm the shunt and 0,25 Ohms the one was in series but both at 1/4 watt,
I believe that make sense why both did not sustain the load.
In my calculations shunt should be at least 20W as this Ohm value gives about one volt drop, the other one the 0,25 ohm should be at least 2W.
What you sugest.
I will test the operation by positioning another ammeter in series with the one in question, i should see normaly half the current on each one, correct? If yes the i can ajust the qauge in question by adding or subtracting the value for resisto in series.
Tks in advance for the replies
Brgds
Dimis

#### AlbertHall

Joined Jun 4, 2014
10,417
300mA in 0.82Ω is 74mW (0.3 x 0.3 x 0.82)
300mA in 0.25Ω is 23mW.