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Need help modifying ac ammeter (shunt)

Kneegrow2

Mar 20, 2018
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So I want my clamp ammeter to read half of what it is currently reading , I can't rewind the coil so I was thinking about using a resistor in parallel as a shunt between the leads. But I'm not an electrical engineer just a hobbyist and they have no idea what value resistor I should use. I read about using a shunt in a homemade analog ammeter but nothing about a digital clamp ammeter.
 

kellys_eye

Jun 25, 2010
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So I want my clamp ammeter
It's a 'clamp' meter so putting a shunt across the leads won't work. You need to find where the clamp detection coil is fitted internally and, even then, getting the required accuracy will be difficult.

Is there some reason why you need to make this change rather than either purchasing an individual meter for the range you require or putting a sticker on the meter display?
 

Kneegrow2

Mar 20, 2018
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Yeah I meant to put them where the leads go into the circuit board. I don't need accuracy. I just want to approximately reduce it by half.
 

Kneegrow2

Mar 20, 2018
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It's not actually a clamp ammeter . But it works off the same principle a primary wire inducts a current in a coil and the circuit measures the current. Im building a prototype and repurposing old equipment. The existing on board microcontroller has a trigger that trips at a certain amperage. The easiest way for me to modify it would be to use a resistor
 

kellys_eye

Jun 25, 2010
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You need to identify where the current transformer windings fix to the board - from the clamp heads - and add a resistor.

The current transformer usually has a resistor across the secondary winding which develops the voltage to be detected/measured - this will need to be doubled in value to half the range reading.

The op-amp that amplifies this voltage may have feedback resistors that you can change more easily but that would mean chasing out the tracks to find out which one it is!
 

Bluejets

Oct 5, 2014
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You could select a higher range and pass the conductor through the pickup multiple times.
 

duke37

Jan 9, 2011
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The current transformer usually has a resistor across the secondary winding which develops the voltage to be detected/measured - this will need to be doubled in value to half the range reading.
Doesn't he want to do the opposite?
Adding a resistor of equal value in parallel will drop the sensitivity by a factor of two.
The wires will need to be able to pass the increased current without overheating.
 

Kneegrow2

Mar 20, 2018
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I've got my clamp ammeter open, there's a 100 ohm resistor 1%, I was going to test on my ammeter before modifying the intended circuit. But guess what? All the Radio Shacks in my town have closed down , there were five of them a month ago. So I need to find something to cannibalize.
 

Kneegrow2

Mar 20, 2018
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Doesn't he want to do the opposite?
Adding a resistor of equal value in parallel will drop the sensitivity by a factor of two.
The wires will need to be able to pass the increased current without overheating.
Not opposite I just need to double the value of the resistor or add it in series with the other one. right because two resistors in parallel halves the resistance?
 

Bluejets

Oct 5, 2014
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As far as I can see, series would be the way as long as one samples off just the one original resistor.
 

duke37

Jan 9, 2011
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So I want my clamp ammeter to read half of what it is currently reading ,
In other words, you want to halve the sensitivity. In order to do this, the burden resistor will need to be halved, not doubled.
Have I got things wrong, yet again?
 

Kneegrow2

Mar 20, 2018
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In other words, you want to halve the sensitivity. In order to do this, the burden resistor will need to be halved, not doubled.
Have I got things wrong, yet again?

Ahhh... so more electricity flows through the coil, so adding a matching one in parallel would cut the resistance in 1/2 and cause more current to flow back through the coil than into the circuit..... I think you're right which means kellys_eye was right originally. I was wrong when I was thinking about doubling the resistance. I want to halve the resistance, so a matching resistor in parallel or replace original with 50 ohm..... shit I'm confused now.
 

duke37

Jan 9, 2011
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The different resistance should not affect the current. This will be the input current divided by the turns ratio of the transformer.
The higher the burden resistance, the higher the voltage drop across the input. Do not get voltages and currents mixed.
 

Kneegrow2

Mar 20, 2018
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let's see if I can ask this as simply as possible. The electrons flow in the coil, the resistor allows electrons to flow back into the coil and bypass the circuit board or analog meter or whatever you're using. So I need less resistance so it's easier for the electrons to flow through the coil bypassing the ammeter circuit?
 

duke37

Jan 9, 2011
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Assume you have a transformer with a turns ratio of 1000. One amp through the primary will give 1/1000 A (1mA) through the secondary. If this current goes through a 100Ω resistor, then a voltage of 1/1000 * 100 (100mV) is developed, this can easilty be measured.

There is a reflected voltage in the primary, in this case 100mV/1000 = 0.1mV which should not affect the current through the primary to any extent.

No current bypasses any part of the circuit.
 

Bluejets

Oct 5, 2014
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Ct is still a transformer regardless and changes will apply as load changes.
As an example, ct's are always shorted when not connected to a measuring circuit so therefore the voltage across the shunt is zero.
Alternatively, if the secondary is left open circuit, voltages can climb into many thousands of volts and create a dangerous situation.
Your assumption that the current remains constant with changing loads would therefore be incorrect.
 

Kneegrow2

Mar 20, 2018
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"if the secondary is left open circuit, voltages can climb into many thousands of volts and create a dangerous situation"


Thank you for this extremely useful information
 
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