What a shunt measures
You cannot measure current directly. What both boards do instead is put a small, known resistor in the way of it and measure the voltage that appears across it — and the size of that resistor, 100 milliohms on one board and 50 on the other, is the compromise the whole design turns on.
Nothing measures current
A voltmeter can touch two points and tell you the difference between them. Nothing can do that for current: to know how much is flowing you have to be in the path of it. Every ammeter ever made works by putting something in the way and measuring the effect.
On both of these boards that something is a very small resistor in the positive wire between your supply and your load: 100 milliohms — a tenth of an ohm — on the TK119, and 50 milliohms on each of the INA3221's three channels. It is the large rectangular part beside the screw terminals, and it is the only part on either board that carries the load's current.
Ohm's law does the rest. A tenth of an ohm carrying half an amp has 50 millivolts across it; a twentieth of an ohm has 25. Either is a voltage, which is the one thing a chip can measure.
The compromise
Move the slider and watch the two bars. A bigger resistor would give the chip more millivolts to work with, and that is genuinely useful — but it is in series with your load, so those millivolts are stolen from the rail the load is trying to run on, and the heat goes up with the square of the current.
At 100 milliohms and 1 A the load loses a tenth of a volt and the board gets rid of a tenth of a watt. At 1 ohm the same current costs a whole volt and a whole watt. That is the entire reason the number is 0.1 and not 1.
Why the two boards chose differently
Switch the figure to the INA3221 and everything halves: half the millivolts, half the heat. That is deliberate, and the reason is in the chips rather than the resistors.
The INA219 on the TK119 has a gain setting that stretches its input to ±320 millivolts, so it can keep the larger resistor and still read up to 3.2 A. The INA3221 has no gain setting and stops at ±163.84 millivolts — on 100 milliohms that would be only 1.64 A. So its board fits 50 milliohms, reaches 3.28 A per channel, and pays with a coarser step. The range you actually have works through both.
What a tolerance buys and what it costs
The TK119's resistor is a Yageo PT2512FK-7W0R1L: 100 milliohms, 1% tolerance, rated 2 watts. A current-sense part rather than an ordinary resistor, because the resistance is the divisor in every reading the board gives you. A 5% resistor would put 5% into every current the chip reports, with nothing to say so.
The tolerance does not go away, though. It is still there, and it is comparable in size to the chip's own error — which is why the error chapter is a chapter, and why calibrating against a known current is worth ten minutes.
Where the voltage comes in
The chip takes a second measurement: the voltage on the load side of the resistor, down to the board's own ground. That is what tells you what the load is actually getting, which is not quite what the supply is giving.
Two measurements, then. The rest — amps, watts — is arithmetic, and two numbers and a product is about which is which and why it matters.
When it does not work
Check that the supply is in the POWER terminal and the load in the LOAD one, and that all four screws are tight. A 100 or 50 mΩ resistor takes almost nothing out of the rail, so if the load has lost several volts the current is going through something other than the shunt — usually a screw that closed on the wire's insulation rather than on the copper.
That is arithmetic, not a fault. The resistor turns current into heat at I² × R, so on the TK119's 0.1 Ω 3 A makes 0.9 W in a part 6 mm long, and on the INA3221's 0.05 Ω the same 3 A makes 0.45 W. The TK119's resistor is rated for 2 W, so it survives, but a hot shunt has drifted in resistance and the reading with it. If it is too hot to touch, you are near the top of what these boards are for.
Because the resistor is in series with your load, so every millivolt across it is a millivolt the load does not get, and the heat goes up with the square of the current. At 1 Ω instead of 0.1 Ω a 2 A load would lose 2 V and the board would dissipate 4 W. A tenth or a twentieth of an ohm is the compromise: enough millivolts for the chip to measure, little enough that the load barely notices.
Both are right about different things if your meter is in series somewhere else in the circuit, and both have their own tolerance. The TK119's shunt is 1%, and the chip adds a fixed zero-point error of about a milliamp — nearer two on the INA3221, because its smaller resistor turns the same microvolts into more milliamps. Compare them at a current of a few hundred milliamps, not a few — at 5 mA the two will disagree by a lot and neither is broken.
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