Initial ResearchWhat is a shunt resistor? The general principle is that measuring the voltage drop across a known value resistor (shunt resistor) will give you current. Since I = V/R, measuring the voltage drop and dividing it by the resistance value will give you the current. This voltage drop across the resistor can also be fed into a differential amplifier, magnifying the voltage drop to a reasonable level (since a shunt resistor tends to be a lower value, so less voltage drop).
The plan is to use a shunt resistor and a current sense amplifier and feed the output to a couple of pins, test points, or a connector (havenāt decided yet). This could then be measured by an oscilloscope and logged so that current monitoring is possible. After some deliberation, I decided that it would be cool to add a seven segment display to the board in order to display the amount of current on the 12V bus or 5V bus. There are two strategies to implement this: Use a specialized IC to convert the analog voltage produced by the shunt resistor +amplifier into an output that can be read by a seven segment display. Program a microcontroller with an integrated ADC to convert the analog voltage into an output. Much of the time, the second solution can be more cost effective and configurable than the first.
Shunt and Amplifier SelectionFirst, letās figure out where to put the shunt resistor: https://www.allaboutcircuits.com/technical-articles/resistive-current-sensing-low-side-versus-high-side-sensing/ To do this, letās talk about CMRR (Common Mode Rejection Ratio). Hereās Googleās Gemini AI engine to help (prompt: Explain CMRR and what a common vs differential mode signal is of an op-amp to me with an analogy like I am 15 years old): Imagine you're at a concert with your friend. The band is the desired signal you want to hear clearly. Common-mode noise is like a bad sound system rumble that affects both of you equally. It might be caused by a loose wire or some electrical gremlins. Differential-mode signal is the actual music from the band. This is what you want to amplify without the rumble.
An op-amp is like a fancy headphone amplifier that boosts the music (differential signal) for you and your friend. Ideally, it should completely ignore the rumble (common-mode noise). In reality, no headphones are perfect, and some rumble might still leak in. Similarly, even a good op-amp will have some CMRR, meaning a small amount of rumble might still be amplified with the music.
Ok, but whatās even the big difference between putting a shunt resistor on the low side vs the high side?  The common mode voltage (VCM) is literally the average of the two voltage inputs into the amplifier. VCM = (V1 + V2) / 2 Thus, when Rshunt is placed on the low side (and since the voltage drop across the shunt is negligible) VCM is far closer to GND Conversely, if Rshunt is placed on the high side, VCM is far closer to Vsupply. This means that using a high side shunt resistor will require an amplifier with a higher CMMR (common mode rejection ratio).
Ok, but then why not just put it on the low-side and go on with our lives? Letās say some system fault happens and the low side of Rshunt becomes grounded. Then, our system load never actually gets power! All of the current flows directly to ground. With a low-side shunt resistor, this would never be detected, as we would just read in a Vshunt of zero or perhaps some floating voltage. With a high-side shunt resistor, however, we read in a voltage of Vsupply, as the entire voltage drop happens across Rshunt. This isnāt really a concern with our application as we donāt really care. If the user of this board happens to detect a fault condition, they will notice it immediately as their board will not get power.
Another issue is ground loops. The gist is that if some components on the system use one ground and some use another, there can be some path for current flow. Even if the grounds arenāt connected, there are many ways things can go wrong (for example, even things like stray electromagnetic fields can induce small currents). In this case, the system load (the device being powered) would have a slightly higher ground reference than our buck converters or USB-PD chip, which means that we would need separate ground planes or paths. On the other hand, if a high-side resistor is used, the grounds of everything is all the same, so no possibility of weird ground shit happening.
High-side shunt resistor is our answer then.
To select a shunt resistor, we also have to keep in mind our amplifier circuitry due to something called amplifier offset voltage. https://www.allaboutcircuits.com/technical-articles/understanding-the-amplifier-offset-voltage-and-output-swing-in-resistive-current-sensing/ Letās say the two inputs to our current sense amplifier are V+ and V-. Ideally, if V+ == V-, then Vout = 0V. However, this is not truly the case, and current sense amplifiers will have some necessary offset on V+ to be applied so that Vout = 0V. 
If Gdiff is differential gain of the amplifier: Next, letās look at error. Error is defined as (Measured Value - Actual Value)/(Actual Value). Measured value is Vshunt + Voffset (not including Gdiff as that is really just a multiplier for ADC measurement purposes) Actual value is just Vshunt since thatās what weāre trying to measure Therefore, Error = ((Vshunt + Voffset) - Vshunt)/(Vshunt) If Vshunt = Voffset , then Error = ((Vshunt + Vshunt) - Vshunt)/Vshunt = Vshunt/Vshunt = 1 = 100%.
This matters because it screws up our accuracy at lower current values (i.e. if Rshunt is fixed, and Voffset is fixed, and Ishunt is low enough, Vshunt <= Voffset which means Voffset/Vshunt > 100%) Thus, we must find our lower bound for Vshunt, i.e. the shunt voltage that we are ok with 100% or greater error because it is probably lower than minimum current of a device connected to the bus. Essentially, how low do we plan to measure? We can try increasing the value of the shunt resistor (Rshunt) in order to increase Vshunt at lower current draws, such that Vshunt stays above Voffset, but that will incur a greater power draw.
This is the optimization we have to run: how small of a Rshunt value can we select before the accuracy catches up to the power savings? Note: On the mouser page Voffset is displayed as Vos for a given current sense amplifier.
5V Current Sense SelectionFirst, the 5V current sense amplifier. Iāve filtered by every Vcm range that includes 5V and selected the cheapest current sense amplifier to start. INAx180 Depending on the selected part, this can be a 1-channel, 2-channel, or 4-channel, which means we could use the 2-channel variant for sensing on the 12V bus too. It all depends on what gain is necessary though. Supply voltage supports 5V
Letās first calculate the minimum and maximum Rshunt which will constrain us.   First, for 20V/V: For 50V/V: For 100V/V: For 200V/V: This is at typical Vsp and Vsn; at maximum Vsp and Vsn, Rshunt_min becomes greater than Rshunt_max which is no bueno. However since this really only affects the upper and lower bounds of current we can sense and not the resolution directly, I think itās probably fine unless weāre going to be measuring a device that draws only 1mA of current.
In order to effectively decide our configuration, letās assume we intend to choose the maximum Rshunt possible in order to maximize accuracy. Thus, weād need to choose the lowest gain option for our current-sense amplifier of 20V/V. Choosing an Rshunt of 80mOhms yields a voltage drop (at max current) of 3A x 80mOhms = 0.24V Pshunt = 3A x 0.24V = 0.72W, 4.8% of maximum power draw of 15W on 5V bus. Percent error at 0.001A on 5V bus: Error = 100uV/(0.001A x 80mOhms) = 1.25 = 125% (100uV is typical offset voltage)
However, weāll choose the exact Rshunt that gets us desired Vout from the 20V/V amplifier, provided itās within the voltage swing range An important note here is that we are using a microcontroller at Vsupply = 3.3V with an LDO, specifically because most of the 5V microcontrollers also have the pin output voltage = 5V which is higher than what our seven-segment display needs. The current is expressed on the microcontroller as a voltage mapping 0 to 3A to 0 to 3.2V (which is 20V/V x 0.24V). The ADC is 12 bits, so that means a value from 0 (0V) to 4096 (3.3V). However, this is a max of 3.2, which equates to a value of about 3972.
12V Current Sense SelectionIf the INAx180 works for the 5V bus, it may work for the 12V bus especially since that requires a lower resolution. Letās do some shunt calculations at the 20V/V gain for the 12V bus. The shunt maximum and minimum resistance would remain the same since Vsupply remains the same for the amplifier (being supplied by the 5V bus). Choosing an Rshunt of 80mOhms yields a voltage drop (at max current) of 8.33A x 80mOhms = 0.666 V. This yields a power drop of 5.55W on the 12V bus, which is designed to supply typically 65W and 100W at max operating conditions. This would mean a percentage use of 5.5-8.5%. However, 0.666V x 20V/V = 13.32V. This is far too high of a voltage for the ADC to handle anyways, and since we want to lower power draw anyways it might be in best interest to choose a smaller shunt resistor. This presumably increases error at lower current points but since this bus is for higher load devices it shouldnāt matter that much. Currently, the percent error at 1mA is 100uV/(0.001A x 80mOhms) = 1.25, or 125%.
If we want 3.2V to be equivalent to max current out, 3.2V/(20V/V) = 0.16V needs to be our shunt voltage at max current 0.16V/8.33A = 19 mOhms. This is below our Rshunt minimum, which means we have a problem. To fix this, we can up the resistance to be above the minimum and then use a voltage divider on the output of the amplifier to get it within measurable range. By using a resistance of 28mOhms, we get a voltage drop of 8.33A x 28mOhms = 0.23324 volts on the resistor, and an output of 0.23324V x 20V/V = 4.6648 volts. Then, we resistor divide by a ratio of 0.685988681 to get to our desired 3.2V.
As shown earlier, 3.2V is approx. a value of 3972 on our ADC. This means a resolution of 8.33A/3972 = 0.0021A = 2.1mA The percent error at 0.002A, our lowest measurable value, would be 100uV/(0.002A x 19mOhms) = 2.63 = 263% This will drop considerably as current increases though, and since the line isnāt really meant for low-power devices anyways Iām not too concerned. Finally, the power draw at max current would be 8.33A x 0.23324V = 1.943W.
We have to select a 28mOhm resistor because we donāt have much wiggle room due to the voltage swing characteristics of this amplifier. Itās settled. Choosing the INA2180A1QDGKRQ1 is in our best interest, the two-channel, 20V/V gain variant of the INAx180
noteNote: I changed this decision to just the one-channel variant after deciding that we donāt need 5V current sense. The new part number is INA180A1QDBVRQ1.
Note: I changed this decision to just the one-channel variant after deciding that we donāt need 5V current sense. The new part number is INA180A1QDBVRQ1.
A Few More OptionsItās actually not settled at all. To satisfy the requirement that we choose the ideal component for the job, weāll have to look at a few more options. Iām not going to go over calculations in great detail, but merely present them here:
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