# Current Sense Amplifier Potentials

I'm trying to implement a current sense amplifier in my application, but I have a question about setting it up with regards to current flow.

Basically, in this case, you put a very small shunt resistor between the source and the load, and the current sense amplifier measures the voltage across it, and outputs a voltage in accordance with the current flow. This is shown below using the INA293. However, this assumes a positive voltage at the load supply with respect to GND, being 0 volts, and most likely, $$\R_{Sense}\$$ is small enough such that the voltage drop is negligible. For my application, I'm considering using it with the LM2673 circuit that outputs a negative voltage shown below. In this case, GND would be the more positive node w.r.t. $$\V_{OUT}\$$, so current will flow from GND to $$\V_{OUT}\$$. So, to rework the current sense circuit, the Load Supply would be GND, GND would be $$\V_{OUT}\$$, and the opamp's GND node would be connected to $$\V_{OUT}\$$. If I were to place the sense resistor between the Load and $$\V_{OUT}\$$, would the current sense amplifier work properly, or is there something that would not cause it to work? EDIT: Proposed Application Circuit: simulate this circuit – Schematic created using CircuitLab

Rough Schematic idea with Comparator OpAmp: simulate this circuit

• what is your application? draw a new schematic to clarify how you plan to use your current sensor? May 18 '20 at 21:16

Based on your description, it seems like you mean that you want to implement something like the circuit below. This is essentially the same circuit, using your switch mode inverter as a virtual ground. As long as $$\ V_{s+} - V_{s-} \$$ is less than 22V (the rated supply voltage of your current sense amplifier) there is nothing wrong with this circuit (or really, anything different from the application diagram from the datasheet that you posted). simulate this circuit – Schematic created using CircuitLab

Note that if you want to connect the output of this current sensor to something else, it will also need to be referenced to $$\ V_{s-}\$$.

• Thanks for your reply. It's almost correct, though to still have a connection to GND for my load, I put the sense resistor after my load. I edited my question to showcase what I'm talking about. Why does the IC have to be referenced to -Vs instead of GND in order for this to work? The voltage difference between the supplies is still within range, but would connecting the IC_GND to the GND node affect the measurements? From the opamp's view, current is still flowing in a positive direction. May 18 '20 at 22:10
• Actually, thinking about how the CS opamp needs to have its GND pin connected to -Vs as you showed, is there a practical limitation as to what -Vs can be? I think I see why it has to be connected. It has to know what the lowest potential is, which is -Vs here otherwise the op-amp won't work. However, the limitations of the supply voltages. INA293 has a min/max supply rating of -0.3 and 22V, but this is based on its GND pin being at 0 V. So, if -Vs = -15 V, then the opamp will see a voltage supply of 20V if +Vs is 5V [5-(-15) = 20]. This is within range, so the opamp will still work properly? May 19 '20 at 16:04
• Your edited circuit is a low side current sensor. You can switch the direction of the current if you use a bidirectional current sensor. You're incorrect that the current would still be flowing in the positive direction if you connect the amp ground pin to 0V. It might help you conceptually to 'redraw' your circuit by re-labelling -Vs as GND. May 19 '20 at 16:44
• Sorry, I miswrote that. When I meant by 'positive direction', I meant that current flows from higher to lower potential. In this case, it flows from GND to -Vs (keeping in mind that -Vs is a negative voltage) through the load first before the sense resistor. For all intents and purposes, -Vs will be negative (going as high as 0 volts), so current will flow like how I describe. Would I still need a bi-directional op-amp or would a uni-directional one still work? May 20 '20 at 15:26
• a unidirectional one is fine, but your input voltages need to be in the range of your supply for your circuit to work properly May 20 '20 at 16:04