Indeed, this is a commonly used circuit; the same fundamental idea is used for pretty much all high-side current sense amplifiers.
The op-amp will output whatever it must to make its inputs equal. As current flows through the load and a voltage starts to appear across the sense resistor, the op-amp gets upset and its output rises. The output will rise until the op-amp is satisfied—when its inputs are equal. This occurs when the same voltage that appeared across the sense resistor now appears across the top 200Ω resistor (i.e., the voltages cancel and the op-amp is left with zero volts across its input). Voltage appears across the 200Ω resistor when current flows through it, and that current
must flow through the 2kΩ resistor beneath the MOSFET. Since the currents are equal, the voltage across the 2kΩ resistor is 10x bigger than the voltage across the 200Ω resistor, and voilà we have a gain of 10x
and a level-shift to ground level.
In this explanation, I didn't even consider the fact that the op-amp has current flowing into its inputs. As shown on page 17 of the datasheet, we see that when the inputs go "over-the-top," the input stage becomes a common-base stage with 14µA flowing into each input (according to the EC table on page 3). When this 14µA current flows through a 200Ω resistor, it causes a 2.8mV offset. To compensate, equal 200Ω resistors are placed at both inputs. The same offset is thus added to both inputs, so they cancel out.
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thunderdantheman said:
so what is the advantage of this method over configuring a traditional diff amp setup?
A diff amp has high sensitivity to component mismatch. Especially when you're measuring a small voltage, you'll want very well-matched resistors. This circuit will yield good accuracy without requiring super-duper resistors. I also wouldn't be surprised if there's a speed advantage.
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thunderdantheman said:
so what is the advantage of this method over configuring a traditional diff amp setup?
A diff amp has high sensitivity to component mismatch. Especially when you're measuring a small voltage, you'll want very well-matched resistors. This circuit will yield good accuracy without requiring super-duper resistors. I also wouldn't be surprised if there's a speed advantage.