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The short answer is yes. You need a shunt resistor with the correct resistor to be able to measure the current.

The INA219 and similar chips are designed to convert a very low voltage over the shunt resistor to a more readable value. The idea is to use a very small shunt resistor to reduce the amount of power dissipation.

In the circuit you have here, as you mentioned, the opamp is to limit the current. No current goes to Opamp inputs. This means all the current goes through the 280R resistor and it is doing the current measurement!

For your circuit, you have two options to measure the current:

Option 1: You are using a UNO board. It has a built-in ADC. You can measure the voltage over the 280R resistor to find out the current. Simply divide the voltage by the resistor to convert it to current (mV/R=mA).

Option 2: If, for any reason, you want to use the INA219, you can divide the 280R into two resistors. For example, let's use a 270R and a 10R in series. Then, by 21mA going through it, the voltage over the 10R would be 210mV. You can connect the Vin+ and Vin- to measure the voltage over this resistor. If you set the INA219 gain to "/8", then it can measure up to 320mV.

The INA219 indicates that the inputs can connect to GND (even less than zero). This means this circuit should work. However, if there was a bias in the measurement, you can swap the place of the 10R with 270R to make sure the input voltages of the INA219 are well far from the GND. But the principal is the same, and you need to measure the voltage of the 10R resistor.

The short answer is yes. You need a shunt resistor with the correct resistor to be able to measure the current.

The INA219 and similar chips are designed to convert a very low voltage over the shunt resistor to a more readable value. The idea is to use a very small shunt resistor to reduce the amount of power dissipation.

In the circuit you have here, as you mentioned, the opamp is to limit the current. No current goes to Opamp inputs. This means all the current goes through the 280R resistor and it is doing the current measurement!

For your circuit, you have two options to measure the current:

Option 1: You are using a UNO board. It has a built-in ADC. You can measure the voltage over the 280R resistor to find out the current. Simply divide the voltage by the resistor to convert it to current (mV/R=mA).

Option 2: If, for any reason, you want to use the INA219, you can divide the 280R into two resistors. For example, let's use a 270R and a 10R in series. Then, by 21mA going through it, the voltage over the 10R would be 210mV. You can connect the Vin+ and Vin- to measure the voltage over this resistor. If you set the INA219 gain to "/8", then it can measure up to 320mV.

The INA219 indicates that the inputs can connect to GND (even less than zero). This means this circuit should work. However, if there was a bias in the measurement, you can swap the place of the 10R with 270R to make sure the input voltages of the INA219 are well far from the GND. But the principal is the same, and you need to measure the voltage of the 10R resistor.

The short answer is yes; you need a shunt resistor with the correct resistor to be able to measure the current.

The INA219 and similar chips are designed to convert a very low voltage over the shunt resistor to a more readable value. The idea is to use a very small shunt resistor to reduce the amount of power dissipation.

In the circuit you have here, as you mentioned, the opamp is to limit the current. No current goes to Opamp inputs. This means all the current goes through the 280R resistor and it is doing the current measurement!

For your circuit, you have two options to measure the current:

Option 1: You are using a UNO board. It has a built-in ADC. You can measure the voltage over the 280R resistor to find out the current. Simply divide the voltage by the resistor to convert it to current (mV/R=mA).

Option 2: If, for any reason, you want to use the INA219, you can divide the 280R into two resistors. For example, let's use a 270R and a 10R in series. Then, by 21mA going through it, the voltage over the 10R would be 210mV. You can connect the Vin+ and Vin- to measure the voltage over this resistor. If you set the INA219 gain to "/8", then it can measure up to 320mV.

The INA219 indicates that the inputs can connect to GND (even less than zero). This means this circuit should work. However, if there was a bias in the measurement, you can swap the place of the 10R with 270R to make sure the input voltages of the INA219 are well far from the GND. But the principal is the same, and you need to measure the voltage of the 10R resistor.

The short answer is yes. You need a shunt resistor with the correct resistor to be able to measure the current.

The INA219 and similar chips are designed to convert a very low voltage over the shunt resistor to a more readable value. The idea is to use a very small shunt resistor to reduce the amount of power dissipation.

In the circuit you have here, as you mentioned, the opamp is to limit the current. No current goes to Opamp inputs. This means all the current goes through the 280R resistor and it is doing the current measurement!

For your circuit, you have two options to measure the current:

Option 1: You are using a UNO board. It has a built-in ADC. You can measure the voltage over the 280R resistor to find out the current. Simply divide the voltage by the resistor to convert it to current (mV/R=mA).

Option 2: If, for any reason, you want to use the INA219, you can divide the 280R into two resistors. For example, let's use a 270R and a 10R in series. Then, by 21mA going through it, the voltage over the 10R would be 210mV. You can connect the Vin+ and Vin- to measure the voltage over this resistor. If you set the INA219 gain to "/8", then it can measure up to 320mV.

The INA219 indicates that the inputs can connect to GND (even less than zero). This means this circuit should work. However, if there was a bias in the measurement, you can swap the place of the 10R with 270R to make sure the input voltages of the INA219 are well far from the GND. But the principal is the same, and you need to measure the voltage of the 10R resistor.

The short answer is yes; you need a shunt resistor with the correct resistor to be able to measure the current.

The INA219 and similar chips are designed to convert a very low voltage over the shunt resistor to a more readable value. The idea is to use a very small shunt resistor to reduce the amount of power dissipation.

In the circuit you have here, as you mentioned, the opamp is to limit the current. No current goes to Opamp inputs. This means all the current goes through the 280R resistor and it is doing the current measurement!

For your circuit, you have two options to measure the current:

Option 1: You are using a UNO board. It has a built-in ADC. You can measure the voltage over the 280R resistor to find out the current. Simply divide the voltage by the resistor to convert it to current (mV/R=mA).

Option 2: If, for any reason, you want to use the INA219, you can divide the 280R into two resistors. For example, let's use a 270R and a 10R in series. Then, by 21mA going through it, the voltage over the 10R would be 210mV. You can connect the Vin+ and Vin- to measure the voltage over this resistor. If you set the INA219 gain to "/8", then it can measure up to 320mV.

The short answer is yes; you need a shunt resistor with the correct resistor to be able to measure the current.

The INA219 and similar chips are designed to convert a very low voltage over the shunt resistor to a more readable value. The idea is to use a very small shunt resistor to reduce the amount of power dissipation.

In the circuit you have here, as you mentioned, the opamp is to limit the current. No current goes to Opamp inputs. This means all the current goes through the 280R resistor and it is doing the current measurement!

For your circuit, you have two options to measure the current:

Option 1: You are using a UNO board. It has a built-in ADC. You can measure the voltage over the 280R resistor to find out the current. Simply divide the voltage by the resistor to convert it to current (mV/R=mA).

Option 2: If, for any reason, you want to use the INA219, you can divide the 280R into two resistors. For example, let's use a 270R and a 10R in series. Then, by 21mA going through it, the voltage over the 10R would be 210mV. You can connect the Vin+ and Vin- to measure the voltage over this resistor. If you set the INA219 gain to "/8", then it can measure up to 320mV.

The INA219 indicates that the inputs can connect to GND (even less than zero). This means this circuit should work. However, if there was a bias in the measurement, you can swap the place of the 10R with 270R to make sure the input voltages of the INA219 are well far from the GND. But the principal is the same, and you need to measure the voltage of the 10R resistor.