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Measure small currents without adding resistive insertion loss

Summary of Measure small currents without adding resistive insertion loss


### Summary This article describes a circuit designed to measure small currents without introducing resistive insertion loss. Unlike traditional methods using probing resistors or single op-amp converters, this design utilizes two operational amplifiers (IC1 and IC2). The circuit sinks the incoming current while simultaneously sourcing an equal outgoing current back to the test circuit. By maintaining equal voltages at the op-amp inputs through matched resistors, the system ensures virtually zero resistance for the circuit under test, allowing for accurate measurement of low-level currents.

Parts used in the Current Measurement Circuit:

  • Probing resistor (RP)
  • Differential amplifier IC1
  • Operational amplifier (current-to-voltage converter configuration)
  • Feedback resistor
  • Input current source (I)
  • Resistor R1
  • Op amp IC2
  • Resistor R3
  • Resistor R4
  • Resistor R2
In most cases, you measure current by converting it into a proportional voltage and then measuring the voltage. Figure 1 shows two typical methods of making the conversion. In one method, you insert a probing resistor, RP, in series with the current path and use differential amplifier IC1 to measure the resulting voltage drop (Figure 1a). A second method is a widely known operational amplifier current-to-voltage converter in which inverted IC1’s output sinks the incoming current through the feedback resistor (Figure 1b).

Measure small currents without adding resistive insertion loss
In the first method, the same current that flows into one node flows from the second node, but a significant voltage drop occurs across probing resistor RP. In the second method, the voltage drop is on the order of tens of microvolts to millivolts, depending on IC1’s quality, but the measured current flows only into the sensing node with no return to the circuit. You can measure only currents flowing to ground.
The circuit in Figure 2 operates in a somewhat similar manner to the one in Figure 1b in that an op amp’s output sinks the incoming measured current. However, the other op amp’s output sources an equal outgoing current back to the circuit under test.

In Figure 2, input current I flows through R1 into the output of IC2, which reduces its voltage by the amount of IR1 relative to the input node. That voltage equals the voltage mean of the op amp’s outputs, which R3 and R4 set at the op amp’s inverting inputs. Consequently, the output of IC1 must rise to a voltage of IR2 relative to the inverting inputs and the equal-voltage noninverting input node of IC2. IC1 sources this current, which returns through R2 to the circuit under test. R1=R2, so the output current is the same as the input current. Because the op amp’s outputs maintain their inputs at equal voltages, the circuit under test has virtually no resistance.

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Quick Solutions to Questions related to Current Measurement Circuit:

  • How is current typically measured according to the article?
    Current is measured by converting it into a proportional voltage and then measuring that voltage.
  • What happens in the first method involving a probing resistor?
    A significant voltage drop occurs across the probing resistor RP inserted in series with the current path.
  • Does the second method allow current to return to the circuit?
    No, the measured current flows only into the sensing node with no return to the circuit.
  • Can the standard second method measure currents flowing away from ground?
    No, you can measure only currents flowing to ground in that specific method.
  • How does the circuit in Figure 2 handle the incoming current?
    An op amp's output sinks the incoming measured current while another op amp sources an equal outgoing current back to the circuit.
  • What condition must be met between R1 and R2 for proper operation?
    R1 must equal R2 so that the output current is the same as the input current.
  • What is the effective resistance seen by the circuit under test in Figure 2?
    The circuit under test has virtually no resistance because the op amp outputs maintain their inputs at equal voltages.
  • What is the main advantage of the circuit described in Figure 2?
    It allows measuring small currents without adding resistive insertion loss.

About The Author

Ibrar Ayyub

I am an experienced technical writer holding a Master's degree in computer science from BZU Multan, Pakistan University. With a background spanning various industries, particularly in home automation and engineering, I have honed my skills in crafting clear and concise content. Proficient in leveraging infographics and diagrams, I strive to simplify complex concepts for readers. My strength lies in thorough research and presenting information in a structured and logical format.

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