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INA110: INA110

Part Number: INA110
Other Parts Discussed in Thread: TIDA-00916, INA597,

Tool/software:

Hi everyone

I am working on an ESC design project for drones, using TI's reference design TIDA-00916 as a guide. I have a question regarding the measurement circuit design.I’ve noticed that when implementing sensorless FOC control for BLDC motors, there are typically two types of current sensing amplifier circuits: one that uses a 1.65V reference voltage, and another that doesn't.

Amplifier circuit with a 1.65V reference voltage.

Amplifier circuit without a 1.65V reference voltage.

I’d like to ask about the differences between these two configurations when used in BLDC motor control. For my project, which involves designing an ESC for drones, which configuration would be more suitable?

  • Hello, 

    These circuits utilize an amplifier in the difference configuration, which amplifies the difference of the two inputs and shifts this amount based on Vref.

    The equation for this would be: Vout = G * (Vin+  -  Vin-) + Vref

    Where G is based on the four resistor's values, and any difference between the four resistor values can introduce error in the gain value.

    Conveniently, we offer integrated solutions of these circuits in the form of our Difference Amplifier products. These chips integrate the below circuit with 'matched' resistors made to achieve its specified gain across temperature variations. 

    The difference between applying a mid-supply Vref (1.65 of 3.3) is whether you expect to sense a bidirectional current or a unidirectional current. 

    When Vref = 0 with supply voltages of 0V and 3.3V, The difference of (Vin+ -  Vin-) can only increase in one direction because the device cannot output a voltage below 0. As a result, you have 0 to 3.3V of unidirectional current sense output swing.

    When Vref = 1.65 with the same supplies, the difference of (Vin+ -  Vin-) will now be centered at 1.65V and swing above or below 1.65 (up to 0V and 3.3V) depending on which Vin is larger. In this case, the current can be bidirectional because the voltage difference can swing both directions. 

    To help you with selecting a device that fits your design, I'd like to know more about your system.
    What amount of gain does your circuit require? What are the specifications of the ADC this is feeding into? What supply rails are available?

    Best,

    Kevin

  • I think your answer is not convincing because I have used a configuration that doesn't require a 1.65VDC Vref to control a BLDC motor and was still able to measure bidirectional current. As shown in the image I sent below.

    Measurement circuits from STMicroelectronics often use this configuration to measure current in BLDC motor control applications.

  • Hello, sorry for the confusion.

    The circuit you have sent does not require a 1.65VDC Vref for measuring bidirectional current because the resistor values divide the 3.3V signal to offset the input signal Vshunt_1 to about mid-supply (1.416V). Notice in your original post the difference between the two circuit's resistor values. The first circuit uses two 1k Ohm resistors and two 33k Ohm resistors while the second circuit uses four different resistor values. See the below equation for my analysis on the circuit without a 1.65VDC Vref:


    R1 = 3.4k, R2 = 13k, R3 = 787, R4 = 8060, Ref = 3.3V, V- = 0 (GND), V+ = Vshunt

    Vout = ( - (V-) * R2/R1 ) + (Ref + ( (V+) - Ref) * R4/(R3+R4) ) *  (R1+R2) / R1

    Vout = ( - 0 * 13k / 3.4k) + ( 3.3 + (Vshunt - 3.3) * (8060 / (8060+787)) ) * (3.4k+13k) /3.4k

    Vout = 0 + (3.3 + (Vshunt - 3.3) * 0.91) * 4.82

    Vout = 15.92 + (Vshunt - 3.3) * 4.39

    Vout = 15.92 + Vshunt * 4.39 - 14.50

    Vout = 1.416 + Vshunt * 4.39

    As a result of the resistor values chosen, this transfer function describes the circuit as having a gain of 4.39V/V and a Vref of 1.416V. Additionally, I would need to know the input common mode voltage of your system to verify my analysis. The tradeoff for implementing a discrete solution like this comes in space and in gain error drift.

    Because both Vref and the gain will depend on these resistors, any changes in resistance over temperature will affect these values. Alternatively, our integrated solutions offer internally connected matched resistors that display excellent performance over temperature compared to discrete resistors. Providing a separate 1.65VDC Vref ensures that Vref will not change in response to resistor variations.

    Please let me know if this clarifies my previous post or if you have any other questions!

    Best,

    Kevin

  • Thank you for your detailed response.

    Notice in your original post the difference between the two circuit's resistor values. The first circuit uses two 1k Ohm resistors and two 33k Ohm resistors while the second circuit uses four different resistor values. See the below equation for my analysis on the circuit without a 1.65VDC Vref:

    That's correct — I am currently evaluating two commonly used current sensing amplifier configurations in motor control systems. The two schematics I provided are illustrative examples and do not represent finalized component values. My objective is to clearly understand the technical differences in operating principles and characteristics between these two amplifier topologies. For my high-power ESC (Electronic Speed Controller) design, where the phase current can reach up to 80A RMS, I would like to assess which configuration is more suitable in terms of current measurement accuracy.

  • Hello, thank you for the clarification on your use-case.

    Topologically, the two circuits are the same. The output voltage transfer function changes based on your resistor values and the reference voltage supplied. It's common for R1 = R3 and R2 = R4 (like the two 1k Ohm resistors and two 33k Ohm resistors) because the matching values simplify the equation to:

    Vout = R2/R1 * (V+ - V-) + Vref

    But this comes at the cost of not being able to provide a mid-supply reference from a 3.3V supply. For more instructional content about the technical principles and characteristics of these topologies, I strongly suggest our Precision Lab Series on Instrumentation Amplifiers here: https://www.ti.com/video/series/precision-labs/ti-precision-labs-instrumentation-amplifiers.html

    Errors in your current measurements will come from the op-amp and from the tolerance and drift of the resistors. Using a difference amplifier minimizes the errors associated with differences between resistors because the resistors are integrated into the device.

    With the goal of sensing up to 80A RMS, you will need to first select an appropriate shunt resistance, which will give you an input voltage range through V = IR. For larger resistor values, a larger input range will be available to the op-amp which provides a better signal-to-noise ratio, but there will be a larger voltage change across the connected load and more power dissipation. For smaller resistor values, there is less power dissipation and less thermal shift, but your signal range is smaller and more prone to noise. This voltage across the shunt resistor will be fed into the difference amplifier, where the signal will be amplified by some amount of gain. Depending on this gain, the output signal may or may not be large enough to cover your ADC's full scale input range, so an additional amplifier may be included before feeding the signal into the ADC.

    Based on your expected input voltage range and input voltage common mode, you can select an amplifier that fits your gain requirements, and the measurement accuracy requirements would presumably be set by the ADC receiving this data. To better assist you in device selection, I'll need more information about your system, but otherwise I can close the thread if you have no more questions. 

    Best,

    Kevin

  • Hi Kevin 

    Thank you for providing me with such a detailed response. I would still like to ask you more questions because of your understanding of amplifier circuits.

    I have tested the circuit without using the 1.65VDC reference in our ESC project design. However, I'm facing an issue. It may be difficult for you to solve, but with your experience, could you please give me some advice?

    ESC for drones with the following specifications:

    Input voltage: 4S-12S (LiPo) 

    RMS current: 60A

    Peak current: 80A

    Rshunt=0.5mOhm

    OPAM: TLV9062IDGKT of TI

    We are currently encountering an issue where the three-phase current becomes increasingly unbalanced as the load varies (current amplitude values are not equal). The imbalance grows more severe at higher current levels. We've observed that the current sensing offset drifts with load changes, which makes it challenging to perform accurate offset compensation through software.

    I have implemented several approaches, but the issue has not been fully resolved.

    1. Increased the gain by adjusting the resistor values

    1. Increased resistor accuracy from 0.5% to 0.1%3.
    2. Increased the input current-limiting resistor

    5.Modified the filter capacitor values

      

    My planned technical approach to resolving the issue.

    My upcoming solution approach is to change the amplifier circuit to use an op-amp with a 1.65VDC reference voltage.

    Do you have any advice for me regarding this issue?

  • Hello, thank you for the additional information about your circuit and for telling me about the solutions you've already tried.

    The variation between the three current amplitudes is possibly caused by the differences in gain between current sense circuits A, B, and C, also known as gain error. Additionally, your gain of about 16 will amplify any differences between the resistors used for circuits A, B, and C so an error of 0.1% is multiplied by your gain. For more precise performance, an integrated solution would better fit your application. 

    If you want to reduce the difference in amplitude, I strongly suggest using a difference amplifier such as INA597:

    https://www.ti.com/product/INA597

    Additionally, I would like some clarifications on your previous post:

     

    We are currently encountering an issue where the three-phase current becomes increasingly unbalanced as the load varies (current amplitude values are not equal).

    Where in the circuit is this measurement being taken? You mention "current sensing offset drift" but the provided oscilloscope screenshot cannot be the amplifier output because the op-amp output cannot swing below the 0V negative power supply. Does this difference appear at the op-amp output?

    Additionally, can you provide more information about your minimum current sensing requirement? What's the smallest difference in current you are trying to measure?

    I greatly appreciate the clarification so I can help you resolve this issue!

    Best,

    Kevin

  • Where in the circuit is this measurement being taken? You mention "current sensing offset drift" but the provided oscilloscope screenshot cannot be the amplifier output because the op-amp output cannot swing below the 0V negative power supply. Does this difference appear at the op-amp output?

    This result is not from the op-amp output — I’m using a three-phase current probe to measure the motor current. The probe is isolated, so it can display both positive and negative values.

    Additionally, can you provide more information about your minimum current sensing requirement? What's the smallest difference in current you are trying to measure?

    I’m measuring current for controlling a BLDC motor, where the load ranges from no load to full load. Because of this wide variation, I’m not sure how to provide an exact answer to your question at the moment.

  • Hello, thank you for answering my questions.

    Have you tried using a higher precision device? If you would like to sample a device with less gain error, please let me know and we will reach out on how to try a device with less gain error. The connections are very similar to your existing circuit, with the resistors connected internally instead. See below:

    Gain error is described as the difference between the observed gain and the expected gain. Because you are expecting equal gain on each current phase, the difference in amplitudes suggest that the lines are experiencing different gains. 

    Does the unequal current amplitude appear at the output of the op-amp? I can't identify the cause of your current imbalance without seeing the other components of the system. If you would like me to look at your schematic, you can post it here or you can accept my friend request and privately message me your design. 

    What circuit is using the INA110? The post title is labelled INA110 but this device has not been mentioned anywhere yet.

    How much variation is in your load? You described that the current measure ranges from no load to full load, but you have no numerical description tied to this measurement. I would greatly appreciate clarification about this.

    Best,

    Kevin

  • Hello,

    Are there any updates on this problem encountered in your circuit? If there are no more questions, I will close the thread.

    Best,

    Kevin

  • Yes, you can close this issue.