• State Resolved
  • Locked Locked
  • Replies 20 replies
  • Answers 4 answers
  • Subscribers 47 subscribers
  • Views 4610 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Current Sensing Amplifier Solution

Rahul
Expert 1005 points

Other Parts Discussed in Thread: INA226, LOG114, OPA192, INA121, INA190, INA191, OPA544, INA592, INA180, INA199

Hi Team,

We are looking for Current sense amplifier for our RF Module Test Jig to measure the sleep current and uplink current[Peak]. Kindly suggest me the suitable workaround solution to build the design.

I Load Min: 1uA [Sleep]

I Load Max: 1A [Peak]

Supply Voltage: 3.3V

or Suggest me the suitable method of implementing the same.

Thanks,

Rahul

  • 0
    TI__Mastermind 32395 points
    Hello Raul,

    Thanks for posting to the E2E-Amplifiers Forum.

    Let me start by saying that you are seeking to measure currents ranging from 1uA to 1A which is a dynamic range of 1 million-to-1. Most amplifiers will swing their output voltage from 50mV to 4.95V assuming a 5-V supply and that 50mV to 4.95V is the output's linear range. This is a dynamic range of 100:1 at best. There are digital current sense amplifiers (like INA226) which can sense unidirectional current with a dynamic range of 8192:1. One of the highest dynamic range amplifiers is LOG114, but I don't think it works for >10mA. You can read more about this device and another potential discrete solution here:
    e2e.ti.com/.../747435 range current sense

    Also, here are two TI Reference Designs For measuring large current ranges, although they won't fulfill your requirement alone.
    http://www.ti.com/tool/TIPD135
    http://www.ti.com/tool/tipd104

    Thus, you most likely will need to split up the current sensing into two circuits (so two amplifiers) or one circuit that changes gain or switches to a different shunt resistor. Or as the previous links show, your could switch in different gain setting resistors for a discrete solution with an op amp like OPA192 or instrumentation amplifier like INA121.

    Here are a list of current sense amplifiers with adjustable gain:
    www.ti.com/.../products.html

    The INA190 is current sense amplifier with very good output dynamic range (~996:1) and can easily sense 1 uA of current with good accuracy over temperature. It also has an enable pin, so it can be put into sleep mode when the RF module is operating normally. See this cookbook circuit:
    www.ti.com/.../sboa336.pdf

    Other options are using a programmable gain or variable gain amplifier (PGA/VGA) here:
    www.ti.com/.../overview.html

    All this being said, you still need to consider what error you can tolerate at both current levels, what is the frequency of the 1A current, and what does your signal chain look like.

    Sincerely,
    Peter Iliya
    Current Sensing Applications
  • 0 in reply to Peter Iliya
    Expert 1005 points
    Hi Peter,
    Very Thanks for your quick response. I would like to go for measuring the 1uA to 104uA using INA190. So below solution would be a perfect choice for first design. I shall implement the design and get back to you at the earliest, so that will have more input to discuss later.
    "The INA190 is current sense amplifier with very good output dynamic range (~996:1) and can easily sense 1 uA of current with good accuracy over temperature. It also has an enable pin, so it can be put into sleep mode when the RF module is operating normally. See this cookbook circuit:
    www.ti.com/.../sboa336.pdf"

    > I will follow the same reference design attached in the link you shared to start with.
    > I will start the design and share you the schematic file for verificaiton
    > Kindly know the capacitor value to be put across the differential pins.

    Thanks,
    Rahul
  • 0 in reply to Rahul
    TI__Mastermind 32395 points
    Rahul,

    I'm glad we could get a solution to solve first half of your system requirements. The INA190 comes with an EVM (evaluation module) you can order it here. This EVM allows for you to solder down a shunt resistor for quick testing.
    www.ti.com/.../ina190evm

    Just about any value of capacitance can be placed across the input differential pins. When the shunt resistance is higher (~ 1kOhm), it best to increase the input capacitor (Cdiff) to 1nF. Higher values could be placed, but this may slow the settling time of your shunt voltage step response. The EVM also has pads to place a Cdiff or other filtering components.

    Also note that if you are just doing unidirectional current sensing, that the INA191 is another option. It is the functionally the same as the INA190, but in a WCSP package and no REF pin.

    Sincerely,
    Peter Iliya
  • 0 in reply to Peter Iliya
    Intellectual 645 points
    Dear Peter,

    We also want to measure the current of a high power opamp driver using OPA544. The resistor shunt is placed at the low side and the current to be measured is around 1mA to 1A. Can we use OPA192 to build a current sensing circuit? We found that OPA192 has very good dc characteristics. Could you provide any example circuit?

    BTW, we also evaluate using INA592 as a low side current sense, can INA592 be used in this application?

    Best

    Benyuan
  • 0 in reply to Benyuan Liu
    TI__Mastermind 32395 points

    Hey Benyuan,

    You might want to create a new thread/post about measuring low-side output current of OPA544 so other groups that cover precision op amps (OPA192) and difference amplifiers (INA592) can help assist you. Including information about the frequency of the sensing current, what your output voltage range should be, and how you want to power the amplifier all need to be included.

    In the meantime I can provide some guidance. The 1mA to 1A will be a slightly difficult dynamic range to achieve with a single amplifier, but is possible. For using an operational amplifier (such as OPA192) for low-side sensing you can refer to the Analog Engineer's Circuit Cookbook: Amplifiers for the standard current sensing techniques.

    You can see all of the potential circuits in the table of contents below.

    Using the INA592 might be difficult since you will probably want to increase the gain from 1/2 to something more manageable.

    You will need to consider what bandwidth you will need from your device. If it is sensing very fast current transients then you may also want a devce with high slew rate and small settling times.

    If the 1mA to 1A currents you are sensing are DC, then you could use the INA226, which is a digital current sensor with I2C output. With a 81.92mΩ shunt resistor, you would be able to sense 1mA to 1A with this single device on a low voltage supply.

    Sincerely,

    Peter Iliya

  • 0 in reply to Peter Iliya
    Intellectual 645 points
    Thank you, I would like to reduce the current range to 100mA to a few amperes, and using OPA192 for current sensing. BTW, the cookbook is great!
  • 0 in reply to Peter Iliya
    Expert 1005 points
    Dear Peter,
    Here i am planning to move for below approaches,
    Point 1: As you mentioned to solve first half of my problem i will go with INA190 sensing amplifier [Status: Board is under design]
    Point 2: What is the next solution for the Higher one, Which means measuring the > 200mA current rating, Kindly tell me the solution
    Point 3: How will i split both/put both the combination in the single circuit board, Will there be any issue for accurate measurement also what is the logical solution to make the circuit.

    Thanks,
    Rahul
  • 0 in reply to Rahul
    TI__Mastermind 32395 points

    Hey Rahul,

    Did the current sensing circuits in the previous links not work for Point 2 (sensing >200mA)? It is very difficult to say what your solution should be without knowing the basic system requirements.

    How much board space is available?
    What error do you need at 200mA?
    Does system need to be low cost?
    What is your power budget?
    What is maximum shunt resistor power dissipation allowed?
    Is high-side or low-side sensing preferred? Is the high-side voltage (VCM) 3.3V? Are there voltage transients that cause VCM > 3.3V? What is the maximum and minimum possible VCM at all times?
    Does point 2’s current sensing amplifier need to be off when system is in sleep mode?
    Are you sensing DC or AC current? Are you trying to measure peak load currents?
    Does current sense amplifier require fast step response times?
    What is the amplifier driving? ADC? Comparator input? How fast is ADC sampling and at what bit resolution?


    Without know these questions I can’t really suggest you a solution or part. Just so you know, you could probably use a second INA190 to measure the 200mA - 5A current. The main restriction to measuring current is either the input offset or the input bias current. When current is in the milli-Amps, then this is much greater than the input bias current of most current sensors, which is in the microamps. The INA180 could also probably work just fine as long as you ensure the power rating of the shunt resistor is high enough for the maximum possible current level.
    --

    For Point 3 once again I don’t know what type of system you are trying to achieve. I will say that placing the shunt resistors needed for “Point 1” and “Point 2” in the same bus line will be problematic. This is because the ~100-Ω to 1-kΩ shunt resistor (Rs1) needed to sense µA of current will be exposed to Amps of current when system is fully functioning. Obviously, this would cause Rs1 to explode.

    Without knowing much about your system you could solve this issue by two FETs around the shunt resistor for Point 2 (let’s call this shunt resistor Rs2 and assume it’s in the milli-Ohms range). When system operating normally the FETs could turn on and make Rs2 the main conduction path for high current. When system goes to sleep, the FETs turn off and make Rs1 the main conduction path for low-current.

    There are many ways to accomplish this, but you will ultimately know best. You could also just place a single FET in parallel with Rs1. For sleep mode, FET would be off and for high-current mode, FET would be on and act as short across Rs1 thus diverting current through FET and not Rs1. In this case, you would want to make sure the drain-to-source leakage current is much smaller than the IB (input bias current) of INA190.
    Hope this helps.
    -Peter

  • 0 in reply to Peter Iliya
    TI__Mastermind 32395 points
    Hey Rahul,

    Just in case you did not know we have a current sensing video training series online here:
    training.ti.com/getting-started-current-sense-amplifiers

    These videos can really help show how the current sense amplifiers operate.

    Best,
    Peter
  • 0 in reply to Peter Iliya
    Expert 1005 points

    Dear Peter,

    Thanks 

  • 0 in reply to Rahul
    Expert 1005 points

    SCHEMATIC1 _ PAGE1.pdfDear Peter,

    Tested my circuit and I am getting the below readings through DMM,

    Shunt Resistor: 0.00Ohm  1%

    Input Voltage: 3.3

    Amplifier Out: 51uV

    This is what i am getting through the designed circuit, PFA.

    Expected Result is below,

    RF Module Sleep Current: 2uA

    Looks like there is a problem with my integration part. Please help to find the actual current which is expected from my Module.(or) Let me know Am i doing any mistake here.

    Thanks,

    Rahul

  • 0 in reply to Rahul
    TI__Mastermind 32395 points

    Hey Rahul,

    I am seeing a couple problems here in the schematic. One, you are probably putting the INA190 into instability by placing too large of a capacitor at the OUT pin. You have the INA190 driving a 0.1uF cap (C3) with no isolation resistance (R2 =0). We spec a maximum capacitive load of 1 nF. I would remove C3 firstly.

    Secondly, if you are trying to measure 2uA with a 2mΩ shunt resistor, then you will not achieve your goal. This results in a 9nV shunt voltage, which the INA190 will never be able to sense given that its input offset voltage is 10uV max. Additionally, 9nV is well underneath the input noise density of 70nV/rtHz. You cannot trust any voltage measurement at the OUT pin which is less than the device's VZL (Zero current output voltage). VZL is specified in the Electrical Characteristics table, section 6.5 in datasheet. For the INA190A1 VZL is 3mV max. So anything less than 3mV is not robust and may not mean anything and this makes the 51uV measurement meaningless.

    You need to increase your shunt resistor to 200 Ω if using the INA190A1 version which has gain of 25V/V. This means at 2uA the input shunt voltage is 400uV which is gained up by 25V/V so INA190A1 OUT pin reads 10mV. You could also choose the INA190A5 (500V/V gain) with a 15-Ω shunt resistor. This means that at 2uA the input shunt voltage is 30uV which is gained up by 500V/V so INA190A5 OUT pin reads 15mV.

    Realistically, the device's input offset (Vos) will add significant error in examples above, especially if you shunt voltage is only 30uV. Most likely if you need more accuracy you need to increase shunt resistance.

    Sincerely,
    Peter

  • 0 in reply to Peter Iliya
    Expert 1005 points
    Dear Peter,
    The suggestion works perfectly. Thanks!.
    Let me tell you the plan, Going for a 2 types of sensing circuit, 1 is Low current measure method and other one is high current [150mA~200mA]
    First level is okay, Now moving on to the next level where I need to measure the current of 150mA ~ 200mA which is a peak current of my RF module. Working out for a the circuit combination to get the proper out from the amplifier INA190A1.
    Please suggest me the appropriate components values to works on the > 150mA value.

    Finally i will add both the circuit in a single board and test both the combination.

    Thanks,
    Rahul
  • 0 in reply to Rahul
    Expert 1005 points
    Dear Peter,
    Tried using the below configuration,

    Shunt Resistor: 1Ohm 1%

    Input Voltage: 3.3

    Amplifier Out: 3.19V

    Calculated the shut voltage: V1 - V2 [0.14641V]

    Current: 146mA

    Please confirm me whether the calculated values is right ?

    Thanks,
    Rahul
  • 0 in reply to Rahul
    TI__Mastermind 32395 points
    Hey Rahul,

    If amplifier output (VOUT) is 3.19V for INA190A1 (25V/V gain), then input shunt voltage (Vshunt) is 3.19V/25 = 127.6mV.

    According to Ohm's Law, this means your sense current is 127.6mV/1Ω = 127.6mA.

    Best,
    Peter
  • 0 in reply to Peter Iliya
    Expert 1005 points
    Dear Peter,
    Can we build a circuit to sense both the upper and lower current using the high gain amplifier, May be 500V [INA190]. need your suggestion to take it further.

    Thanks,
    Rahul
  • 0 in reply to Rahul
    Expert 1005 points
    Any further update
  • 0 in reply to Rahul
    TI__Mastermind 32395 points

    Hey Rahul,

    I am unsure of what you are trying to accomplish. I will assume you are still trying to measure 150mA to 200mA and this requires a very simple circuit when using a current sense amplifier. Previously you had said VS=3.3V and shunt resistor was 1Ω; however, this will not work with 3.3V supply for many amplifiers. In order to scale your shunt resistor for INA190 (or any other amplifier such as INA199 or INA180), you must determine VS and the maximum amount of error allowed at minimum current). I have included links for how to calculate error in a previous thread, but here they are again.

    https://training.ti.com/getting-started-current-sense-amplifiers-session-5-understanding-different-types-error-current-shunt?cu=456802

    Here is an example for INA190. Note that even with the lowest gain option (25V/V) I needed to reduce shunt resistor to 500mΩ so the VOUT signal does not saturate at 200mA load current. Please consider downloading our TINA Simulator along with the current sense amplifier model of your choosing.

     INA190A1_simulation.TSC

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Expert 1005 points

    Dear Peter,

    Thanks for all your comments.

    Let me understand you clearly. I have a solution called Test Jig set up for my RF module.

    During the RFmodule test, I need to measure the current during each functional test.

    One is Peak current measure & Other one is Sleep current measurement.

    I have achieved both using INA190A amplfier circuit constructing individually.

    Actual Aim,

    > I need to club both the solution in to a single test process. Kindly help me. 

  • 0 in reply to Rahul
    TI__Mastermind 32395 points

    Hey Rahul,

    Could you just place both circuits on the same RF module? What would happen if you took both of your circuit's shunt resistors and placed them in series on your on RF module.

    This is the most straight-forward, easy way to accomplish what you are asking for. However, there will most likely be a problem when your module is running at peak current (200mA) and this current passes through your large shunt resistor of 15-Ω. The first problem is that this might blow up the shunt resistor if it not properly rated. So in this case it would need a power rating of at least R*I^2 = 15Ω*(0.2A^2) = 0.6W.

    The second problem is that this shunt resistor would drop your RF voltage by 3 Volts = 15Ω*0.2A and thus downstream circuitry might not see the voltage they need to see.

    This is why I recommended using the highest gain INA190A5 (500V/V) to measure the sleep current so you minimize this shunt resistance as much as possible. However, you can only reduce the shunt resistance so much before you saturate the INA190A5 output pin. Once Vsense = Rshunt*sleep_current becomes to small the INA190A5 output will not be in the linear region and its output will slam into the ground. This is why we provide the Zero-current output voltage spec, so you know how large the VOUT should be when part is ground referenced (VREF=0V).

    You could minimize the shunt resistance even further, if you provide a reference voltage to the INA190A5 REF pin. Doing this offsets the output voltage (VOUT) by the REF voltage. VOUT is always equal to this equation:

    VOUT = Vsense*Gain + VREF

    where Vsense = VIN+ - VIN-

    While this would require an extra voltage source and circuitry, providing a reference voltage allows you measure smaller voltages since you can read the INA190's output voltage differentially. That is to say the ADC will read the VOUT with respect to VREF and not with respect to GND.

    Overall, the input offset (Vos) of INA190 will still limit the accuracy of measuring the sleep current; however, you could program into your system an offset calibration procedure where Vos can be measured by the ADC, recorded into memory, and the subtracted off from all measurements there on out.