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Current carying capablity of ADC pins for TMS320F28027

Ashutosh Pailwan1
Expert 2480 points

Other Parts Discussed in Thread: TMS320F28027

I am using TMS320F28027 .

1. I am using current transformer to sense AC current 

2. The Current Transformer is of ratio 5 Amp / 20 milliamp.

3. Then I have use current to voltage Active amplifier 

4. In this I am biasing Non inverting terminal with DC voltage by 1.65 volt.

5. Now as in electrical specification of GPIO/AIO pins of TMS320F28027 is said 

a) High level output source current -4 mAmp

b) Low level output sink current 4 mAmp

6. In view of it if my secondary of current transformer ( current sourcee) is 20 mAmp max ( Iin ), wont it damage the GPIO/AIO pins used for ADC conversion. Eventhough, the output voltage of active amplifier is is within limits of 0 to 3.3 Volt i.e Vout.

As I have aprehension that as ADC pin of TMS320F28027 is connected to Vout of amplifer. however, current from input source is 20 mAmp it will enter into ADC pin too and damage the TMS320F28027. Whethre it will happen like this????

  • 0
    Guru 192875 points
    Hi,

    20mA is way to high 4mA is all the ADC pin of controller can Sink In. Yes you'll have to convert current into voltage for example:
    0-50A to 0-3.3V or -50A to 50A to 0 to 3.3V.

    Regards,
    Gautam
  • 0 in reply to Gautam Iyer
    Expert 2480 points

    Gautam I mean to say.

    I am using current to voltage converter using OPAMP.

    But as Current trnasformer, which is of ratio 5 Amp/ 20 millAMP act as current source. This current source is applied acorss current to voltage converter 0 to 3.3 volt. Then output of current to voltage converter is connected to ADC pin.

    Now as Current trnasformer is current source the current will be same i.e 20 mAmp it will flow through OPAMP negative feedback to ADC pin.

    AM I correct???

    If yes can it damage the controller

  • 0 in reply to Ashutosh Pailwan1
    Guru 192875 points
    Generally you've a Series resistor from the output of Opamp to ADC input.
  • 0 in reply to Gautam Iyer
    Expert 2480 points
    You are confusing me Or I am able to pose my question in proper way.

    As I understand from google . The ADC pins are high impedance pins with leakge current of 5uAmp and input capcitance of 10pF. Thus, I interpret as, the current source after conversion to voltage by using Active current to voltage converter won't push more than 5uAMp even if current source is 20 mAmp. It will sense the voltage output of current to voltage converter with mere input of 5uAMp.
    I am connecting 82.5 ohms in feedeback circut of oamp with nonverting terminal connected to DC bais of 1.65 volt.
    AM I right.

    Then why resitor is required in OPAMP.
  • 0 in reply to Ashutosh Pailwan1
    TI__Guru 60865 points

    Hi Ashutosh,

    In the circuit above, the input current entering the node next to the negative node of the op-amp is a "virtual ground".  The potential should be close to 0V due to the negative feedback, but no current goes into the negative node, so the current flowing back through the feedback resistor must exactly equal the input current to satisfy Kirchhoff's current law.  This gives a positive voltage at the output of the op-amp (for positive input current) and this voltage doesn't depend on the load placed on the op-amp output by the ADC.  Since the ADC input is high impedance, minimal current flows into the ADC input. 


    As long as the voltage stays within the ADC input range, you shouldn't have any issues.  You should be concerned, however, that if the input current is higher than expected, or negative, the output voltage of the op-amp could exceed the ADC input range.  In this case, the ADC input diodes will turn on, and the ADC input will no longer be high impedance.  Current will begin flowing into the ADC input in this case, and that current could be much higher than whatever is flowing into the op-amp input.  For example, if the feedback resistor is 150 ohms, and the input current is -4mA, then the op-amp output voltage is -0.6V, which is outside of the ADC absolute input voltage range.  If the ADC input resistance of the diodes after they have turned on in this case is 15 ohms (just an arbitrary low impedance number that I picked, it probably isn't this bad) then the ADC input will see -40mA! Basically the amount of current driven into the ADC will depend only on whatever voltage is seen at the ADC input is as well as whatever current the op-amp can drive (not whatever current is being driven into the op-amp input - this current is completely decoupled). 

    If you know the input current will be out of range, but not by much, then you can reduce the feedback resistance and trade a little bit of ADC range to ensure the signal will always stay within the safe ADC input range.  If it could go out of range, but only for a very brief time, then you can add some series resistance at the output of the op-amp to the ADC input to limit the amount of current that could flow into the ADC (but don't make this too high, or the ADC input won't settle in time).