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OPA818: High speed current source (upto 50Mhz) to voltage translation

Tarun Kumar Kundrapu
Prodigy 60 points
Part Number: OPA818
Other Parts Discussed in Thread: TINA-TI, OPA858

Hi ,

I am new to the transimpedance amplifiers. I have a constant current source(1mA) with speed upto 50Mhz. can i use this OPA818 as voltage translation to about 3V, by forcing 3V at IN+. ( circuit in image)

Some help would be great.

thanks.

B.R,

Tarun

  • 0
    TI__Expert 6050 points

    Hi Tarun,

    You are correct that by setting IN+ to 3V, when the current input into IN- is 0A the output voltage will be 3V. Then when a current is input, the output voltage will change in proportion to the current and the feedback resistor (standard of transimpedance amplifiers [TIA]). I have provided a simulation in TINA-TI showing this behavior, below:

    OPA858_TIA_E2E.TSC

    A few comments about the circuit you provided above:

    • While the circuit you provided above is a proper TIA circuit, the OPA818 is not a good choice if you are trying to center your output voltage at 3V. You will see in the datasheet that the most positive input voltage is the positive supply voltage (Vs+) - 3.2V, meaning the input voltage cannot exceed 1.8V. Therefore, you would not be able to set IN+ at 3V (this can be seen in simulation). Because of this, I chose the OPA858 in my example design as that allows for a higher input voltage range. Here is the snippet from the OPA818 datasheet:

    • The feedback resistor that you chose will cause the output to saturate because the output would exceed the negative output swing limit. The change in output voltage is proportional to the input current and the feedback resistor through Ohm's Law. Therefore, the output voltage would swing 3.3V if a 3.3kohm feedback resistor was used and 1mA current was input (V = IR -->  V=[1mA]*[3.3kohm]=3.3V). In my example, I used 1.5kohm so that the output would not be saturated, but that can be adjusted based on your application needs
    • The feedback capacitance can be calculated based on the feedback resistor, total input capacitance, and closed loop bandwidth needed. We have a TIA calculator (found here) that will compute the values needed for stability.

    In general, we have some really good resources for frequently asked questions and common design considerations for TIAs here. You can also reference Section 10.1.1 of the OPA858 datasheet as it also provides some helpful design information.

    Please let me know if you have any further questions.

    Thanks,

    Nick

  • 0 in reply to Nick Nauman
    Prodigy 60 points

    thanks Nick for the detailed explanation. i tried to implement the same circuit with opa818 and i see a satuarated output. At the moment i am ok even if the IN+<1.8 or at least 1.2V. can u help me this. thanks

  • 0 in reply to Tarun Kumar Kundrapu
    TI__Expert 6050 points

    Hi Tarun,

    There is something majorly important to consider that I did not catch when I looked at your original circuit. The OPA818 needs a minimum power supply voltage of 6V, so you will not be able to use 5V single supply as you have shown in your circuit.

    Also, because you are sourcing current into the op amp, the output voltage will swing in the negative direction from the voltage bias on the IN+ pin. Therefore, you could be seeing saturation as the output voltage swing low is 1.2V about the negative power supply rail, which does not allow for a large output swing if you choose a voltage bias on the IN+ pin that is close to the negative power supply rail. If you use +/-5V power supply rails, there is a much larger output headroom. 

    OPA818_TIA_E2E.TSC

    Thanks,

    Nick

  • 0 in reply to Nick Nauman
    Prodigy 60 points

    thanks, Nick for the detailed overview. I have tried this hardware and it works as expected. Do we have any opamp available (without using a negative supply (-VCC), that can do the voltage translation from current where my low level will be under 1V and high level would be voltage >1 till 3V from  TI? thanks again. and I wish you a nice weekend.

  • 0 in reply to Tarun Kumar Kundrapu
    TI__Expert 6050 points

    Hi Tarun,

    Can you clarify a few points for your application?

    1. What is the input current range you are referring to and what do you consider as low level current and high level current?
    2. Are you wanting IN+ biased at 1V (based on the description you gave)?

    Thanks,

    Nick

  • 0 in reply to Nick Nauman
    Prodigy 60 points

    Hi Nick,

    1) i can control my current source from about 10 u A to 500uA for the high current level, & 0ua for the current low level. 2) I want to feed this current to the opamp to translate into voltage. For the low current level  I  would like to translate to voltage in between "0V till 1.2V"-> which I read as low and for the current high level translated to between 1.2V to 3.3V which I read is as high level. Thanks.

  • 0 in reply to Tarun Kumar Kundrapu
    TI__Expert 6050 points

    Hi Tarun,

    We do not have a one device solution that can handle this. To handle the low current level of 0-10uA, the selected Rf should be 120kohm to get a voltage swing of 1.2V. To handle the high current level of 10uA to 500uA, the selected Rf should be about 4.3kohm to get a voltage swing of 2.1V (like a 1.2V to 3.3V swing). 

    Because of this, you will see that you would need to switch between gains depending on which current level you are trying to translate. And this would require knowing which current level you are reading. Is this a capability of your application?

    Thanks,

    Nick