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Current source with OPA381

Perhirin Steven
Intellectual 380 points
Other Parts Discussed in Thread: OPA381, OPA350, OPA340, OPA380

Hi,

I would like to create a courent source which could provide 25 mA as constant output current. I read the datasheet(OPA381) and it says that the current typ is 10 mA. I couldn't find how much is the ouput current max. I already have the schematics of my circuit that include an Amplifier. I would like to know what Operational amplifier Low power could support 25 mA as output current?

Thank you for taking time.

Steven

  • 0
    Guru 21975 points

    Steven,

    It would help to better understand your requirements:

    Do you want a constant current or is it to be controlled by a varying voltage?

    Sourcing or sinking current?

    What is the required compliance voltage range?

    The OPA381 is a special amplifier intended for transimpedance applications. Was there some special reason for selecting this amplifier?

    With 25mA output current, why is low power important?

    If you can provide your schematic it would be very helpful.

    Regards, Bruce.

  • 0 in reply to Bruce Trump
    Intellectual 380 points

    Hi Bruce,

    I want to have a constant current, I would like 3.3V input voltage and 25 mA output current. And I would like also use 3.3v supply. I need a low power, because it is a embedded system.

    The final circuit will be a Tee bias for a Laser. The Amplifier that I am looking for will provide the constant current.

    I have some OPA381 in stock, that why I mentioned it.

    The schematic is on this link :

    http://forums.futura-sciences.com/attachments/electronique/1761-source-de-courant-sourcecourantpulseesymmetrique2oegif

    In my schematics, I won't use R2 (so not Vref)


    Please let me know if you require any further information

    Steven

  • 0 in reply to Perhirin Steven
    Guru 21975 points

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    Steven,

    Low power and microPower op amps are not generally designed to provide output current of 25mA. There many compromises in designing an op amp for high output current and the natural assumption is that microPower applications would not be driving 25mA.

    At 3.3V operation, driving a laser diode (don’t know the maximum voltage at the diode) little compliance voltage remains for the drive circuitry. The op amp must be capable of swinging quite close to the positive rail at 25mA output. Probably our best op amp for this behavior is the OPA350. It is capable of 25mA output and swinging within 0.5V of the rail, though this capability is not assured by production testing. The OPA350 draws approximately 6mA quiescent current.

    There may be an alternative approach:  It may be possible to provide a source of current through a “helper resistor” connected from your 3.3V supply to the output of the op amp. This resistor is chosen to provide the majority of the load current. The op amp them must only provide the balance of the current. The Howland current source circuit you have shown can source and sink current, so the helper resistor can be chosen to make the nominal current supplied by the op amp near zero. The approach relies on the fact that the load voltage is reasonably constant (including temperature variation). The op amp must be able to supply the remaining current throughout all possible load voltage variations that might occur. Optimizing and choosing the op amp for this type of circuit would require in-depth understanding of the possible load voltage variation.

     Regards, Bruce.

  • 0 in reply to Bruce Trump
    Intellectual 380 points

    Hi Bruce,

    Thank you for all this information

    I wish you a godd week.

    Regards,Steven.

  • 0 in reply to Perhirin Steven
    Guru 21975 points

    Steven,

    I simulated an example of the approach I suggested using the OPA340 op amp. A 2.5V reference voltage is used to create a 25mA output current in the modified Howland current source. A 1.8V voltage is assumed to be the nominal load voltage, shown with current meter in series. R6 is the "helper" resistor which supplies 25mA at 1.8V load voltage. As the load voltage varies (due to temperature or other variations) the current supplied by R6 changes. The op amp supplies or sinks the difference current to maintain a constant 25mA output. The circuit is not tested or optimized but does demonstrate the basic approach.

    If the nominal load voltage is high (becoming closer to 3.3V) this approach becomes less effective. At high load voltages, variation in load voltage causes a larger change in the current supplied by R6. This requires a much larger change in current supplied or sinked by the op amp.

    Regards, Bruce.

  • 0 in reply to Bruce Trump
    Intellectual 380 points

    Hi Bruce,

    Thank you for your schematics. I understand better how use the Helper resistor.

    In my application, V2 is 3.3V and VS is 1.1V. I am using Calc (exel) to get resistor values with AM1 at 25mA. My schematic is simplest :

    It is like yours but without R6 and R4. I set R1 and R3 as high values (20k)(because I want to avoid lost of current to the ground) and R2 (in Ohms) controls the current through V2. If I want to provide 10 mA current with OPA380, I set R2 as (V2-Vs)/R2 = 25mA - 10 mA.

    Then I need to optimize it. I want to have the least lost as possible : (Vs1*AM1) / ((V2*I2)+(V1*I1)) => closest of 1. I feel that I'm getting closer of the solution :).

    Thank you again for your help.

    Regards,

    Steven