• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Answers 1 answer
  • Subscribers 47 subscribers
  • Views 582 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.

OPA928: Appropriate choice of load resistance to estimate junction temperature

Camille Leclerc
Prodigy 120 points
Part Number: OPA928


TLDR: How do I choose an appropriate Rload value for the output of an op amp other than using the value from the datasheet?

Hello,

I am in the process of designing a highly sensitive op-amp system with the OPA928 to amplify a small modulated (1kHz to 100 kHz sampling frequency) photodiode current.

As this is my first time designing an analog circuit onto a PCB, I followed the TI “Precision labs series: Op amps” to try and educate myself on designing/simulating op-amp circuits.

Something I want to double check is the estimated junction temperature (TJ) for my trans impedance amplifier stage to ensure it is within the specified operating temperature of -40°C < TJ < 85°C. (To ensure the input bias current is minimal).

As I expect the largest current produced by my photodiode to be ~12 nA peak-to-peak, my intuition tells me that TJ will be fairly small… still want to double check. My calculations showed that the estimated TJ for my first stage will be ~25.2 °C… which is great!

Using the equations provided in the “power and Temperature” lecture of the aforementioned video series, specifically:

TJ = (PQ + PAC_avg_max)*ϑJA + TA

Where PAC_avg_max is

And RL = RLOAD||(RRf1 + RS) based on my circuit below.

My calculations showed that the estimated TJ for my first stage will be ~25.2 °C… which is great (but also seems too good to be true)! However, I have some question regarding Rload.

Apologies if this is a mundane question, but how do I determine an appropriate value for Rload other than using the value from the OPA928 datasheet? From my search online, Rload is supposed to represent the equivalent resistance of what the output of the op amp will be connect to… is that right? Since I plan to couple the output of this op amp to another “Stage 2” op amp, is Rload the equivalent resistance of the feedback loop of the “Stage 2” op-amp, or its input impedance?

 

Link to video: www.ti.com/.../ti-precision-labs-op-amps.html

  • +1
    TI__Mastermind 25648 points

    Camille,

    1. You are correct.  The load for your first stage will be the input impedance of your second stage.  I don't see a schematic for the second stage, so I cannot say for certain what the input impedance is.  For a standard inverting amplifier, the input impedance is the gain set resistor (Rin = Rg, where G = -Rf / Rg).  For a non-inverting amplifier the input impedance is the op amp input impedance (this is very high for example 100MΩ).
    2. You are correct that the effective load includes the feedback network RL = Rlaod || (Rf1 + Rs).  In your case the the feedback network is quite large, so the load is mainly the second stage input impedance.
    3. Assuming the second stage input impedance is 10kΩ, you can calculate the maximum AC power with the equation you mentioned.  With a 3V supply voltage the max AC power should be: P = 2(3V)^2 / (pi^2 * 10kohm) = 0.182mW.  Pq = (Vcc - Vee) x Iq(max) = (3V - (-3V)) x (400uA)= 2.4mW.  Thus, in this case the normal quiescent current contributes most of the power.  Total AC power including Pq is 2.58mW.  Self heating is 2.48mW x 113.9C/W = 0.3C.  Thus, your assessment for temperature looks correct.
    4. In general, signal amplifiers driving a moderate load will not have a lot of self-heating.  This type of calculation is more useful for higher power type amplifiers driving low impedance loads.  Also, for amplifiers with higher supply voltage you may see some self heating.  In your case the supply voltage is relatively low and the load current is also low.
    5. I normally don't do power calculations for amplifiers with low voltage supplies driving load resistances of 1k or greater.  As you do power calculations, you will develop an intuitive understanding of when to expect self-heating.
    6. Besides self-heating the load resistance can limit the output swing.  In your case a 10kΩ load is fine.  I just mention this as you are looking through the TI precision labs content learning about these different topics.  Normally, when I see a load less that 10k ohms, I check to see if the load will impact the output swing range of the device.  The OPA928 data sheet specifies output swing for a 10k and 1k load (note that swing is better for the 10k load).

    I hope that information above helps you.  I'm glad to see you are reviewing the TI precision labs.

    Best regards, Art

  • 0 in reply to Art Kay
    Prodigy 120 points

    You are a well of knowledge, Art

    Thanks again for your continued support!

    Best,

    Cam

  • 0 in reply to Camille Leclerc
    TI__Mastermind 25648 points

    Happy to help!,  Art