• State Resolved
  • Locked Locked
  • Replies 2 replies
  • Subscribers 50 subscribers
  • Views 553 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

OPA827: The output voltage question for OPA827

Mickey Zhang
Genius 12080 points
Part Number: OPA827

Hi team,

I have a question for 5.1. TI Precision Labs - Op Amps: Bandwidth 1.  In this video, the input signal frequency is 200kHZ, the closed gain

is set to 100V/V. The GBW for OPA827 is 22MHZ. So, 200kHZ*100V/V=20MHZ<22MHZ, So, the 200KHZ input signal can be 

normally amplified 100 times. That is  when the input signal is 200KHZ, the closed gain should be 100V/V, correct?

But in TINA circuit, it shows the closed is 37.75dB when the input signal is 200KHZ. Would you explain this?

  • 0
    TI__Genius 12080 points
    Hi team,

    Would you explain this?
  • 0 in reply to Mickey Zhang
    TI__Expert 6275 points
    The gain bandwidth formula is: GBW = Gain * BW
    So for G=1 on OPA827, GBW = 22MHz . G * BW = 22MHz. 1 * 22MHZ = 22MHz. For a -20dB/decade Aol slope like on OPA827 GBW = constant
    So now for G=100 on OPA827: 22MHZ = 100 * 220kHz
    So closed loop gain of 100 will have a closed loop Vout/Vin -3dB frequency at 220kHz.
    Simulated Measure closed loop AC gain at 100 shows closed loop gain of 37.7dB at 200kHz which makes sense since at 220kHz we would predict 37db (or -3dB down from 40dB)