2
\$\begingroup\$

My circuit is given below.

enter image description here

I used this equation for calculating the bandwidth of the TIA.

enter image description here

My RF is 200 kΩ and Cd is 1.3 pF (diode) + 9 pF (opamp input capacitance).

After calculation I am getting a value of 11 MHz and simulated value is 800Khz;.

May I know why such a difference?

EDIT:1

As Andy and Vincent suggested I removed the feedback capacitor and re run my simulation.Please see the response below.May I know is this expected.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Why is wired TLV7031 as a non-inverter with "positive" feedback ? \$\endgroup\$ Commented Jun 25 at 9:03
  • \$\begingroup\$ TLV7031 is a comparator with hysteresis. \$\endgroup\$ Commented Jun 25 at 9:06
  • \$\begingroup\$ ok ... I thought it was an "error" ... \$\endgroup\$ Commented Jun 25 at 9:11
  • 1
    \$\begingroup\$ @Confused Bonnie Baker and Jim Todsen each wrote many good application notes on transimpedance amplifiers and integrators. I worked with both. Jim was always helpful and running out in front to help. Bonnie was "very dry" by comparison. But very competent. I enjoyed both as helpful people while at Burr Brown. Read their application notes. Their writing is gnood and will help you work out bandwidth and noise issues. I'm too busy to review your writing and refresh my thoughts. So I must leave you there for now. \$\endgroup\$ Commented Jun 25 at 10:35

2 Answers 2

Reset to default
2
\$\begingroup\$

C1 and RF created a low-pass filter.

$$ \frac{1}{2.\pi .200k.1pF}=795\:774Hz$$

Try removing C1 from the simulation.

- Edited following clarification of the OP's question -

Here are the results I obtained in a simulation with C1. The gain is 106.02dB, which corresponds to a current/voltage conversion of 200,000 and a cutoff frequency of 800kHz.

enter image description here

Without C1, the cutoff frequency is 11.32MHz.

enter image description here

Curve GAIN_I_to_V = VOUT/AM1

- Edited following a question from the OP in comments -

If you look at the ROC (Rate Of Closure), the cross between the open-loop gain and the noise gain of your OPA301, you'll see that the presence of Cf reduces the ROC = 20dB = stable system.

enter image description here

Ressources: https://www.youtube.com/watch?v=U6tP78BLr_A https://www.ti.com/video/6216778063001

\$\endgroup\$
5
  • \$\begingroup\$ I re run the simulation after removing feedback capacitor.I updated the question with the new response.could you please check \$\endgroup\$ Commented Jun 25 at 11:14
  • \$\begingroup\$ @Confused, I just edited my reply to answer your question. \$\endgroup\$ Commented Jun 26 at 9:28
  • \$\begingroup\$ Thank you,but without Cf the curve I obtained is entirely different \$\endgroup\$ Commented Jun 26 at 9:49
  • 1
    \$\begingroup\$ Without Cf, the phase margin is too small, resulting in instability (overshoot). With Cf you improve the phase margin, and you simply don't see the overshoot anymore. Run the simulations and look at the phase. \$\endgroup\$ Commented Jun 26 at 11:45
  • 1
    \$\begingroup\$ @Confused, I added a clarification in my reply \$\endgroup\$ Commented Jun 26 at 13:25
2
\$\begingroup\$

After calculation I am getting a value of 11 MHz and simulated value is 800Khz

The formula that you have used is being totally overridden by the feedback capacitor. It with the feedback resistor are the dominant components that define the banwidth: -

$$BW = \dfrac{1}{2\pi R_F C_F}$$

\$\endgroup\$
3
  • \$\begingroup\$ I updated the question can you please check \$\endgroup\$ Commented Jun 25 at 11:15
  • 1
    \$\begingroup\$ @Confused before I reply I'd like to point out that you are evolving the question into a different area now so please be aware that there is only so much help that can be given in comments. At some point in the near future it might be better to ask a new question..... more to come.... \$\endgroup\$ Commented Jun 25 at 12:31
  • \$\begingroup\$ @Confused what you see in your edited image is exactly what I would expect to see. If you did a much wider spectral analysis I would expect the peak that is beginning to form reach a maximum than return back to low levels as frequency rises. This is what happens with TIAs and, it is the reason why we must carefully select the feedback capacitor value in order to minimize the peak. Standard TIA theory basically and it relates to what we call noise-gain. \$\endgroup\$ Commented Jun 25 at 12:34

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.