LTC6268-10 1khz bandwith application

Hi Cesur:  If your pulse is 10ns, then you probably want 100MHz bandwidth as a minimum, to capture that pulse.  This is true even if the repetition rate of the pulse is only 1kHz.     

Hi,

Thank you for your answer.  Yes, it is true, I want 100 Mhz bandwidth to capture 10 ns pulse. But repetition is 1khz. Does the noise be too much to work with? The online simulation program offers 24.9 k feedback resistance and a 100 fF feedback capacitor to get 96 Mhz bandwidth. I guess the minimum output of the trans-impedance amplifier will be 10 mV when the laser light comes from 10 meters. Does the noise of the opamp can prevent me to read that signal because the simulator says fast opamp is not proper for low-frequency application.

Thank you.

Hi Cesur:  If you let me know the PD capacitance, then I will be able to calculate the LTC6268 output noise in a bandwidth of 100MHz with 24.9k gain.

Thank you, 2.2 pF at 20V reverse bias.

Regards

Hi cesur and Glen,

I'm thinking you can use LTspice to simulate your circuit.

LTspice shows about 100MHz BW with the conditions you've specified. The RMS noise over the entire bandwidth simulates to around 2.9mV_RMS, as shown below.

Please take a look.

Simulation file:

7838.LTC6268-10 TIA EZ 4_13_20.asc

Regards,

Hooman

Hi,
I have a question about calculating the input current value from the photodiode to LTC6268-10. I have 2ns pulse laser power of 10mW and wavelength of 630nm, photodiode of sensitivity of 0.24A/W and an active area of φ0.1mm. In the photodiode circuit wizard, the formula is shown in the figure. How to calculate the light intensity valued based on this data?

Hi Arunthathi,

Sorry about the late reply!

The Analog photodiode wizard tool computes the peak diode current, in the example you've attached, simply as the following assuming 100mW/m^2 Peak light intensity:

https://tools.analog.com/en/photodiode/

• Peak Light Intensity * Sensitivity * Active Area  = 3.6uA peak current
• 100 mW/m^2 * 360 mA/W * 100 mm^2 * (1/1000^2) m^2/mm^2 = 3.6uA peak current

In your particular 630nm laser example, I don't think the 2ns pulse width matters. The computation will yield your photodiode current when the pulse is active (2ns duration).

The tricky bit is figuring out how much of your 10mW laser gets to your photodiode.

I found this calculator program which computes the incident power (mW/m^2) depending on divergence angle and the diameter at aperture:

https://www.laserworld.com/en/laserworld-toolbox/divergence-calculator.html#divergence

If I use your laser (10mW) with a 30 deg. (523mrad) divergence and an aperture diameter of 3mm (made up numbers), I get the following:

Note: There is something wrong with the program in that if you enter 0.1m distance, the report at the bottom just shows "0m" instead. That's why I used 1m.

The resulting power W/m^2 is very small because I had to use 1m distance with a 10mW laser. But you can play with the program and use the numbers in your application.

Regards,

Hooman