I am designing a transimpedance amplifier using LTC6268-10 from Analog Devices. The photodiode that I use is S9055-01 from Hamamatsu and the laser source that I use has a power of 10mW with the wavelength of 660nm. The laser is a pulsed source with frequency of 50KHz and the pulse width is 1ns. By using the responsivity value(0.35A/W) of the photodiode and the laser power value(10mW), I have calculated the output current of the photodiode to be 3.5mA using below formula

photodiode current = responsivity X laser power

The output voltage from the amplifier should be 1.3V. So I have used these current and voltage values to determine the feedback resistor and capacitor values using the formulae below

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The frequency value that I have used in the formula to find the capacitance value is 1GHz since the pulse width of the laser source is 1ns. The input source capacitance i.e. the junction capacitance of the photodiode is 0.5pF. The common mode capacitance and differential mode capacitance of LTC6268-10 are 0.45pF and 0.1pF respectively. So the Ci value is 1.05pF(0.5 + 0.45 + 0.1). The resistance and capacitance values that I've got from the calculation are 371Ω and 0.428pF respectively. Also the GBW value that I've got from the formula is 3.45Ghz which is below the maximum GBW of LTC6268-10. Are these values correct? Or should I modify the calculation to get the correct values? I have attached the datasheets of LTC6268-10 and Hamamatsu S9055-01 below

LTC6268-10 datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/626810f.pdf

Hamamatsu S9055-01: https://www.hamamatsu.com/content/dam/hamamatsu-photonics/sites/documents/99_SALES_LIBRARY/ssd/s9055_series_kpin1065e.pdf

Thanks in advance!


1 Answer 1


First, your laser pulse will not be a rectangular pulse. It will probably be a Gaussian with a nominal 1 nsec FWHM spec. So I hope you're not counting on a 1 GHz ADC to collect your data. In the worst case two successive samples will bracket the peak.

The GBW of the op amp is adequate, but you've overlooked the issue of slew rate. At a slew rate of 1500 V/usec, your output will take just about 1 nsec simply to to go from 0 to 1.3 volts, and your falling edge will be even slower. Since your waveform is not a rectangular pulse, and your TIA gain is much less than that used for the slew rate specs, your final output pulse will be at least a factor of 2 less than you think.

Your responsivity figure is correct as far as it goes. However, you need to look closely at the data sheet. Look at the photograph. See that tiny little speck in the middle of the package? That's the photodiode and you you need to focus ALL your laser beam on that spot. The data sheet indicates a PD area of 0.1 to 0.2 mm (and I assume they intended that to be mm squared). That says your laser beam will need to be focussed down to a spot with a diameter of about 300 microns. Have you taken that into account?

Assuming you have concentrated your light properly, you are looking at considerable power (10 mW / 3.5 mA). Under these circumstances, you should not be using a TIA at all. Getting accurate measurements at Ghz frequencies is not for the faint of heart - or the inexperienced. For instance, your capacitance values assume the PD connects to the amp with infinitesimally short PD leads connected directly to the pins of the op amp. I assume you'll be mounting these on a pcb, and ANY trace length will have an effect.

Your feedback resistor should be about 370 ohms, which ought to be an indicator that a TIA isn't called for - TIAs are useful when you need both high gain and high bandwidth, and your gain is very low.

You then do your capacitance calculations correctly, as far as I can see, but you've missed an important point about op amps. Op amps work by using negative feedback, and in the process the effective gain of the OA is reduced. The more precision you want, the more gain has to be "wasted". You are proposing to use a 3.5 Ghz GBW op amp at 1 GHz.

At these frequencies and power levels, you are much better off doing something like


simulate this circuit – Schematic created using CircuitLab

You won't get a perfect result (not enough op amp gain), but you'll be able to separate the PD from the op amp and connect with standard 50-ohm coax.

The small voltage swing on the PD (about 0.2 V) means that the bias voltage on the PD will not vary significantly with signal level.


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