-2
\$\begingroup\$

here i have designed a TIA circuit with the suitable values for the feedback resistance and feedback capacitance and using the FDS100, to achieve a bandwidth around 43 MHz, the simulation is running good,

schematic

simulate this circuit – Schematic created using CircuitLab

and now the PCB is working, i want to measure the Signal to noise ratio (SNR), also the rise time and the bandwidth when i am using a blue LED, so i want to know how can i do that ? Thank you.

\$\endgroup\$
4
  • \$\begingroup\$ To use a photodiode, you need to apply either 0 bias (for photovoltaic mode) or apply positive bias to the cathode (for photoconductive mode). Your circuit applies negative bias to the cathode which just makes the device act like a normal forward-biased diode. \$\endgroup\$
    – The Photon
    Commented Apr 5, 2017 at 15:51
  • \$\begingroup\$ Second, getting 40 MHz bandwidth from an op-amp TIA requires very careful attention to details including the photodiode selection and board layout to minimize parasitics. FDS100 looks like a very large device for this application, but I haven't checked all the math out on this. \$\endgroup\$
    – The Photon
    Commented Apr 5, 2017 at 15:57
  • \$\begingroup\$ well, i am using a reverse bias voltage on the photo diode, and i know that i could use different photo diode or even different amplifier, but i must work on these two devices, the bandwidth is not important to me, all i want is to see if i will get this bandwidth as in the simulation, and what my SNR will be, so i can decide whether i should use a different photo diode with different Rf,Cf values or not. \$\endgroup\$
    – Hamzeh Abu Tabanjah
    Commented Apr 5, 2017 at 16:08
  • \$\begingroup\$ If you are using reverse bias on the photodiode, then please edit your schematic to show the actual circuit you are asking about. Because your circuit shows "-v bias" on the cathode and +0 V on the anode, which most people will interpret as forward bias. \$\endgroup\$
    – The Photon
    Commented Apr 5, 2017 at 16:09

1 Answer 1

Reset to default
0
\$\begingroup\$

This is all pretty straightforward.

First, signal to noise. Get an RMS AC meter with a bandwidth of about 10 MHz, and measure the TIA output when the detector is dark. (Make sure to account for the meter's equivalent noise. AC meters are notorious for picking up crap in the ether.) This will give you your noise spec. Then signal to noise ratio will obviously depend on your signal.

Next, get an oscilloscope, and build the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Mount the LED and PD as close as possible, with a light shield around them.

Now drive the LED from a function generator, and with a pulse or square wave (watch out for reverse biasing the LED) you can determine the rise time of the LED and PD together, which should be less than 100 nsec. Make sure that the PD is not being overdriven if your LED is very bright. Look for maximum current on the data sheet (10 mA is a good first assumption), then make sure that the output to the scope does not exceed the iR product of the max current and the load resistor.

Now connect the LED to your PD/TIA and measure the output. You should see the rise time is much slower.

\$\endgroup\$
4
  • \$\begingroup\$ well, that was pretty clear, and does not sound so hard to do, at least i hope so... i have designed a test bed or lets call it a black box to prevent any light to get inside so the noise will be minimum from other light sources, but today i was discussing with my supervisor and he mentioned FFT over a certain bandwidth to see the impulse response of the the system, so we can get an idea about the link budget, so that is my concern now. Thank you for your help and if you have an idea about what i am talking about it will be so nice to tell me. \$\endgroup\$
    – Hamzeh Abu Tabanjah
    Commented Apr 5, 2017 at 15:59
  • \$\begingroup\$ @HamzehAbuTabanjah - Sure. An impulse has a flat spectrum, so all frequencies are excited. So an FFT of the response to a pulse will give the overall frequency response of the system. But, you first have to characterize the response of the PD, and the circuit I've given will do that for you, since the 50 ohm load will respond to very high frequencies even with appreciable diode capacitance. You'll need to determine the LED pulse width in order to get an appreciable response. For instance, I'll bet a 10 nsec drive pulse will produce no measurable response at all. \$\endgroup\$
    – WhatRoughBeast
    Commented Apr 5, 2017 at 20:10
  • \$\begingroup\$ So it means that from knowing the pulse width of the LED i will be able to know the sampling frequency and the number of tabs that i need for the FFT, if i got that correctly! ? \$\endgroup\$
    – Hamzeh Abu Tabanjah
    Commented Apr 6, 2017 at 7:14
  • \$\begingroup\$ @HamzehAbuTabanjah - No, it means that you need to understand the response time of the LED/PD well enough to produce a usable pulse. You should talk to your supervisor. \$\endgroup\$
    – WhatRoughBeast
    Commented Apr 6, 2017 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.