I am working on my master thesis these days, and I am having a serious problem. I am using FDS100 photodiode with LT6268-10 Amp to design a circuit. The required bandwidth for me is 10MHz, and I am using photodiode circuit design wizard to help me with my design. My problem is that when I use the equations to calculate the feedback resistance and capacitance then use these numbers in the design I get a frequency response around 23MHz, which is higher than the bandwidth required. Let's say I have a current around 30uA from the photodiode, and I want the peak voltage value to be 1V, so




which is really low, and I am not sure if we can even manage to get it in real life. I am not sure if my methods to calculate these values are correct, but there is a lot of them and they all should give the same results in the end. How do I calculate the equations correctly?

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    \$\begingroup\$ 480fF is practical value. \$\endgroup\$ – User323693 Mar 16 '17 at 16:01
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    \$\begingroup\$ i am not sure about that, because it's too low and i get this warning from the website;Warning: The feedback capacitance, Cf, needed for your design is only 479 fF. At this low level, board capacitance will have a significant effect on the peaking (Q) of this design. The actual value of Cf should be determined empirically with your specific PCB board. Because the value is so low, the tool has not converted the capacitance into a standard value. And the most important thing for me is the bandwidth or frequency respone which i am looking for 10MHz, but i am getting higher than that. \$\endgroup\$ – Hamzeh Abu Tabanjah Mar 16 '17 at 16:06
  • \$\begingroup\$ physical layout , resistor size, ground planes, bias voltage, all affect actual capacitance. Verify each part, don't assume. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 16 '17 at 16:37
  • \$\begingroup\$ Datasheet shows 40pF. \$\endgroup\$ – analogsystemsrf Mar 17 '17 at 3:52
  • \$\begingroup\$ which data sheet says that... \$\endgroup\$ – Hamzeh Abu Tabanjah Mar 17 '17 at 9:02

At some point, you apply the theory, and use an adjustable component to achieve a desired result (especially when you're running out of time). A test jig with swept frequency response, or transient rise-time measurement will be required. Few variable capacitors can accommodate your target of 0.5 pf. Stray capacitance variations will likely dominate your calculation efforts anyway.
Your desired bandwidth is less than that of your circuit, so if you wish to limit bandwidth with this capacitor, its value will be higher than 0.5pf.
In this kind of circuit, a small variable capacitor can be constructed from two insulated wires (teflon coated wires would be good):
variable capacitor Capacitance is adjusted by clipping off sections of the open end, being careful to ensure the open wire ends do not short together. Of course, you can reduce its capacitance by clipping, but you cannot increase capacitance. At 25 MHz., twisted length should be a small fraction of a wavelength, so transmission-line effects should be insignificant, and you have mostly a capacitive reactance between wire ends. Just keep it away from other components.

You'll still have to decide if stray reactance variability or temperature effects will detract too much from your desired result. And if this is a product, is the cost of a test jig, and adjustment procedure worth it?

  • \$\begingroup\$ To be honest i didn't understand what you are telling me totally, but my problem is still in simulation, i mean if my calculations are correct ad the methods i am using are correct, why the simulation is not working correctly!! because as far as i know the BW=1/(2piRfCf), and with the values i have i get around 10MHz but the simulation show a lot more than that. \$\endgroup\$ – Hamzeh Abu Tabanjah Mar 16 '17 at 18:45
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    \$\begingroup\$ RfCf product is very crude - a very rough approximation, but a good place to start. Many other circuit capacitances, and op-amp characteristics affect high-frequency response (it can even oscillate). Your design wizard should ask for these, but some capacitances you can only estimate. \$\endgroup\$ – glen_geek Mar 16 '17 at 19:23

If it turns out that the bandwidth is 23 MHz and requires a capacitor of 0.5 pF then use a higher value to get a 3 dB bandwidth of 10 MHz. Try raising the capacitance to maybe 0.8 pF. These are quite easy values to get hold of but you need to go surface mount to do the job properly. Basically, the capacitance of a 0603 component (whether it's R, C or L) will be about 0.1 pF when mounted on reasonable copper pads.

I'll also point out that the Analog Devices photodiode wizard can be used with different capacitor values and it will churn-out the new frequency response.

If you are really having problems getting such low value capacitors you can use a 10:1 resistive voltage divider on the op-amp output and use a 5 pF capacitor for feeding back.

  • \$\begingroup\$ I tried to do that and i am still getting around 18MHZ, the main problem for me is that i need to verify the equations i am using to calculate these values, but since i am not getting the same results in this circuit wizard. Yeah i can use a second stage AMP with a low pass filter, but my supervisor is telling me that i need to use only one stage AMP to get this BW!!! \$\endgroup\$ – Hamzeh Abu Tabanjah Mar 16 '17 at 16:21
  • \$\begingroup\$ analog.com/media/en/analog-dialogue/volume-47/number-2/articles/… this is the methods i am using to do my work for now, which are the same the this circuit wizard is using... \$\endgroup\$ – Hamzeh Abu Tabanjah Mar 16 '17 at 16:22
  • \$\begingroup\$ LTSpice (made by the same people who make the op-amp you use) is a big option. See what it comes up with and do make sure you model your photodiode's capacitance accurately. \$\endgroup\$ – Andy aka Mar 16 '17 at 16:26

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