It will be one of the FDXXX Series photodiodes from Fermionics. Within these series a photodiode with the largest area whose capacitance will permit a bandwidth of 40MHz will be chosen.
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For this design I intend to use 0402 (imperial) size parts to minimize stray capacitance and inductances at 40MHz, and if I can, up to 100MHz. will this make a difference at 100MHz? Or will 0603, or even 0805 work as well at 100MHz?
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With care Tektronix used 1/4 watt through-hole parts up to 200MHz overall BW. Sure, higher-impedance nodes have capacitance issues; low-impedance branches have inductance issues. Most places - it's the device reactances, or block-interconnect reactances that cause problems.
-F
4817-1_4817-2.pdf
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Maybe. The larger packages will have a higher parasitic capacitance. Whethe r that matters depends on how high the resistance is. I have used a larger package to use just the parasitic capacitance across the feedback resistor. You may need to cut out the planes below the high frequency parts to preve nt parasitic capacitance from that also. But if you are adding a capacitor of a few picofarads across the resistor, the fractional picofarad addition from the part won't matter much. You can test the effect of different paras itic capacitances with spice simulations.
Does the parasitic capacitance to the PCB have a significant effect, or is it all end-to-end? ...Jim Thompson
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Yes. The pad to the plane and then back to the other pad can have lower a impedance than the resistor. It also acts like a capacitor to ground. I often cut holes in the plane around TIAs and I never worked at frequencies as high as 100MHz.
It's super educational to try putting realistic strays in the circuit simul ation and seeing what they do. For a 4 layer board with ground on L2, figur e 0.15 to 0.25 pF per pad. Adjacent pads don't couple as much as you'd thin k because there's air on one side and the ground plane on the other.
Resistors have about 0.05 to 0.1 pF in parallel, depending on the ground co nfiguration.
I generally put a bootstrapped pour under the summing node.
Cheers
Phil Hobbs
I've seen a 100MHz receiver built with 3/4" parts between tag strip and vacuum tubes.
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If I were to place a cutout beneath high frequency parts wouldn't this cause a disturbance in the image plane effect? Ref Section 17.2.2.3, page 7, of:
Or would this not matter when the components are small enough, and close together enough, to be considered lumped?
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The current plan is to use your front end topology you show in Figure 9, Page 4 at:
The question is where is the best place to connect the output resistor of the error integrator (R3 Figure 12)? In your front end schematic is it best connected to A1's summing junction (inverting input)? Or the emitter of Q1?
I can see how the R3 connection will sacrifice bandwidth either way.
The schematic in your book is slightly different from the one you give in the PDF you linked to. In your book you have in Figure 18.12 a transistor with its collector and base connected to the +15V supply and
2Mohm resistor. Is that there to compensate for the base emitter voltage drop on Q1? If so wouldn't be best to make that transistor the same as Q1?-- To respond to me directly remove sj. from the my email address's domain name. This is a spam jammer.
On 09/01/2017 07:35 PM, Artist wrote: > On 7/11/2017 1:09 PM, snipped-for-privacy@gmail.com wrote: >> Good TIAs at that bandwidth are not that easy to design. I suggest spending most of your time on a fast, quiet DC-coupled TIA and filter afterwards. >> >> You might find this article helpful. >> > > The current plan is to use your front end topology you show in Figure
9, Page 4 at:Why bother? Are you working at very low supply voltages or super wide ranges of photocurrent, or something like that? By the time the photocurrent gets as far as dropping 50 mV across the feedback resistor, you're already in the shot noise limit.
Need more info. If you want to reject DC, why do that instead of using a series capacitor between the first and second stages? There are sometimes reasons, of course, but a series cap is pretty simple and trouble-free.
2Mohm resistor. Is that there to compensate for the base emitter voltage drop on Q1? If so wouldn't be best to make that transistor the same as Q1?Yes. But then it's only a first-order compensation, so it doesn't matter very much which transistor you use.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Because the AC modulation is only 1% of the DC level. If I can optimize the TIA feedback resistor for only this AC there is a potential 100 to 1 noise reduction to be gained over a design where the feedback resistor has to be optimized for the AC plus the DC.
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If Iphoto * Rf > 200 mV, there's only 1 dB to be gained by going to the limit Rf -> infinity, because the shot noise power is already 4 times the Johnson noise power. You can't separate the signal from the shot noise unless you go to spooky quantum measurements, and even there it doesn't usually get you that much. (*)
If you have a very wide range of photocurrents, so that you can't keep within the range 200 mV < (Rf * Iphoto) < VDD, you just have to pay the price in voltage swing to avoid dying of Johnson noise at the lower limit. It's not unheard-of to run a current-nulling servo off a 250V supply.
Cheers
Phil Hobbs
(*) (There's a scheme called 'quantum illumination' that I don't understand well, but which bids fair to help quite a lot in special situations.)
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Another approach is to segment the resistor in the nulling loop and use a d iode in parallel with all but one section, forming a breakpoint servo amp. The good news is that you don't need a 250V supply. The bad news is that the servo bandwidth depends on the signal level.
Cheers
Phil Hobbs
Rf * Iphoto for the AC component is expected to be roughly 25mV if the Rf were to be optimized for AC + DC.
In an effort to reduce lost bandwidth due to R3 from the error integrator to the virtual ground suppose I do any of: substitute R3 for a current source such, as the Howland Current source:
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That's not the right way of thinking about it--the shot noise is white, and depends only on the DC. If I_DC R_L > 200 mV, you've only got 1 dB of room to improve, independent of frequency, regardless of how fancy a circuit yo u use.
Cheers
Phil Hobbs
Today I found out that the TIA is to be used over a wide variety of illumination levels on the photodiode. I have to design so as not to saturate at the highest expected power level. For the lowest illumination level it very well could be that the DC level of Rf * Iphoto will be under 200mV.
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That makes sense. In that case, the two normal ways of proceeding are: (1) use a current feedback loop running off a much higher supply voltage, or (2) Use a CFB loop at normal voltages but with a nonlinear feedback resistance cobbled together from diodes or diodes plus resistors.
Option 1 requires a separate supply, and the range may be impractically large. OTOH the loop BW stays constant.
Option 2 relaxes the supply voltage, but the loop bandwidth changes with signal level.
Cheers
Phil Hobbs
The output has to be linear so option 2 is out.
Why would a much higher power supply be required for option 1? Are there any options for current sources other than than the ones I mentioned?
It is permissible for there to be a second power supply in this design.
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As I said, it doesn't matter how complicated you make the circuit. The issu e is that for a DC-coupled amplifier there is no escape the thermal noise o f whatever resistor you use to remove the DC, whether it's the feedback res istor in a TIA or the emitter resistor in a current source. Do the algebra and you'll see.
In an AC-coupled amp, you can use transformers and/or capacitors for feedba ck instead of resistors, but not at baseband.
You can also use range switching.
Cheers
Phil Hobbs
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