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I am a mechanical engineering student and I am working on a research project where we want to use a reverse biased segmented photodiode (OSI SPOT2DMI series) to detect deflections from a high speed laser. We are trying to detect very small variations in the laser signal so there is a large noise signal to deal with. Up until a month ago I had very little knowledge on photodiodes and I have never designed nor built a circuit board before so please forgive me if I am too vague or use the wrong terminology. The design I currently have utilizes four high speed op amps (OPA680 & OPA650). Two OPA680 op amps are used as transimpedance amplifiers for current to voltage conversion of each of the current signals from the segmented photodiode and another 680 is used to generate a sum signal. The OPA650 op amp is used to generate a difference signal which will later be connected to a lock-in amplifier. The laser signal we are studying is around 10MHz so according to the datasheets for each of these op amps they will meet the speed requirements.

My problem is that I don't understand how to connect the power supplies for the op amps. I am going to use a 12v battery for the reverse bias voltage and I would like to use a battery supplied voltage for the op amps instead of a bench top power supply. I am planning on using this prototyping board and soldering on two female BNC pcb mount connectors to extract the sum and difference signals. The OPA 680 requires a +5v and -5v power supply with a max rating of +-6.5Vdc. The OPA650 has a specified operating voltage of +-5V and a max rating of +-5.5V. I don't know much about circuits but I am assuming there is some way you can split a 12Vdc battery supply into +6V and -6V supplies, which would work for the 680, but not the 650. Somebody mentioned to me that I could usse a diode, which usually has a specified voltage drop, to reduce the voltage from the battery, but I am not sure if that will work/how to do that. If there is a way for me to use a 12V battery and split it into +6V and -6V I am also unsure of how I set up my ground bus on the circuit board and how this will affect the bias voltage since the positive terminal of that battery will be connected to the ground. If you would like to see a crude sketch of my circuit I can post it and any help is appreciated!

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Yes, there is a way to split it. Some people just use two biggish equal resistors across the supply, and use the middle as ground (it really is a reference), and the negative terminal would be your negative, and the positive, your positive.

There are some disadvantages, not the least of which is the lack of a low-impedance ground that can sink current

A better approach is a chip like the TLE2426. Just read the data sheet. An alternative is to just use two DC-to-DC converters, one to +6, and one to -6, off your 12 volts.

Also, once you have +/- 6V, you could use the appropriate voltage regulators to bring that down to +/- 5V. For example, http://www.digikey.com/product-detail/en/MC78L05BP-AP/MC78L05BP-APMSCT-ND/804696 and http://www.digikey.com/product-detail/en/MC79L05BP-AP/MC79L05BP-APMSCT-ND/804701. I haven't verified the drop outs (i.e., how much voltage you need to feed them to get your required output voltage), or asked about your current requirements, but your search term would be LDO regulator if these don't suit.

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  • \$\begingroup\$ If I was to use a dual output DC-DC converter to convert the 12V to +5V and -5V I would still be able to supply a 12V bias to the photodiode, right? I am thinking about using one 12V battery to supply to the DC-DC converter and then running the 2 outputs (-5V & +5V) to the supply busses that run throughout the board. I then want use another 12V battery to supply the 12V bias to the cathode of the photodiode. \$\endgroup\$ – elegrookie Jul 6 '15 at 15:49
  • \$\begingroup\$ @elegrookieyup,yes, you can use the 12V to bias the diode. \$\endgroup\$ – Scott Seidman Jul 6 '15 at 15:50
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It is perfectly possible to do what you want, except that if you do it you will lose the ability to provide a -12 volt bias. Your circuit will look like

schematic

simulate this circuit – Schematic created using CircuitLab

Note that all of the components to the left of the op amp must be isolated from the rest or the circuit. The op amp itself should have a decent current output capability, at least as much as the other op amps are expected to draw in total.

Like I say, you will lose the ability to produce a 12 volt bias on your PDs - only -6 volts is available. This should not be real problem, as it ought to increase your PD capacitance by 30% or so, and if you're that tight on bandwidth you're likely in trouble anyways.

Speaking of which, can you please provide the exact PD model you're using, and the expected optical power. Likewise, can you tell us the value of the feedback resistor you're planning to use in your TIA. I get a slightly bad feeling about your design. High-speed PD amps can be rather trickier than you seem to think.

EDIT - Also, can you tell us if (as seems to be the case) you are trying both for precision intensity measurement AND beam position. If this is so, you should be aware that this is probably not a good idea unless you go to a quad-cell detector with beam positioning to center the laser beam.

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  • \$\begingroup\$ Hoping for easy results with high speed trans-impedance amplifiers on a breadboard or strip board is also optimistic. I have use dead bug methods on a copper clad board and had to add shielding to prevent hand movements from causing irritating effect. You would also like to add two LOW drop linear regulators +5V and -5V to the split supplies you make and make sure the battery is fully charged when you do your testing. \$\endgroup\$ – KalleMP Jul 3 '15 at 17:51
  • \$\begingroup\$ @elegrookie - Please see edit as well. \$\endgroup\$ – WhatRoughBeast Jul 3 '15 at 18:00

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