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i am very new to power supply design

i have a photo diode whose peak current is 150mA and bias voltage of 12V, photo diode being reverse biased.

i have a input voltage coming 28V from battery, so i would require a 28V to 12V regulator to bias the photo diode

my doubt is, what should be maximum output current that can be expected while selecting an LDO or DC-DC(do i need it?)?

as the current is generated by the photo-diode itself even if any high current output is un-necessary i feel, so a normal LDO like TPS79801-Q1 with Imax being 50mA should fit in my application ?

also i am less familiar about heat that would be generated across the LDO in conversion of 28V to 12V, as the current is very low i dont need to worry that much about power dissipated ?

EDIT: datasheet of photodiode

EDIT2: pulse width 10ns-150ns and PRF of 20us-1S

EDIT3: kindly put your comments at similar question of same problem to know behavior of photo diode in saturation

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  • \$\begingroup\$ It seems unlikely to me that you only need biasing for the photodiode. No doubt there will be an amplifier as well as photodiodes usually give very weak signals. The 150 mA peak current is very likely the maximum forward current (where's that photodiode's datasheet ?). That current is not going to flow when you're using the photodiode as a light sensor because then you'll operate the diode in reverse mode. I STRONGLY recommend that you browse the Internet looking how others do this and learn from that. You're fixated on the biasing while that is only a small part of the whole. \$\endgroup\$ – Bimpelrekkie Aug 22 '16 at 10:21
  • \$\begingroup\$ updated the datasheet, the diode is followed by a TIA anyway, i did not find enough data about power sources used in photo diode biasing so i have posted this question \$\endgroup\$ – kakeh Aug 22 '16 at 10:38
  • \$\begingroup\$ And that proves why a datasheet is essential because this is not your standard photodiode ! It does not state what the current requirement is. Where does the 150 mA come from ? It is not in the datasheet. If the TIA uses a 12 V supply, I suggest you use that supply. \$\endgroup\$ – Bimpelrekkie Aug 22 '16 at 11:10
  • \$\begingroup\$ That's not a standard photodiode at all! You should ask the manufacturer for a reference circuit design. \$\endgroup\$ – pjc50 Aug 22 '16 at 11:19
  • \$\begingroup\$ the 150mA is an calculated value, from max incident power of saturation, not displayed in datsheet \$\endgroup\$ – kakeh Aug 22 '16 at 13:02
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1) If you are converting 28 volts to 12 volts, you need to stop talking about LDOs. LDO stands for "low drop out", and refers to a regulator which can operate with low voltage across it. "Low voltage" is generally used for less than about 2 volts, although there is no universal standard. Since any regulator you use will have an input-output voltage of 16 volts, this is obviously a non-starter.

2) In principle, at a current of 150 mA, a linear regulator will dissipate 16 x 0.15m or 2.4 watts. For continuous use, your suggested part will die. A dc-dc converter seems the obvious choice in this case.

3) Now we get to the important stuff. The stuff you haven't mentioned. How, exactly, are you going to use this? Note that the data sheet suggests it for use as an optical equivalent of an RWR. In these conditions, it will almost never receive much illumination. But let's say it does. At 12 volts and 150 mA, peak dissipation in the detector array will be 2 watts, and I really doubt that that package can handle that. So, are you going to use it for transient detection, or do you expect continuous use? If it's continuous, you are misusing the part and it will probably fail. At these levels, of course, there is no need for a TIA - does this feed into your previous questions about fast pulses with very wide dynamic range? In principle, you can look at the amplitude of the reference outputs and disconnect the high sensitivity outputs when necessary.

EDIT - In a comment you state that "i use a TIA followed by photo diode , the opamp saturates at input 2.5mA itself". Now, I assume you meant "TIA following each photodiode". Even so, your requirement of 150 mA makes no sense. The high-sensitivity channels will be saturated and putting out 20 mA or so, so you'll need optical attenuation to keep from overdriving your TIAs. 6 outputs times 2.5 mA is 15 mA total.

The preceding, of course, could be wrong, since I could be misinterpreting your setup, operation and requirements. Which is why I asked, "How, exactly, are you going to use this?".

Stop playing coy. Stop trying to parcel out information in dribs and drabs. You are willfully keeping us in the dark, thinking we can help you if we don't know what your problem is, and your limited disclosure makes no sense.

Furthermore, if you look closely at the data sheet, it gives a diode rise time of 5 nsec. This will make the device largely unusable for pulse widths under 10 nsec.

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    \$\begingroup\$ LDO doesn't mean that the regulator can only handle small drops, just that it is capable of them. TI, for example, calls all of their linear regulators "LDO", even ones that can do 40+V drops. \$\endgroup\$ – mbrig Aug 23 '16 at 2:09
  • \$\begingroup\$ i would very rare operate in maximum incidence power, and more over the pulse is of very short time 10ns and repeating at rage of 20us, and i use a TIA followed by photo diode , the opamp saturates at input 2.5mA itself \$\endgroup\$ – kakeh Aug 23 '16 at 8:22
  • \$\begingroup\$ @kakeh - See edit. \$\endgroup\$ – WhatRoughBeast Aug 23 '16 at 12:34
  • \$\begingroup\$ i thought datasheet spoke a lot, my requirement is not 150mA, i cannot stop the photo diode gettting exposed to high intensity light such as laser which will generate such high current, so basically i dont know what power supply i have to choose in this situation, that is why i have selected a lowest possible current output dcdc, is it like just only bias is required in this case ?, it also raised a question on behaviour of PD in saturation here:electronics.stackexchange.com/questions/251984/… \$\endgroup\$ – kakeh Aug 23 '16 at 17:23
  • \$\begingroup\$ i know i dont require high current output, but i not able find the lowest possible current output in market,which one choose is also a dilema, if a high current ouput one is chosen, any problems ? \$\endgroup\$ – kakeh Aug 23 '16 at 17:26

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