- Adventures with the Optronics MicroFire camera
- 29 Jan 2022 03:37:48 pm
- Last edited by Botboy3000 on 31 Oct 2023 05:29:02 pm; edited 1 time in total
In an attempt to find a cheap DSLR camera to use for astrophotography, I took a trip to the university surplus sale. Sadly, there were no DSLR's to be found. As I was heading out the door though, this thing caught my eye. A cast aluminum mystery item. Intrigued, I picked it up off the shelf and gave it a look. To my amazement, a CCD sensor looked back at me! Immediately I checked the price tag. $3! I paid the clerk and took it home. Upon further inspection, it looks to be a microscope camera from 2002. After some online research, it seems that this thing sold for over $7,000 when it came out. It seemed I had gotten the bargain of a lifetime.
However, this of course came with a few caveats.
1. It uses FireWire for data transfer. Not a huge problem for me, my computer has a FireWire card
2. It was missing its DIN 5 power supply.
Upon opening up the aluminum casing, I found a big, fat Xilinx FPGA at the heart of this camera, as well as nicely labeled test pads (labeled with voltages!) and using a multimeter, I was able to deduce the power supply connections by finding conductivity between the pins and labeled test pads:
Code:
So that was nice, but I have no idea what kind of power requirements any power supply I would make would need to be able to meet. From some googling, it sounds like CCD's are pretty power hungry, but I would like to get some hard numbers.
I really would rather not open up the CCD sensor housing part of the case so that I don't damage the sensor and don't get any dust on it. Looking through Optronics MicroFire documentation, the only thing I can find is that it's likely 1600 x 1200 pixels (although the aspect ratio doesn't seem square enough for that, take a look at the pictures below, maybe I'm wrong) but no sensor part number that I've found so far.
Plus there's the issue of the weird voltages that I'd have to supply. +-17V isn't exactly something you see every day, and 5.8V is also something I've never heard of anything using until now.
So, if anyone is knowledgeable on CCD sensor chips and the power they require, please let me know! I will be trying to build a power supply for this and I would like to get it right
The final issue is what software does it use? Well, Optronics seems to have included a copy of their own "PictureFrame 2.2" with each camera they sold. Sadly, I don't have that. Hopefully the firewire control and transfer is generic enough that some other software would also work, but believe it or not I'm having some trouble finding FireWire camera software in this day and age. Not sure though! I have to get the thing powered on first before that becomes a problem.
And as promised, here are some pictures! (Click to enlarge)
EDIT: I was looking at various CCD datasheets and the voltages they all needed were 14.5 - 15.5, 15V nominal, not 17V. So I wondered if there were some internal voltage regulators and sure enough, glancing at the photo again, the top of the PCB sure looked like a bunch of voltage regulators. So some probing with a multimeter and a little bit of googling proved that yes, the +- 17V is only there so that the circuit can convert it to +- 15V (and also 10V). Here is the connection diagram I drew up real quick:
So according to the LT1129 and LT1175 datasheets, I should be able to just give it voltage that will cover the dropout voltage. In the case of the LT1129, up to 30V and in the case of the LT1175 (the negative voltage regulator) up to 20V if I'm reading that one correctly. The 5.8V goes into a few more voltage regulators, and the 800mV above 5V just seems to be there to cover the dropout again. So, it seems like the +-17V and +5.8V are more of guidelines since they go straight into regulators. So I could use +- 18V rails and a 6V supply and it should be fine I think. Hopefully those I will be able to find instead of build myself!
However, this of course came with a few caveats.
1. It uses FireWire for data transfer. Not a huge problem for me, my computer has a FireWire card
2. It was missing its DIN 5 power supply.
Upon opening up the aluminum casing, I found a big, fat Xilinx FPGA at the heart of this camera, as well as nicely labeled test pads (labeled with voltages!) and using a multimeter, I was able to deduce the power supply connections by finding conductivity between the pins and labeled test pads:
Code:
_______
/ o \ GND
| o | +17V
| o | GND
| o | -17V
\ o / +5.8V
*******
So that was nice, but I have no idea what kind of power requirements any power supply I would make would need to be able to meet. From some googling, it sounds like CCD's are pretty power hungry, but I would like to get some hard numbers.
I really would rather not open up the CCD sensor housing part of the case so that I don't damage the sensor and don't get any dust on it. Looking through Optronics MicroFire documentation, the only thing I can find is that it's likely 1600 x 1200 pixels (although the aspect ratio doesn't seem square enough for that, take a look at the pictures below, maybe I'm wrong) but no sensor part number that I've found so far.
Plus there's the issue of the weird voltages that I'd have to supply. +-17V isn't exactly something you see every day, and 5.8V is also something I've never heard of anything using until now.
So, if anyone is knowledgeable on CCD sensor chips and the power they require, please let me know! I will be trying to build a power supply for this and I would like to get it right
The final issue is what software does it use? Well, Optronics seems to have included a copy of their own "PictureFrame 2.2" with each camera they sold. Sadly, I don't have that. Hopefully the firewire control and transfer is generic enough that some other software would also work, but believe it or not I'm having some trouble finding FireWire camera software in this day and age. Not sure though! I have to get the thing powered on first before that becomes a problem.
And as promised, here are some pictures! (Click to enlarge)
EDIT: I was looking at various CCD datasheets and the voltages they all needed were 14.5 - 15.5, 15V nominal, not 17V. So I wondered if there were some internal voltage regulators and sure enough, glancing at the photo again, the top of the PCB sure looked like a bunch of voltage regulators. So some probing with a multimeter and a little bit of googling proved that yes, the +- 17V is only there so that the circuit can convert it to +- 15V (and also 10V). Here is the connection diagram I drew up real quick:
So according to the LT1129 and LT1175 datasheets, I should be able to just give it voltage that will cover the dropout voltage. In the case of the LT1129, up to 30V and in the case of the LT1175 (the negative voltage regulator) up to 20V if I'm reading that one correctly. The 5.8V goes into a few more voltage regulators, and the 800mV above 5V just seems to be there to cover the dropout again. So, it seems like the +-17V and +5.8V are more of guidelines since they go straight into regulators. So I could use +- 18V rails and a 6V supply and it should be fine I think. Hopefully those I will be able to find instead of build myself!