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I have Microsoft HD3000 lifecam studio webcam. I have some noise, when operating it in low light conditions. I would like to minimize noise as much as possible, detaching infrared filter and using it as a grayscale imaging device.

As a software engineer i could go simple way: I could suppress noise to some degree purely with software means. Applying some lowpass filtering between frames to suppress high frequency noise, suming UP R+G+B channels to get grayscale image, or further - decreasing its resolution 2 times (via adding 4 nearest pixels' brightneses, so gaining extra bit depth in exchange for resolution) But that's not point of question.

But i think i could achieve some SNR improvements in hardware circuitry itself. I suspect noisy ATX PSU is main reason of such noise (400W of constant load, constant writes to hard drives, so on) So some noise could possibly go to web cam, messing up with exposure process itself.

So i would like to try to power my webcam from separate power source. (Dual 18650 li-ion batteries in series) I suppose to connect web camera USB power pins to a li-ion accumulators, via LM7805 1A linear regulator. So now you got the idea, and my questions are as follows:

  1. Can be "1A" considered suffient current?
  2. Is it safe to operate camera in such way?
  3. Is it safe to connect just D+ and D- pins to PC?
  4. Could i achieve some noise suppression this way? (If not - why?)
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    \$\begingroup\$ How does detaching an infrared filter make it a greyscale imaging device? I do not understand. \$\endgroup\$ – yoyo_fun Aug 15 '17 at 22:29
  • \$\begingroup\$ I don't think it even has a ir filter. \$\endgroup\$ – user3528438 Aug 15 '17 at 22:32
  • \$\begingroup\$ @yoyo_fun , Colors will be messed up when you detach filter. Thus, making webcam a bit unusable in RGB mode. But that operation dramatically increases its light sensitivity, several times. Then (in software) i sum up R+G+B values of each pixel, in order to get grayscale pixel value \$\endgroup\$ – user149105 Aug 15 '17 at 22:34
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    \$\begingroup\$ Although the only way to verify this is to artificially inject noise into the power and see to which point image is noticeably more noisy (or find literature that has done this), but my gut feeling tells me not to waste time on it but turn to other approaches like using larger aperture lens, using more advanced and larger sensors, IR active illumination, software processing, and even cooling the sensor helps a lot. \$\endgroup\$ – user3528438 Aug 15 '17 at 22:43
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    \$\begingroup\$ You absolutely need to post an image from the camera showing the noise. It's most probably sensor noise due to the low light and nothing to do with the power. I use such sensor noise as a true random number generator. \$\endgroup\$ – Paul Uszak Aug 15 '17 at 23:19
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But i think i could achieve some SNR improvements in hardware circuitry itself. I suspect noisy ATX PSU is main reason of such noise

Your suspicion is highly unfounded. The USB VBUS comes from a separate power source (either a dedicated 5V supply or a standby rail +5VSB), which is not loaded at all by memory or drives. This is in accord with Intel guidance. Second, it is bypassed by a 120 uF capacitor at each motherboard connector. Third, the power goes over a long cable, with has significant inductance, and then terminated with another capacitor at the device end, forming another filter. Then this input voltage goes through local power regulator, which usually has substantial ripple suppression.

Now, to your questions:

1: I can assure you that the HD3000 doesn't take more than 500 mA, as per USB standard port. So 1A is clearly enough;

2: Is it safe to feed the port from an external power? Why not? Millions and millions USB devices are self-powered.

3: Can you connect only D+ and D-? Yes, as long as you connect the ground wire as well, I hope it was a simple omission in your question.

4: can you acheve noise reduction? Highly unlikely, see above. You might get somewhat lower noise if you chill the sensor with a Peltier cooler, but this would be a challenge.

To summarize, image sensing is a pretty intense area of hardware engineering. If a sensor does show certain noise under certain low-light conditions, this is pretty much it. The image sensors are not just frame grabbers, they have intensive DSP mid-processing with embedded MCUs. For HS HD-grade USB webcams the image is transmitted already in compressed (H.264 or else) format, so doing decoding-encoding won't improve much. Please take a look at some introductory articles as Wikipedia before asking more questions.

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  • \$\begingroup\$ First, the USB VBUS comes from a separate standby rail (+5VSB), which is not loaded at all by memory or drives. Citation required. Most usb ports are not on the stand by rail... \$\endgroup\$ – Passerby Aug 16 '17 at 1:39
  • \$\begingroup\$ Standby usb ports are not standard. Most pcs have none or only a few are standby. Some pcs you have to turn the feature one, and that really means the pc is in a sleep mode, not actually off. The ones that do stay on when the pc is "off" typically have a little battery symbol on it too. "Off" for a computer simply means some standby state. Computers and tv's and radios and most appliances are only really off when you pull the cord. \$\endgroup\$ – Passerby Aug 16 '17 at 2:16
  • \$\begingroup\$ @Passerby, you are correct. The official guideline from the founding father of USB - Intel - calls for a switch between +5V and +5VSB rails, usb.org/developers/docs/whitepapers/… However, I haven't seen one, it adds significant cost. I will accommodate this info into my answer. \$\endgroup\$ – Ale..chenski Aug 16 '17 at 2:27
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The easiest way to figure out how much current that your webcam draws is to plug it through a device that looks like this:

enter image description here

These will work with any USB device that has a Type A male cable which connected to the unit in a pass through manner. You can purchase this and similar devices on Amazon and eBay for approximately 10USD.

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  • \$\begingroup\$ =) Seems you know market well.. What camera models can you recommend? I need grayscale, high-depth (eg 16 bit per pixel), moderate resolution (eg 1280x800), high framerate (60-120fps) camera, that can operate in low light conditions. Lens is somewhat near fisheye. Preferably, ethernet device, self-powered with separate power(5-12v, not PoE). Do you know one? \$\endgroup\$ – user149105 Aug 17 '17 at 16:32
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    \$\begingroup\$ @xakepp35, if you are looking at 16 bits per pixel and low noise, you are looking into a $2,000 - $20,000 product, not into a $30 consumer toy. Something like this one, highpointscientific.com/… \$\endgroup\$ – Ale..chenski Aug 17 '17 at 17:18
  • \$\begingroup\$ @ali-chen great example, but i have another goals. i an in need for realtime robotics vision, not astro imaging. so i'd go for a lower resolution, 1 megapixel at most, and higher frame rate, like 120 hz (8.3 ms per frame => so no exposure => so low read noise requirement). it needs to operate it in low light condition, eg at night, and be able to capture fast motion. probably i will end up with 12-14 bit depths. also it needs to be low-lag device, to operate realtime, and to drive some motion decisions from a robot. and a bit cheaper. what could it be? \$\endgroup\$ – user149105 Aug 17 '17 at 21:00
  • \$\begingroup\$ @ali-chen sure, for experiments i could use a webcam, but for a production device, it should have some enormous data rate, 1280*800*2bpp*120hz is a whooping 234.375 mbytes per second. what cameras and interfaces are capable of such operation/transmission? \$\endgroup\$ – user149105 Aug 17 '17 at 21:04
  • \$\begingroup\$ @xakepp35, USB3.0 can handle this. And, as I already said, HD-class image stream is usually compressed. But again, if you are up to machine vision and other industrial applications, a $30 toy won't cut it, period. This is not an area for dilettantes. Compare with this: aegis-elec.com/camera/usb-3-0.html \$\endgroup\$ – Ale..chenski Aug 17 '17 at 22:11
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Can be "1A" considered suffient current?

Maybe. Look at the label, it may say how much current is used. Or use a usb current meter. Most commercial usb devices will stick to the 500 mA limit. The Microsoft camera should at least.

Is it safe to operate camera in such way?

Sure. There are many self powered usb devices that do not use the host power. Simply disconnect the 5V line.

Is it safe to connect just D+ and D- pins to PC?

No. You will need ground as well. Both grounds must be connected.

Could i achieve some noise suppression this way? (If not - why?)

You could. You will find most of the noise is the cmos sensor being starved for light, but you may see a non - negligible difference.

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Can be "1A" considered suffient current?

Yes, the most USB can possibly supply for a single device is 900 mA, but it is unlikely that your camera uses more than 150 mA continuous. Maybe up to 400 mA peak.

Is it safe to operate camera in such way?

Yes.

Is it safe to connect just D+ and D- pins to PC?

You'll need ground too, but with those three connections it won't break anything, except maybe adherence to the USB standard. But since USB is a host-centric protocol (i.e. The host initiates all communications) it will work fine. People will tell you that the USB power pin is necessary even for devices which don't use it for power (the devices will still use it to know when the host bus has shut down or been disabled) but for experiments it will work fine. It's just not exactly in spec anymore.

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    \$\begingroup\$ There is no "maybe" about connecting the ground. It is a "must", period. The USB signaling is not strictly differential, many protocol states use single-ended signaling, like "USB_RESET", or Chirp-j/k to negotiate HS mode. Each of FS/LS USB packets must be terminated with EOP (SE0) as well. \$\endgroup\$ – Ale..chenski Aug 16 '17 at 3:59
  • \$\begingroup\$ @AliChen I agree. I didn't say maybe. \$\endgroup\$ – Blair Fonville Aug 16 '17 at 4:00
  • \$\begingroup\$ @AliChen Maybe the way I worded it was unclear. I didn't mean to say that working without a ground won't break anything; it absolutely will. I've worked with USB enough to know that the signals need the reference ground. I'll fix the wording. Thanks. \$\endgroup\$ – Blair Fonville Aug 16 '17 at 4:09

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