# Tag Info

25

Well, it has the potential to work. You would have to line the inside with RF absorbing material, otherwise the incoming waves would just bounce all over the place. Using copper plates to detect the RF power probably isn't the best idea. I would recommend using actual wifi antennas for that purpose, each one connected to an LNA and 2.4 GHz bandpass ...

16

Try this circuit: simulate this circuit – Schematic created using CircuitLab The transistor (non critical, any general purpose NPN will likely work) amplifies the weak photocurrent from the LED used as a photodiode. Make sure you get the polarity right on the LED or you will damage the LED and the transistor. Connect a 1K in series with the LED if ...

11

You might have some luck with this approach that Greg Charvat demonstrates using an LED radio detector and long-exposure photography. The obscura idea is interesting, but getting RF to behave that way sounds... a little crazy ha! It would be awesome if you could account for and control all of the re-radiation and reflection that would probably happen. If ...

10

Unfortunately you're going to run up against a limit in terms of diffraction. We know that (at least for optical pinholes), the ideal focal length for a given pinhole radius s is s^2/λ, and the spot size at this distance is about 0.6 s From these, we can determine that for a given resolution n with a 'normal' field-of-view (think of n as the width or ...

9

Yes, a LED can be used as a sensor. It gives output current proportional to light intensity, similar to a solar cell. But because its area is very small, current is also very small. You can measure photocurrent with a digital voltmeter. Most meters have input resistance of 10 MEGohms, and you will be measuring current through this resistance. So if your ...

6

Diffraction through a small, wavelength-sized hole will just fill the area behind it. Pinhole lenses for light have the same problem. Your idea would work if you scaled it up, say you used a football stadium with a metal roof, made a 10 x 10 m hole in the roof, and placed sensors on the field. Not practical. Why not consider a single pixel camera? use a ...

6

This is called an image sensor. This particular image sensor in your picture is a CMOS image sensor, commonly found in pretty much anywhere a small to medium sized camera is used, as well in some big and/or professional gear. The camera in your phone is almost certainly a CMOS image sensor. It is probably also the type of sensor captured the image you ...

6

Fortunately, 20kHz isn't high frequency. Just about any photodiode should be able to handle 20kHz with no problem, especially if you negative bias the anode to reduce the capacitance and thereby reduce the rise time. Here's a fairly plain photodiode from Vishay, available for under a dollar from DigiKey: http://www.vishay.com/docs/83471/tefd4300.pdf The ...

6

You don't need a separate capacitor. From the lab description you linked to: The  physically  separated  anode  and  cathode  behave  as  half  of  a  cylindrical  capacitor;  the incoming  light  acts  as  a  current  source  to  charge  up  this  capacitor  (C)  with  electrons.    The total  charge  (Q)  and  potential  difference  on  this  capacitor  ...

6

Ica (what they are the "collector light current") is the current through the collector and out the emitter vs. Vce (collector to emitter voltage) at different intensity of light impinging on the device. So, for example, if you had a Vce of 10V and shined a light of 0.1 mW/cm2 on it, you should expect an Ica of about 0.3 mA (more or less).

5

I just wanted to post and mention that the suggestion made by @tomnexus is quite workable. I just finished the first tests of a similar rig. My setup uses a satellite dish with LNB, a satellite finder (to pickup the signal strength,) an Arduino, and a little software on a PC. The Arduino controls a couple of servos and reads the signal strength from the ...

5

The general term would be an image sensor, the underlying technology would either be a CMOS sensor (a.k.a. active pixel sensor), or a charge coupled device (CCD). I have no idea which of the two it is based on your image.

4

I'd call it an image sensor: - You can also call it a pixel array. See this site for more details

4

Many photodiodes, particularly silicon-based ones, will not emit signifcant light, even when driven in quadrant I. Operation in quadrant IV (or along the I=0 axis) is normally called photovoltaic mode, not "solar cell". The distinction you're making between "active" and "passive" elements is not the way these terms are normally defined. For example, a ...

4

No, that doesn't exist. Visible light has a wavelength much larger than your particles, so the waves will go right around them unimpeded

4

Usually the labels are as follow: A - Anode, this is the LED anode (+) K - Cathode, this is the LED cathode (-) E - Emittor, phototransistor Emittor C - Collector, phototransistor Collector. The photo transistor base is basically the light from the LED.

4

"Daylight" is full of various wavelengths of light, including infrared. In daylight, your sensor will "see" infrared from the mishmash that is sunlight, and react the same as if it were seeing the reflected infrared from its emitter. To make things worse, the detector can also "see" normal light as well as infrared. It has a ...

4

"I'm interested in estimating Ic based on the Irradiance I calculated (~0.5mW/cm^2)" Ic mainly depends on: How precise is the estimated value (mW/cm^2) or what is expected range The value of the resistor connected to the collector (Rc) The DC voltage source connected to the collector resistor (Vcc) Assuming Vcc is given by your circuit, by ...

3

Initially, I'd try a narrow angle visible light LED and phototransistor, and set up the phototransistor in 'line of sight', maybe inside a tube to reduce the chances of refraction or reflection from the droplet effect the sensor. Something cheap and simple like an old pen body The rationale is the visible light LED would make it easy to set up. Infrared (...

3

Try it! Fix the LDR to your desk with tape, connect your multimeter on resistance mode to the LEDs, fix a desk lamp above the LDR and gradually shade it with a piece of card sliding it across the surface of the LDR. You should see the resistance gradually increase. Why? simulate this circuit – Schematic created using CircuitLab Figure 1. Imagine the ...

3

The following should work for you. You may have to switch the motor wires around to find the right polarity though. If there is an obstruction the relay will flip back and cause the thing to reverse till the switch clears then it will re-engage and should continue to try to close. Relay may buzz a bit though if the obstruction remains, significantly ...

3

The equations shown are all derived from the same formula which you already gave: $$t_r = \frac{0.35}{B}$$ A single-pole system has the frequency response $$H(s) = \frac{A}{1 + \tau\cdot s}$$ This sytem has a pole at $$p_d = \frac{1}{\tau} \Rightarrow BW = \frac{1}{2\pi\cdot \tau}$$ The transient step response of this system can be calculated as h_{...

3

The AD500-9-8015 has the amplifier built in, you need to supply the appropriate Vcc which in the datasheet can range from 4.5V to 11V for proper operation. The bias pin needs 100 to 240V: This detector requires +4.5 V to +11 V voltage supply for the amplifier and a high voltage supply (100-240 V) for the APD. The internal APD follows the gain curve ...

3

GS Tube doesn't make the FEU49B, or any other tube. They sell new old stock Soviet produced vacuum tubes. The datasheets are therefore difficult to read - I know English and German, but no Russian. As I understand your question, you are looking for the pins for "Photo cathode" and "Modulator," and can't find them on the pins that go into ...

3

A typical approach would be to use a photodiode with a TIA (trans-impedance amplifier). Most small photodiodes are easily capable of 1 kHz operation. The TIA can just be an opamp and a few resistors. Without knowing more about your light source and what you are doing with the signal afterwards, that is about all the advice I can give.

2

Visible or IR won't make much difference, and you've got answers that go into the electronics better than I could, but don't forget the optics. Essentially everything has to line up very well. If you're using an LED you'll probably want to have a small hole in some foil in front of of it so that the droplet covers the hole as it falls past. A matching ...

2

I'd build a photointerruptor based on an IRLED pointing at a silicon photodiode across a gap through which the water droplet would fall. The diode would be operated in reverse-bias mode and would be illuminated all the lime by the IRLED so that when the water droplet fell through the beam it'd momentarily shut off the IR incident on the photodiode, ...

2

This is an absorption spectrum of water: Source (Kebes at English Wikipedia [ CC BY-SA 3.0 or GFDL], via Wikimedia Commons) It shows the absorption coefficient $\mu$ for the formula of the intensity $I=I_0\cdot e^{-\mu\cdot x}$, where $x$ is the "thickness" of the layer of water. Common lasers and LEDs are in the wavelength range of 350nm to 850nm, ...

2

The forward current is the absolute maximum rating. Just about any current will suffice to power the LED. Minimize the current for your application, generally 5 to 10 mA is sufficient for firing the LED. The reverse voltage should be limited to prevent damage to the device. Although I don't see any reverse protection in the symbol, just keep the maximum ...

2

While it's a little bit tricky to compare the specs of an LDR to those of another light-sensing device, such as a photodiode, phototransistor, or solar cell (because they are very different devices), I'll try and summarize the main differences here: Response time: from the LDR datasheet, it has response times in the 10's of ms (20-40ms). Photodiodes and ...

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