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10

There is no danger to the LED from removing the voltage source V1 or shorting it to ground. The LED itself limits the voltage across the capacitor, i.e. if it was safe while the voltage source was still connected, it's also at least as safe after it has been disconnected. However, if you allow C1 to charge up to greater than the LED forward voltage and then ...


2

12 Mbd over 45 km with a single wavelength with multi-mode fiber will just not happen, due to the physics of light propagation. Really, really good multimode (OM4/OM5) fiber (that's mainly about consistent geometry) has some 3 GHz·km bandwidth-distance-product at 850 nm, and something like 500 MHz·km at 1300 nm. That means that after 45 km, you get 500/45 ...


2

The chosen opamp MCP6241 is quite suitable for this application as it is rail-to-rail input/ouput, and Vcc as low as 1.8v. You have two options to handle it: 1) add a resistor between Vcc and inverting-pin of opamp. Because when the + pin is zero, the - pin tries to reach zero as well but because of input-offset, an error appears on the output. so you have ...


1

Assuming the motors are on the same power supply and not on separate power switches, LEDs in parallel will only tell you the power supply is active. You need to detect some current flow (motor active) but not too much current flow (motor stalled or bogged down) - and you want to do it without losing very much voltage in the detector ( 2V for a LED would only ...


1

As you have guessed, such a low resistor value means that any variation in LED Vf, which will happen as it heats up, will translate into a substantial change in current. Since LED forward voltage decreases as they get hotter, more current leads to more current. It might smoke, or it might not, but you'll have to fiddle with it, measure current when hot, try ...


1

A typical string of LED holiday lights uses either resistors or capacitors to limit the current, as well as a reverse-blocking diode, as shown in the below schematic. Capacitors tend to be less common for some reason. Perhaps it's cost-related, but I wouldn't expect a capacitor to cost a whole lot more. Anyway, if you can isolate one of the canisters and ...


1

3V on the LED is what you'd expect from a white LED. Forget about the 9V supply. Instead use a USB "charger" ie a 5V 1A power supply. This should be much more useful as you'll be able to plug the light on a powerbank, on a cellphone charger in the car, etc. Also the resistor power dissipation will be manageable. Cut a USB cable and connect to the ...


1

R1 and C1 to taste. I have no idea what the general consensus is on which opamp is suitable for breadboard/protoboard casual use, but it seems like the consensus is that the 741 belongs in a museum. The MCP601 is a readily-available single-supply rail-to-rail (more on that in a second) opamp that gives 22mA Isc at 5V supply, comes in a DIP-8 package, and ...


1

Clearly that circuit not work well with near zero reference voltage. Maybe input voltage offset of the opamp. The resistant value is very sensitive if you choose to low then the offset voltage can cause more current at zero voltage ref. as I = V/R. if you select too high value then the current will satterate at low value. the maximum voltage across resistor ...


1

As others have mentioned, photoelectric effect (solar cells) and photoemission (LEDs) are pretty much inverse processes. My high-school-physics mental model is like this: Electrons orbit a nucleus in shells. The physical size of those shells dictate how much energy the electrons have (and somewhat depend on the type of atom). Thus, to move from one shell ...


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