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8

You seem to have used a 22Ω resistor for the LED instead of the 220Ω shown in your schematic, and the link between the LED and that resistor is missing (fortunately for that poor LED). The graph of figure 10 on page 10 of this LM358 datasheet, it can be seen that the output is unable to swing to the positive supply voltage, when sourcing current. If your ...


4

Since the entire circuit is in the DC domain all you may need is a voltmeter to check voltages at the op amp terminals. What is the current required to drive the buzzer? Can the op amp drive it? Did you ohm-check the traces you separated on your universal PCB between pins 1-8, 2-7, 3-6 and 4-5 of the DIL package? (Did you remove all the copper there?)


4

The ground pin is tied to neutral at the breaker box, so it is perfectly possible for noise injected into the neutral wire to appear on the ground wire as well. Beyond that, most recent (not slow) powerline ethernet adapters use 2x2 MIMO, and utilize both line to ground and neutral to ground as the two MIMO pairs. If your adapter has 3 prongs, you can ...


3

@metacolin is generally right, but my first action would be to throw away the LED PSU and get a better one. The powerline ethernet works by injecting high-frequency currents into the power line. It takes care not to inject much, because for other devices this accounts for lowering the power quality. This is why it is easily affected by powerline noise. The ...


3

The LED is not connected to resistor, it is disconnected. On the other hand, it looks like there is a solder bridge which shorts the LED directly to buzzer pin. There might be other solder bridges as well at the op amp inputs, which may be the reason nothing works. It is also impossible to say if you have LEDs and the photodiode connected with correct ...


1

Your method 2 is correct. You might find it useful to learn about load lines. These are useful when the device (the LED) doesn't behave in a linear fashion like a resistor. Figure 1. Various loadlines for a 5 V supply and various LED types. Image source: Loadline resistance graphic tool. Each loadline is drawn from 5 V, 0 mA to 0 V and the current ...


1

When calculating the current through an LED and a resistor in series, you need to take into account the voltage drop across the LED. So method 2 is the correct one.


1

Microcontroller with enough I/Os can do that. If you want something more elegant and don't feel like writing code, you could program a Silego Greenpak to do this. The resulting circuit would be really tiny, and you could eliminate those diodes.


1

Your signal averages at 0VDC, it does not rectify and filter the input signal. Then the average brightness is with its input at 0VDC. You need a "peak detector" which rectifies the signal and filters the resulting DC voltage that represents the loudness of the input signal. Here is my peak detector:


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