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Nice to be mentioned in dispatches!
The other point Klaus didn't mention but is relevant at low voltages is how 'closed' the switch has to be. A triac will still drop around 0.6V when fully conducting and that could be significant. There are other possibilities that might work but more...
... but check the 'bias T' works over the bandwidth you need and can carry enough current.
Perhaps a simpler alternative is to simply add back to back Zeners in series with the motor so it doesn't operate until >12V is present then reverse the polarity to change motor direction. A bridge...
Easy with ICs, doable with discrete components.
You need a good RF isolation filter to keep the DC and RF apart, a 12V regulator to power the pre-amp and something to detect 15V and 24V. The voltage detectors could be say a 13V and 22V Zener diodes with pull-down resistors to ground and series...
No, I'm suggesting the 3V3 to R4 and pin 3 of R2 is connected to the MCU side of jumper P4 so when the jumper is removed it also removes 3V3 from the potentiometer and NRST pin. In other words, disconnect 3V3 from R4/P2 and connect that point to the top of C4/C5/C6/C7 instead.
Brian.
Assuming the potentiometer is on P2, as explained before the reason is there are two diode junctions on most MCU pins. The diode is a 'feature' of the silicon design but for practical purposes it looks like a diode between ground (VSS) and input and a diode between input and VDD. The diodes are...
As a guess, the potentiometer is feeding a voltage in when the jumpers are not fitted and the pin protection diodes are conducting it through to the VDD/VDDA lines.
If that is the case, there is a risk to the MCU of reverse voltage damage. The supply to the potentiometer should also be...
The threshold is set at half supply voltage by the equal values of R2 and R6.
The principle of operation is that the voltage on U1-1 pin 2 is set very close to the threshold and the additional voltage from the microphone takes it over the threshold and makes the comparator change state. There...
True, the voltage will still reach the same peak but the average power will be halved. It is the power that matters, the mass of the filament makes the temperature follow the average voltage rather than instantaneous peaks. By eliminating one half of the AC cycle, the effective power is also...
One capacitor in series with each lamp. The lamp will appear as a resistor of approximately 22 Ohms, the inductor will appear as about 10.2 Ohms in series with it (ignoring its wire resistance) to drop the voltage. The advantage of using an inductor is lower power dissipation than a real...
You need an equivalent series resistance of 10.2 Ohms. You could use a series capacitor but the value has to be about 250uF so finding one rated at >120V and class X2 would be very hard. However a series inductor of 27mH rated at >3A is more practical, smaller and cheaper. I have assumed...
The 1K resistor at the top of the schematic probably should be across gate and source pins, not gate and ground. With high voltage on the drain you are probably exceeding the drain-gate limit.
Brian.
Seems like you are measuring the pulse width ratio rather then RMS. If it is a 0V to 3.3V 'logic level' signal, treating it as a digital signal and using a simple MCU would probably be easier. You can use a timer to measure frequency (if needed) and the width of the pulses within each cycle...
A circuit diagram showing what the connections actually do would help. I for one am not going to track down the boards information from the picture to see what the connections are for.
Brian.
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