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The best way to use a flux concentrated DC Hall is like this Blue is ferrite or iron, black is the two Hall devices. First, note there is no milling out of pockets with a Dremel for the sensor to sit inside. This way, there is a large consistent air-gap to lower the overall permeability of the ring, to prevent it saturating with small currents. You've not ...


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Figure 1. The common on one pole of the switch is connected to the upper or lower tag depending on switch position. The other pole of the switch is the same and is electrically isolated from the first. Connect the monitor + and - to the common tags. Connect one power source to the top tags, observing polarity. Connect the other power source to the bottom ...


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The user "rdtsc" in has identified that this "pump" is actually a valve, in a comment. You shouldn't use a resistor with it. The valve is designed to be connected directly to 6V, and at this voltage, it will use about 220mA (assuming those are the correct numbers). The resistance of the valve itself is what you calculated (22.72 ohms). If you put a 22.72 ...


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That's a solenoid, not a pump. It basically allows or disallows airflow from an external source of pressure, it's like a mechanical on/off switch. If you hear a 'thunk' it's working correctly. To address some of your questions in the comments: People don't use motors with resistors because it's grossly and unnecessarily inefficient. The power you waste ...


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Possible reasons: Many of those adapters are unregulated which means it might give 9 V when fully loaded but a higher voltage when lightly loaded. You used an AC adapter. The alternating current alternates the polarity and that or may have caused damage. You used one with polarity reversed. Check the label. Normally they are centre-pin positive. You ...


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There's plenty of mains AC powered equipment which behave as yours. They have metallic shield to reduce electromagnetic interference. That shield is connected directly or with a big capacitor to the output which is considered to be the signal or output voltage GND. In addition it's connected to the protective earth wire of mains AC input. Finally they have ...


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If I told you that there is 5 mA flowing through the inductor just before the switch opens can you see that? It’s based on the assumption that the switch has been closed for a long time. Do you understand that bit? Can you also see that as the switch opens, in that instant there will still be 5 mA flowing in the inductor? Based on that, you can calculate ...


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As a general note, always be wary of parts without datasheets! It looks very similar to this switch (Digikey link) . You can see how this type of switch is generally wired in the bottom right "Schematics" section. In short, each of the 2 rows of pins are identical, just different sets of the same connections. The middle pin is your common (always part of ...


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The main problem is you have C1 in the wrong place. It should be put just after the rectifier to smooth out the DC voltage, but you have it across the Zener diode which will cause a large current spike each time the transistor is turned on. A secondary issue is that your scope is not measuring the output voltage. To see what the output waveform looks like ...


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The biggest potential issue is that your negative supply is only negative with respect to that wonky quasi-split-supply point and usually that's not what you want in a DC-coupled circuit like a regulator that needs to provide an output and accept an input relative to ground. It might be okay enough for an AC-coupled (both input and output) amplifier of ...


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My initial approach might look like this. simulate this circuit – Schematic created using CircuitLab


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The regulator needs to have a capacitor, around 0.33uF nearby between input and ground. Also a small capacitor 0.1uF between output and ground is good practice. The regulator could be unstable without it, and without a little bit of load on the output. That may explain the current draw, and resulting voltage drop across R22. Speaking of R22, normally it is ...


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Hm, no. That motor was designed for exactly the shape of magnet used. You'll practically have no chance replacing that with something stronger. Then you'll change the cabling, i.e. you'll rip out all the hard-lacquered copper and wind something new. Which requires you to get a different cutout to hold your coils. At that point, you're not replacing ...


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