# Tag Info

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While the bare relay linked in the comments is the device the tutorial is using, the board you have already purchased should work just fine if wired correctly: Edit: Note that I only connected the ground of the sensor to the second GND terminal of the relay board to keep the diagram tidy. The positive and negative inputs to the sensor can be connected ...

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According to the linked ad my comment was incorrect. Input control signal voltage: 0V - 0.5V Low stage (relay is OFF), 0.5V - 2.5V (unknown state). 2.5V - 24V High state (relay is ON). <-- It takes a 12 V control signal. Input control signal high state current: 2.5V: 0.1mA 5V: 0.35mA 12V: 1.1mA <-- It will draw 1.1 mA on the control input. 20V: 1....

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My MOSFET was broken after all, I got a replacement and now it works fine!

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If I am understanding you correctly, you have 2 different voltages that you want to add together? So for example you have 3V and 2V and you want to add them to produce 5V? For this, you can use a summing amplifier configuration. There are 2 types. Inverting and non-inverting (pretty self explanatory). The non-inverting summing amplifier looks like this: ...

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This appnote describes, in detail, 2- and 4-quadrant motor control with regenerative braking for each direction. Link: https://www.roboteq.com/index.php/docman/motor-controllers-documents-and-files/documentation/application-notes/application-notes-1/33-an70614-1/file The short answer is you need at least 2 FETs to effect regenerative braking in the forward ...

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Voltage is not energy. It's only power (over time resulting in energy) if it flows through some resistance. When the upper switch is closed, current builds up due to the EMF on the motor coils, in some period of time (maybe tens of microseconds) it builds up to some peak current. You then open the upper switch and the current flows backwards through a diode ...

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You are using the wrong FET for your application. The IRFZ 44 is a "standard" gate drive FET, and will not reliably turn on with 5 volts of gate drive. If you go to a data sheet, you will find that Vgs(th) (gate turn-on voltage) is 2 to 4 volts. However, this only guarantees 250 uA of current, which is much too small, like 1/1000 of what you need. The fact ...

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> Is it possible to patch together a DC system with these random materials? YES PV1, 2, 4 appear to be close enough to 15V Vmp to work together. PV3 may be - measure Vo/c = out unloaded in full sun. If Voc > 15V it can also be used. Assume for now that PV3 = 15V, 300 mA. ___________________________________ Panel combining: It seems like that combining ...

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Quick answer: You need to connect two wires to the (+) side of your power supply, and two wires to the (-) side. If your physical supply has an output cable with only one pair of wires, you'll need to splice the wires together somehow. Connect all the (+) wires together in one group, and the (-) wires in another group. There are lots of ways to do this. ...

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I think there are some wires coming from the photocell that get attached to the power supply, and some wires go from the power supply to the relay. It sounds like there are black, white, and red wires already connected to the photocell...the black goes to the + side of the 12V supply, the white goes to the - side of the 12V supply, and the red wire is the ...

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I've learnt the meaning of firing angle only in the case when the input supply is AC. Figure 1. The upper trace shows the trigger delayed close to the end of the cycle. The resultant effective voltage is low. The lower trace shows the trigger close to the start of the cycle. This will result in close to full voltage. The relationship between phase angle ...

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That drawing is discussing a DC → three-phase AC inverter. The circuit is switching the MOSFETs off and on in a sequence that will generate three-phase AC. The angle measurements are referring to the timing of one output phase (Phase A).

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Well, it is an odd way of saying it needs 5V DC in the datasheet, but that seems to be what they mean based on the picture of the device itself. The connector is standardized, yes, and it is called a "DC barrel jack", or some variation of that. There is different sizes of the DC jack available, measured in inside diameter, outside diameter and length of the ...

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It needs 5V 1A power supply. It does not say what connector and polarity. Officially HDMI sink devices may not consume more than 10mA (50mA when turned off) and sources must limit current to 500mA, so this HDMI switch violates the specs in many ways and simply can't work properly without external power supply.

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Most brushless DC motors used for RC hobby use are Delta wired. Delta wiring provides a higher KV value and lower torque by a factor of 1.7 when compared to Wye connection.

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The base-emitter current is a fraction of the collector-emitter current, but a non-zero fraction. For a larger collector-emitter current you need a large base-emitter current. The RF energy might be too weak to supply the larger base current required if you were to try and maximize your collector-emitter current in a single stage. The 1 megaohm resistor is ...

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Because trying to measure the voltage changes it! Your meter has a finite input resistance (usually 10 megohm). Pin 2 is not connected to any of the other pins, so there is no reason to expect it to have any particular voltage relative to them. When you connect your meter between 1 and 2 it pulls the voltage between 1 and 2 down to zero. Similarly when you ...

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I am struggling my self at same situation as yours. What I have identified is that you have to make sure you power supply has current enough to cover more current than your whole system need. I am planning to build up a 8 cam cctv and I definitely will use a terminal block to connect all power wires into it and in the other side just two wires from the ...

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A simple constant current source can be made with a few components, to ensure long working life for the laser. The red LED (not the laser) has a constant voltage drop of about 1.4V and serves as a voltage reference for the resistor R1 which sets the current. Alternately a pair of 1N4148 can also be used in series. The base-emitter drop of 0.7V is subtracted ...

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First, the goal is not to reduce the voltage for the laser diode (just like an LED). The diode itself will determine the forward voltage drop across it (2.4Vin this case). This is not a choice. you don't get to choose to run the diode at a higher or lower voltage. The diode will do whatever it can to make the voltage across it 2.4V when you apply a voltage ...

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Your two metal blocks (assuming they're insulated from each other) make up a capacitor. Yes, current will flow as the capacitor charges. Then when you remove the source, the charges will find some path, however high the resistance, and the capacitor will discharge at a rate determined by the capacitance and the resistance.

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There will be a brief flow of current when you first connect the battery to the metal blocks. That current will charge the capacitance between the blocks. Once that capacitance (and other stray capacitance) is charged, there will be no further flow. If there is no leakage current between the blocks, the charge on the capacitor formed by the blocks will ...

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It depends on the total current and what the cable is rated for Let's say each 12 V LED strip consumes 12 W, then each LED strip has 1 A flowing through it. For each LED strip, the ground wire needs to carry 1 A, from left to right in your schematic it will show 1 A, 2 A, 3 A and 4 A at the bottom. Given my 12 W example you need a cable that is rated for ...

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If the motor 3.25 A locked rotor current and the 6 A power supply output current rating are reliable, that should be ok. A lot of power supplies have over-current protection that is quite sensitive and shut down when a motor is switched on. I would be concerned about the motor current at full speed. Your 5.9 V test would have only driven the motor at half ...

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Stall current is(theoretically) the maximum that the motor will draw, so your power supply should be more than sufficient for your use. Some power supplies don't use very large output filter capacitors, though, so it never hurts to add an additional somewhat-largish capacitor across the output of the power supply and/or input of the motor, at least a 16 ...

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Only the manufacturer knows why these connectors are rated for 12V usage. If you use them at 24V you are on your own. The connectors might fail immediately or might never give you any trouble. If they fail they could fail into a safe mode, simply cutting off the flow of current, or they could catch fire. We can give you an opinion about using the connectors ...

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Looks like it might be a female cable version of a snap and lock DC power connector Edit: perhaps it's some sort of custom version of the KPJX-CM-4S?

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So I pulled the cover off one of these a couple of months ago with the same plan in mind, this is our second door, and the batteries only last 3-4 months. The system seems to run the batteries both in parallel and in series. The 6v of batteries in series seem to run the electronics and the 3v in parallel seem to run the rfid loop, so supplying just 6v DC ...

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The capacitor will charge to the voltage you have on the load (including the wiring). The resistor R1 will create an additional voltage drop (that's the whole purpose of a shunt resistor). So on the right side of the resistor you will have the load voltage $V_{load} = V_{C1}$ and on the left side of the resistor you will have $V_{load} + V_{R1}$. This ...

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The GPIO pin of your micro-controller is an high-impedance input. It will allow little to no current to flow through the circuit. On average, the capacitor will not let the current go through either. As drawn, your circuit will only measure the current used to charge or discharge the capacitor, a pulse that will last: $$\tau = R \cdot C = 150 ns$$ (the RC ...

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The answer is, it depends on what’s determining the DC current. Any AC current will also flow though those circuit elements. For example, if the DC current is being driven by a high quality current source, that source will adapt to cancel the effect of the transformer. A more extreme example: take the case of $I_{DC} = 0$ because it’s an open circuit. ...

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Yes you can do this. The transformer must work at the frequencies of interest and you must keep the DC current through the secondary well below the rated peak current or core saturation will cause unwanted distortion of the ripple.

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First, there will be just one current in the circuit you show. The current into the coil must be the same as the current out of the cell. You can do pretty much exactly what you have drawn. Use a transformer with the secondary winding connected as shown, in series with your dc source. You can use a function generator or other sine wave source connected to ...

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