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1

But I'm a little bit confused about the output voltage of the relay. From what I understand from the datasheet of the relay, I'll need a 9V to drive the relay, but what will be the output of the relay? Try this GIF animation from this site: - It uses a 6 volt coil and switches 120 volts AC. Or maybe this is more your style: - Same coil drive voltage as ...


0

The coil and the switch parts of the relay are isolated from each other electrically. The relay can't have output voltage, you're meant to wire it like a switch into your 24V motor circuit with its own 24V power supply. The coil still needs to be powered by 9V (according to the relay specs), no matter the motor voltage.


1

With a peak voltage > 325V to a boost cap into a UF4004 less than 1 Ohm might be excessive capacitance. 100 nF is around 13 ohms at 100kHz and lower at the harmonics resulting in a peak current of 32A at 100Khz. But for the 1st pulse from a square wave, the current will depend on the ESR of the cap and series R for RC risetime. Ic=CdV/dt-V/ESR Consider ...


2

At the ends of the coil, 1/2 of the enamel is stripped off. That is not the correct way to make this type of motor. All of the insulation must be stripped on on one end of the coil and half on the other. Once you have corrected that, the motor should work, but you may need to give it a push. Once you have basic motor function, you can experiment to find the ...


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Yes. The motors will only pull the current they need, as shown by their voltage vs current graph (or the values you listed). You can connect them directly to the 3.7~4.2V battery only and their strength will change as the battery drains. You can use a series resistor like you would with an led if you want to limit the power drawn. Or you can connect them to ...


2

A resistor divider does deliver a voltage similar to the one you desire, just like you say and found out. But it also presents a series resistance, too. That series resistance then stands in the way of your motor. Here's a circuit that is the equivalent to your constructed circuit: simulate this circuit – Schematic created using CircuitLab It's that ...


2

But with no losses in efficiency this angular velocity should remain constant, right? No. Efficiency and speed relation to voltage are not the same thing. Let's consider an ideal DC motor wound with superconducting wire, with a wound field, that can be either in series with or parallel to the armature, driving a load torque, or twice that load torque. The ...


3

Yes your analysis is correct. In a 100% efficient motor the output speed would not change with load applied. As you varied the torque the current pulled from the electrical supply would change in direct proportion to the torque to maintain the relationship Pmech = Pelec. You can actually see from this how an real motor works, because the above statement ...


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I agree that the angular velocity of the system is lower with the disc attached because of the conservation of angular momentum. No. Conservation of angular momentum has nothing to do with it. Mechanical power = torque * angular speed in rad/s The 'torque' you are calculating by (presumably) motor power / speed, assuming the motor to be 100% efficient, is ...


2

When nothing is attached to the motor shaft, the measured power is entirely losses in the motor. One part of those losses is the power lost due to the resistance of the copper windings. That loss is called the copper loss or I squared R loss. There is also power lost in the iron due to hysteresis and eddy currents, but that might be negligible in some motors....


1

with 48V PWM pm on a brushed motor you are going to get 4 times the resistive and eddy current heating that you would have gotten with direct 12V drive. that might be ok. or might not. With 48V PWM on a 12V brushless motor it will break immediately. I'd reccomend a DC-DC converter, one with an output current limit may be able to start the pump without ...


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I was thinking of fitting a diode on the positive wire to the fans to prevent back voltage from the fans to the rest of the car. Would that work ... I bet it works, even if it seems a little overkill (big diode even if inexpensive), but powering the fan (and only the fan) directly from the battery would be a cleaner solution.


3

Probably the best solution is to power the fan directly from the battery (preferably through a fuse), instead of from the ignition. If you don't want the fan to ever be on when the ignition is off, just power the fan controller and relay coil from ignition. Like this:


1

For option 1, is there any precaution to be taken because the load is inductive rather than resistive? Not really, as long as the DC-DC converter can handle the motor´s starting current which can be pretty high. For option 2 : is it OK to apply such a high voltage on a 12V motor if the average remains voltage within specs? Is it OK if the motor is a ...


0

It is not possible to draw 5A with a 7824. In addition, it is not possible to regulate with just 1.8 V more. These regulators have a dropout limit of 2 V (typical.) Incidentally, series regulators are the worst choice as the excess voltage has to be dissipated as heat. Perhaps you can proceed more skillfully and not regulate the voltage, as this should be ...


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If a voltage regulator was the only feasible solution I'd be looking at a low-drop-out buck regulator like the one below using the LT8638S (for example): - The example circuit above is for a 12 volt regulator but it looks like it should be good for a 24 volt output at up to 5 amps without much trouble. I'm not recommending that device in particularly (...


5

Here's a bootstrapped H bridge driver circuit that should work OK: - Image taken from here. Focus on this part: - C1, C2 and D1 create a voltage at VB that is several volts higher than the 12 volt supply and, this ensures that MOSFET Q1 can receive a proper gate-source voltage and turn the MOSFET on into full conduction. Output HO will switch to a high ...


2

As mentioned before, this controller seems to be a SCR bridge. The SCR´s will be fired in a determined angle from zero crossing to get the desired set point in the controller. A simple equivalent of dc motor is a R + L and a EMF voltage proportional to speed. The "average" 90 V in the waveform is the EMF voltage. When a SCR is fired, you can see ...


2

With no flux, how can there be any torque on the conductor? The flux doesn't have to "cut through a conductor" to either induce voltage or produce torque or movement; it's quite sufficient (and sometimes preferable for other reasons) to have the flux totally surround the conductors and not come into contact at all. Comparison: An ideal transformer ...


2

How can speed not change with load? If Coil R = 0 and mechanical load increases, so does electrical current in order to generate the Torque to maintain the same speed. Motors have a constant V/RPM ratio for no load. (or kV/RPM) Yet conductors have losses so speed loss depends on the current loop resistance. (unless cryogenically cooled)


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