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5

I am not endorsing wiring the L298Ns in parallel, but simply discussing the issues involved, and how it might be accomplished. As others have pointed out, an easier solution is to purchase a more robust driver. Simply wiring the L298Ns in parallel will cause problems. A partial schematic of one of the motors, and one of the drivers, driven in one direction ...


4

You need to specify height or weight to get an answer. Here I have assumed a height. Motors efficiency is somewhere around 90-95%. I doubt that what you are attempting to do is practical. Potential energy of water reservoirs has been used to store energy on a large scale. [formula from wikipedia]


3

I would have to go with a servo motor. Possibly on a worm gear driven rack and pinion. The servo control system, I would include absolute positioning sensors, (opto wheels) so that the motor control optic wheel is fallows the steering wheel optic wheel with precision. I wouldn't recommend a potentiometer as a command potentiometer, because of mechanical ...


3

Is it simply a matter of "half the voltage = twice the motor inrush" Yes it is. For a given load power, the current taken by a load in the US will be about 2.1x higher than an equivalent load in the UK. This not only applies to continuous load current but inrush current too.


2

What causes the motor to burn up is mainly current (which is proportional to torque in DC motors). Running at higher RPMs means higher frequency currents running through the motor since the currents change direction more frequently through the windings due to commutation, and this induces eddy losses in the core which causes additional heating but this is ...


2

simulate this circuit – Schematic created using CircuitLab Figure 1. Modified schematic. The "sinewave" is PWM unless it has been filtered in some way (which you don't want in this application). When the PWM goes high (5 V) R1 provides almost 10 mA to the base of Q1. This should be enough to drive it into saturation so that the collector (...


2

What causes waste heat in the motor? Is it purely proportional to the current used to drive it, or is it current and voltage? In other words, if I drive it at 10A/10V, will it get just as hot as if I drive it at 10A/20V? It is both. Current causes a loss in the winding resistance, commonly called 'copper loss', which is proportional to current squared. In ...


2

I'm confused though as on the datasheet there is no voltage mentioned, only a max phase current of 2.1A and same holding torque. But on the line right under the amps/phase of 2.1A, there's a line that says the coil resistance is \$1.6\Omega\$. \$2.1\mathrm A \cdot 1.6\Omega = 3.36 \mathrm V\$, so there you go. I know that voltage applied to a stepper, ...


2

For the electrical energy, $$\begin{aligned} 100\ \text {kWh} &= 100 \times 3600 \ \text {kWs} \\ &= 360,000 \ \text {kWs} \\ &= 36 \times 10^4\ \text {kJ} \\ &= 36 \times 10^7\ \text {J} \end{aligned}$$ (Note proper capitalisation of SI prefixes and units.) Potential energy is given by \$ E = mgh \$. Rounding g to 10 m/s2 we get \$ mh = ...


2

There are two questions here 1) what does rated current mean Does rated current represent the stall, the starting or the rated power point?. This all depends on the motor in question. From this particular motor we know the rated power (2200W) rated velocity ( 4500rpm) and the flux linkage ( 0.0976Wb) From the rated power and rated velocity the torque at this ...


2

It's not optimal but there is a way to run small 3-phase induction motors at reduced torque from a single phase supply. It's called "Steinmetz connection" (Steinmetz being one of the developers of said motors). In short you can use a capacitor to create a 'fake' phase from the supply line. Search the web for formulas and details.


2

Your question doesn't make it clear but I am assuming that you want to motorise the steering rather than provide power to drive the vehicle. Stepper Most stepper systems don't have any position feedback. This means that the control system won't know if it has lost position due to a knock or high resistance while driving. On power-up the steering would need ...


2

Will the following two scenarios give different results on the wear/safety of a consumer-grade electric motor: Yes, but what difference there will be is dependent on too many variables to say which of your two scenarios will result in more or less 'wear/safety', or whether the difference will be significant. For example:- A has 20 times more starts and ...


1

Usually motors are capable of putting out constant torque over a wide speed range. But power increases with RPM. That is because of the mathematical relationship between power and speed and torque. Power = torque * speed * k. But if we convert torque to newton-meters and speed to radians/sec then k will be equal to 1. This means that we can easily calculate ...


1

A motor comparsion chart like below would give you a rough idea of the torque vs speed characteristic. The above chart is from the following Q&A where you can find more useful info: How to control DC motor speed by motor driver with PWM input? Asked 9 months ago Active 3 days ago Viewed 915 times You might also find the above post's Appendix B - PWM ...


1

You probably have a ground loop, and it's causing false signalling from the Arduino to the motor driver. Basically, if you look at your diagram, you need to mentally substitute a string of resistors and inductors for all of the wires. The longer the wire is, the larger the resistance and inductance -- and ten meters is entirely too long for a single-ended ...


1

How can I determine for certain which pulse width corresponds to 0° and 180° using Arduino? You cannot. To begin with, no hobby servo, even expensive ones, have exact 180° travel. Cheap servos almost guaranteed to have it different. Also, some cheap servo electronics depends on frequency, so 0 position at 60Hz can be different from 0 at 65Hz. if I were to ...


1

As far as I can read the motor specs, it requires a 6uF capacitor. Your 1uF capacitor that is salvaged from a microwave oven is therefore not suitable.


1

Is it okay to multiplex these drivers with pulling all the GNDs HIGH No! According to the L298 datasheet, input pins must not go more than 0.3 V below ground. Another possible problem is that each motor driver would need its own fully isolated power supply to keep the grounds separate. If the motor drivers had opto-coupled inputs then you could do it, ...


1

All torque is proportional to current BEMF reduces applied V from velocity so holding torque reduces slightly with rising step rates. Voltage affects current from I= V/DCR(=1.6ohm) V=L(=3mH)dI/dt or I as the integral of voltage.


1

As has been suggested, datasheets often not only explain how to use the chip but also offer example circuits. There are also websites that go into great detail on how to use the L293 such as All You Need to Know About L293D In any case, the L293D is a set of half-H bridges and there are some general concepts that are true for all half-H bridges. Think of a ...


1

I think you have slightly misunderstood the wiring schematic. Typically pin 8 and 16 are the main supplies that will be driving the motors. The PWM signal from the arduno will often be connected to pin 1 or pin 9, depending on witch H-bridge you are using TI's driver comes with a wiring schemattic you can use: https://www.ti.com/product/L293D


1

I was thinking of purchasing a supercapacitor bank and connecting it in parallel to the converter's output. My understanding is that the converter will slowly charge up the caps initially, and once the motors gain momentum, it will draw the in-rush current from the capacitor bank. Once the motor starts turning, a back-EMF is generated in the motor due to ...


1

Using a supercapacitor is reasonable. Just make sure your power supply doesn’t complain about: charging current during the initial start-up back-powering when PSU is off Why the concern? The capacitors will discharge through the power supply when primary power is removed. Diode isolation can prevent this. As far as charging current, a large supercap bank ...


1

A few thoughts: There's no capacitor in there. Remember that the AC voltage drops to zero twice every cycle of the mains. A "capacitive dropper", using a capacitor to limit the current, would generate a lot less heat than a big resistor. It's a very common circuit design, often used for driving LED lamps. The circuit may generate an apparently ...


1

Spike closed the other question for lack of clarity. You may find 6 uF film caps like this or close to it in any machine parts distributor. Schematics for “Start and Run” caps are easily findable and intended to compensate for the poor power factor in each state. The start cap also reduces the grid voltage drop by boosting a power factor corrected current ...


1

Your first question was answered very well by @DKNguyen. Is there any way to estimate the maximum current/power the motor can safely take if this is not published by the manufacturer? Yes and no. All the RC motors pretty much use the same magnets, and the same construction. So similar-weight motors have similar power-handling capabilities. While I ...


1

I answered a similar question regarding optical encoder accuracy here: PWM accuracy vs speed control accuracy Minimum resolvable angle: \begin{equation} \theta(1) = \frac{360}{4000}=0.09deg\\ \end{equation} Sampling Period: \begin{equation} T=0.001s \end{equation} Expected error bounds: \begin{equation} \omega_{error} = \frac{\theta(1)}{T} = \frac{0.09}{0....


1

TL;DR timing There are 8 6 combinations of 3 phase vector control ABC ACB BCA BAC CAB CBA The correct two vectors must be swapped to match the independent phase reversal for 2 motor phases. Only 1 sequence is correct in each direction. Your bad direction may have the wrong combination. I verified this on a 3ph 100 W fan recently which I modified to be ...


1

Speed At 4.8V 0.25 sec / 60° Speed At 6.0V 0.22 sec / 60° At 4.8V if it takes 0.25sec for a 60 degrees shift then for a full shift (360 degrees) we will need 6*0.25 = 1.5s which is the period of the shift. f = 1/T = 1/1.25 = 0.8Hz The same way for 6.0V We don't have enough info for the duty cycle of the motor.


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