Replacing 6v Dc Motor with 12v Dc Motor for Better Speed [closed]

You are confusing voltage and power. The voltage a motor is designed to run at means nothing directly about how fast it can spin or how much power it can put out.To move your gocart faster, you need more power. That means a motor that can produce that power at the right combination of torque and speed, and a battery that can deliver that power. This has nothing directly to do with the voltage the motor is intended to run at. To some extent, higher power motors will require higher voltages because to get the same power at lower voltage requires higher current, which can cause some problem once you get up to a few 10s of amps for something like a gocart motor.So the answer is to get the right motor. This has to be able to deliver a certain power at your desired speed. There is probably some gearing in your gocart. By moving it forward manually a measured amount and counting the number of turns the motor makes over that travel distance, you can figure out how fast the motor will spin at your desired speed. Get a motor rated for the desired power at that speed. This could be a 6, 12, 24 volt or something else motor. It does not matter as long as you can supply that voltage at the required power, plus a bit to cover loss. Since you already have a 12 V battery, I would start by looking for a 12 V motor.

1. In a miniature jet engine, if I power the compressor from a DC motor do I still need to include a turbine?

What the other two said

2. What is the difference between armature voltage control and field controlled separately excited DC motor?

Variations in the magnetic field or flux are due to the changes in the field voltage as well as armature current. Resistance can be added to drop the voltage and further there is a decrease in the armature current. So, you can say that the major difference that you will be found between the armature voltage control and field controlled separately excited DC motor is changing or bringing an adjustment in two distinct sources.

3. How to control speed of dc motor using servoblaster

You can control the speed of a standard DC motor using PWM.PWM works my splitting each second into many time slots. The number of time slots per second is the PWM frequency. During each time slot the signal will be high then low. The percentage of the time slot that the signal is high is the PWM duty cycle. It's a gross simplification but for the sake of argument full speed is a 100% duty cycle, half speed is a 50% duty cycle, etc. Servoblaster generates servo signals. These have a PWM frequency of 50Hz. So each time slot is 20 milliseconds long. Servos typically accept signals in the range 1 to 2 milliseconds long. That is roughly the range which servoblaster will output. That means you can vary the dutycycle between 5-10%. I doubt that will be enough to even start a DC motor turning.You really need to be using a proper PWM generator. My pigpio will generate servo signals and PWM signals. pigpio is pre-installed with recent versions of Raspbian

4. How to create PWM DC-motor speed controller ?

For drawing your circuit and possibly designing a PCB if you choose , There is a free version for hobbyists which will allow you to draw a nice schematic easily. There is also a limited BOM feature that lists information about the component you placed on your schematic. The BOM creates a table of data or an excel spread sheet. Component listings (BOM - Bill Of Material) are easy to do, they are just a table of data that contain the designation of a component (R1, C2 etc), its value (100K , 0.1uf etc) and a part number from a source to buy the part. (DigiKey: 401-2000-ND or Newark: 5789K11 etc) PWM is easily done using a micro controller to provide the PWM . A variable resistor is used as an input to an ADC on the microcontroller as a means of changing speed. If you nee to boost out put power the micro can drive a single transistor or an H-bridge. Take a look at an arduino as a simple solution to get you started.

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The two resistors (R10 and R11) provide a voltage that the processor uses to detect when the motor is stalled.When running normally, the voltage at the junction of R10 and R11 will be around half the supply voltage. Probably have to subtract voltage drops across the transistors from that. When the motor stalls, it becomes basically a short circuit. The voltage at R10 and R11 will drop.The processor is probably using an ADC to monitor the voltage. The engineers might also have been real tricky and arranged it so that they could use a digit input to detect a stall. They could jigger it so that the normal operating voltage is above the level for "high" on the processor, and below the level for "low" when the motor stalls.C1 smooths the stall detect signal. It forms a low pass filter together with R10 and R11. The cutoff will probably be way lower than the PWM frequency. It also cleans up "trash" from the motor brushes.I am not going to try to guess the real voltages present at R10 and R11. That's going to depend on the transistors, the power supply, the motor, and how "hard" the processor is driving the transistors (whether they are in saturation or not. )Not everything you would need to calculate is there, and I am not sure I could do it anyway.Low voltage when stalled, somewhat higher when running normally1. Can I use a resistor to slow down a DC motor?You CAN use a resistor, but understand all you are doing is dumping power out the resistor to drop the voltage to the motor.If you want to go really slow, the resistor method will probably cause the motor to stall way before you reach your desired RPM.Using PWM ensures you get pulses of full torque, which allows you to drive the motor to really slow speeds.2. If I have 10 3000Farad capacitors at 2.7V wired in series. What would happen if I hooked up a dc motor?The voltage would decay very quickly. The motor current at 24 volts will be 4(746)/24=124 amps. The total capacitance is 300 F. dV/dt=I/C=124/300=.41 volts/sec. This is a rough approximation to figure out how long the caps will power the motor, but you can see it wo not run very long3. Overvolting a DC motor while keeping power constantThe since the speed is directly proportional to voltage, the motor will attempt to run at 133% of rated speed. If you are driving something like a fan for centrifugal pump, the torque load on the motor would increase to 1.33 X 1.33 = 1.77 pr 177% of rated load. The increased load will prevent the motor from going that fast, but it is still likely to be overloaded and overheat rather quickly. The driven equipment might not fare too well either. Look for or design a speed controller that will limit he voltage to 36 volts.It is also possible that the commutator will have arcing among the segments with increased voltage. The winding insulation will probably not have a problem.Re commentWith traction applications, the load torque will increase with speed to the extent that the vehicle is subject to aerodynamic drag. Increasing the voltage does not increase the torque capability of the motor. Therefore, any load that requires more torque to operate at a higher speed has the potential of overload the motor if it is operated above rated speed. With a traction application, I would think that you would need a speed controller. I see that you are intending to use a chopper. That should be configured to adjust and limit the speed and limit the current. Since it controls the voltage by controlling the duty cycle, but the peak voltage would still be 48 V. There may be no problem with the commutator, but you should look to see if there is any problem.4. Why is speed of a DC motor inversely proportional to the armature current?Because every running DC motor is a DC generator too.If you see the construction of a permanent magnet DC motor and generator, it will be the same. When you apply DC power to such a motor, the armature starts turning providing mechanical power. At the same time, the armature windings moving in the magnetic feild (essentially, a generator of sorts) generates an emf too, but in reverse polarity to the applied voltage. This emf opposes the applied voltage and reduces the current flow through the armature coils. The faster the armature revolves, the more the back emf generated will be. This reduces the armature current.It confirms to the law of conservation of energy too.When the rotating motor shaft has no applied load, the only energy spent will be as resistance and frictional heat through the coils and bearings. Since very little energy will be spent this way, only that much energy will be drawn from the power source. If the power source is of fixed voltage type, less energy drawn means low current flow throught the circuit.Why is speed of a DC motor inversely proportional to the armature current?.
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