proportional gain on the step response of a position control DC motor?
85674
If your question is about the effect of a change in Proportional gain on the step response of the Position Control of a DC motor, Then the Answer is that the change will result in a change in the speed of response to a step input.
398218
Successive approximation A/D's work similar to a binary search algorithm.  The result can be very accurate if the A/D is designed correctly and the rate of the analog signal is relatively slow compared to the clock of the A/D, or a sample and hold block is used.  \nhere is an example of the operation (for this example a 4bit and that the input voltage is 1.2V):\n1.  using a DAC apply 1000 as a reference the  "1" is the test bit (say 1000 =1V)\n2.  Compare analog signal to this reference (result true) therefore the test bit is kept\n3.  Now apply the next test bit and the kept bit 1100 (the one in the 4's place is the new test bit and the one in the 8's place is the kept bit 1100= 1.5V)\n4.  Compare analog signal to this reference (result is false 1.2V <1.5V) therefore test bit is not kept\n5.  Now apply the next test bit and the kept bit 1010 (the one in the 2's place is the new test bit and the "10" in the 8,4's place is the kept bits 1010= 1.25V)\n6. Compare analog signal to this reference (result is false 1.2V <1.25V) therefore test bit is not kept.\n7. Now apply the next test bit and the kept bit 1001 (the one in the 1's place is the new test bit and the "100" in the 8,4,2's place is the kept bits 1001= 1.125V)\n8. Compare analog signal to this reference (result is true 1.2V >1.125V) therefore test bit is kept.\nResult is 1001\n\nHope this helps
807450
For a wye-connected machine:\nThe phase voltage (measured between a-b, b-c, or a-c) is (SQRT(3))* the line voltage (a-n, b-n, or c-n). In this case, connect your invertor across a-n, b-n, c-n (this may mean breaking the "common" neutral connection).\n\nFor a delta-connected machine:\nThe phase voltage equals the line voltage. In this case, connect your invertor across a-b, b-c, and a-c.\n\nWhich polarity you choose for the three connections will determine rotation - make sure your resultant rotation matches that on the machine nameplate, or you'll likely burn it up due to insufficient cooling...
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You probably want alpha, the common base current gain. In general, current gain is designated "beta," "alpha," or "gamma," for transistors connected in the common emitter, common base, or common collector configurations, respectively.  Beta = di(c)/di(b), the rate of change of collector current with respect to change in base current.  Alpha =di(c)/di(e), rate of change of collector current with respect to change in emitter current.  Gamma = di(e)/di(b), rate of change of emitter current with respect to change in base current.  Also, alpha = (beta)/(beta - 1), beta = (alpha)/(alpha - 1), and gamma = beta + 1.  For a detailed explanation, check out : http://www.tpub.com/neets/book7/25f.htm\n\nAnother possible answer: "amplification factor" for any amplifier circuit, regardless of configuration, or whether it uses tubes, BJTs, FETs, etc, refers to its voltage gain, dv(out)/dv(in), the change in output voltage with respect to a change in input voltage.
312735
Try a damping circuit.
149491
This is a "rule of thumb" that is the average of a lot of kinetic data.  To blindly apply it is unwise, but as a zero order estimate, it should work OK.  Detailed measurements must be done on your system of interest to find out true kinetic parameters and their sensitivity.
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The voltage at which the attractive forces between the capacitor plates overpowers the fasteners.\n\nTake a look\n\nwww.math.ubc.ca/~ward/papers_ps/memspap.pdf
755501
you need to regulate 12 v down to 5v.\n\nYou could use a common 555 timer, and adjust the delay with a potentiometer for the Resistor in the RC network.
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d: Subjected to a small voltage.
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http://www.allaboutcircuits.com/vol_6/chpt_7/5.html\nhttp://hyperphysics.phy-astr.gsu.edu/Hbase/electronic/nandlatch.html
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Easy!\nIf the input is the same(1.1kW) it depends on their performance.
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Depends how the  maschine is calibrated
275517
- A DC brushless electromotor is mainly a step-motor with a rotor made out of permanent magnet, and a stator made out of coils (mostly 4 coils, but it depends on the type of step-motor). As this is a step-motor, it needs some kind of regulator, which is made on a little electronic board. So... a DC brushless motor is made out of 3 things: \n\n1 Electronic Circuit Board \n2 Coils \n3 Permanent Magnet \n\n- Electronic Circuit Board and Coils: \n\n- This part is doing the trick - rotating the permanent magnet. There are in most every case only 4 coils in a normal fan, grouped as 2+2. The first group (let's call them the North-South group) is connected to one electrical circuit on the electronic board. And the second - East-West - is connected to the other electrical circuit. \n\n- On the electronic circuit board is located one small device that is most responsible for this trick. It is called Hall-probe (or Hall-sonde), and its location is just under the permanent magnet. \n\n- It's used to detect the south pole of the permanent magnet. The Hall-probe regulates two different electric circuits driven by two transistors, and depending on the pole it detects, it drives the proper circuit. With this type of regulation, it is possible to create a rotating magnetic field, and rotate the permanent maget located around the coils. \n\n- A permanent magnet has two-poles - North and South. Let's look what happens when the fan is powered. \n\n- Depending on the location of the south pole, Hall-probe drives one of the circuits. So, let's say that Hall-probe is sensing the south pole. It drives one of the circuits (let's say north-south circuit). \n\n- When the north and south coils are powered, they produce the magnetic field. This field is of the same direction in both coils, but... :) the Magnet hasn't got the same direction of the magnetic field in both poles. It looks like one coil is pushing the north pole away, and on the other hand, the second coil is pulling the south pole to itself. \n\n- Now, when the magnet is moved, the south pole goes away from the Hall-probe, and as it isn't sensing it anymore, it drives the second transistor (second circuit). With that, east-west coils are turning on, doing the same job as north-south coils did before. \n\n- So, basically the trick was to produce the rotating magnetic field. And it was done very cleverly! :) As coils aren't made as a moving part, it is obvious that there are no brushes, so that's the part where the name came from - 'DC brushless'. \n\n- Permanent Magnet: \n\n- This is the mostly known part of the electromotor. I think that I don't need to explain much about it. It has two poles - north and south, and using the external magnetic field it is possible to move the magnet around. In a fan, the magnet is circuit-shaped and it rotatrd around its axis. \n-------------
567501
a DC motor loaded with a semi-circle iron piece in its shaft
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
497883
Hey lets make it simple...\n\nwhenever u use only proportional controller and tune it to good gain then also u will have an offset error i.e ur Process value will never reach the setpoint, so if u introduce I (Integral) controller it will reduce this offset error or even makes it zero, but u gota tune it porperly, If u increase more time valve of I controller they system will become slow so be care ful if want more details then refer to  http://en.wikipedia.org/wiki/PID_controller
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i=C(dV/dt)\n\nwhere\n i= the instanious current through the cap\nC= the cap siz in farads\ndV= the change in voltage\ndt= the change in time\n\nI use this equation in my op-amp class where were are doing PWM power supplies and motor drivers right now.