indulis

75% duty cycle is 75% duty cycle and frequency has nothing to do with what the voltage will be. The average voltage will be the duty cycle multiplied by the input voltage. Motors are wound such that they have pole pairs. I have seen up to a 12 pole DC motor. The easy way to find out how many poles your motor is to count the "strips" on the commutator ring and divide by 2 (at least I think that will tell you, not 100% sure). There is a correlation between the number of poles and what the optimum PWM frequency is. Most small motors are PM (permanent magnet) The motor controller I built for my wind tunnel uses a PM 24VDC fan, and I had the "adjust" the swithching frequency to make the fan turn at really low speeds (low duty cycle...~3%).

Like many 3 terminal regulators, there is a reference voltage between the output pin and the adjust pin. In the case of the LM317, it's 1.25V

Ahhhhh yes....... I should have checked the data sheet first. Yup, since the reference voltage is between the output pin and adjust pin (not adj pin to common... I guess I've been around TL431's too long), any resistance in series will screw up the load regulation (not line regulation). As for the capacitor C1..... I have had circuits with a LM338 oscillate under certain line/load conditions without out it ("conditionally stable").

The .22 ohm resistor does not ruin the LM317's voltage regulation. D5 is another story as far as regulation... it's forward voltage will change with current and temperature. The point that is regulated is the "node" from which the feedback is taken. In this case the "output" side of the .22 ohm. The .22 ohm also helps a little with power dissipation. As the current increases, some of the power that would have been dissipated in the regulator, is now dissipated in the resistor. Also, don't be to fast to get rid of the capacitor C1 on the "adj pin". These regulators can and will oscillate sometimes without it!!! If you leave it, the worst that can happen is the transient response will be a little slower.

A power supplies output is only the voltage potential between the terminals. It's polarity is determined by which terminal is used as the reference.

The company I work for recently dumped PCAD in favor of PADS.... I liked both, but prefer PADS. I suppose it's what your use too. I did find it easier to make "intelegent" schematics in PADS when linking to the layout. As for simulators... Orcad, MicroSim, Intusoft and Workbench. My vote goes to MicroSim and Intusoft, but quite often I have convergence problems with both. Like anything else, you get what you pay for. The simulation is only as good as the models. I've yet to see anyone with a good overall model set. Most seem to be good in one or two areas, but poor in others..... for example, they have a great MOSFET and PWM selection, but the magnetics models stink!!! Intusoft is kinda neat in that they have a magnetics design program and the model can be linked to their simulation program.

I don't think it would be a wise thing to take a 180-200VAC signal, rectify it and try to charge a battery without a regulator!!! The batteries internal resistance is probably what's holding the voltage down. As for the diode voltage rating.... you are interested in something called PVI (peak inverse voltage). It is the max voltage that can be applied in a reversed bias condition before the junction will break down.

Can you be a little more specific about your problem, and describe the power supply further? In general, switching power supplies take a DC voltage and "chop-it" into a square wave. In the case of an off-line switcher, the AC is rectified, filtered and then "chopped". In a DC-DC it is just chopped directly. I this day and age, virtually all "chopping" is done with MOSFET's. Somewhere in the power supply there will be one or more magnetic elements depending on what type (topology) of power supply it is. If you have a 10VDC voltage and switch it on and off such that it is on for 50% of the time and off for 50% of the time, rectify and filter that signal, you will have a 5V output (assuming an ideal diode). So on-time divided by off-time plus on-time (period) gives you what is referred to as duty cycle, and duty cycle multiplied by the input voltage will give the you your output voltage. A feedback circuit is used to change the duty cycle such that the output voltage remains constant over variations in line and load. As for troubleshooting, more info about the supply is needed in order to suggest "a plan of attack".

First, let me reiterate that the concept I am referring to, is applicable to transient circuit analysis (not steady-state), and by definition only holds true for the instant in time that you are interested. Again, take a simple series RC circuit with a source and switch. To make it simple, let

Mr Klampfer Explain to me what you don't understand about the following statement? "While applying principals of transient circuit analyisis, capacitors become "SHORTS" with a series voltage source and inductors become "OPENS" with a parallel current source. "

14mA thru 5K is 70V. Is that what you meant to say??

"Of course the capacitor's voltage is at ground the moment power is applied to the circuit...." That's not the point.... you said, "You were taught wrong." He wasn't, and I just gave a simple example of how a capacitor can "appear" to look like a "SHORT" (short being the key word) to ground. While applying principals of transient circuit analyisis, capacitors become "SHORTS" with a series voltage source and inductors become "OPENS" with a parallel current source. Walid's application of those principals just wasn't 100% correct.

He wasn't "exactly" taught wrong, you just have to remember the conditions under which it is true!! Take a simple series RC circuit (no initial conditions i.e. 0 volts on the cap)with a DC source and a switch. If you close the switch at time "zero" the capacitor at that instant indeed looks like and behaves like a short!!!!!!! If you have a charged capacitor and you apply a transient, you would model it as a discharged capacitor in series with a voltage source that is equal in value to the capacitor voltage prior to the transient!!! In this case such a lage value for R4 it doesn't matter much, but a diode in series would block R4 from being in the discharge equation. If R4 was 100K, IT WOULD MAKE A BIG DIFFERENCE!!!

From what I remember, and that was close to 30 years ago, it's just like writing mesh equations..... if you use the same conventions everywhere, the worst that can happen is the result, relative to your chosen polarity/direction will be negative.

Without going into the differential equations, the solution and general form for charging is: Vcap=Vapplied*(1-e^-t/RC) and for discharge is: V=Vcap*(e^-t/RC) on the discharge, don't forget to include the forward voltage of the diode.

Be careful here……… mechanical and electrical resonance ARE NOT THE SAME THING!!! Electrical resonance is a second order phenomena (2 reactive elements). As for the dampening effect of a shorted voice coil, how about the induced field in the coil opposing the magnets flux. If you removed the magnet and shorted the coil… what do you think would happen??? NOTHING, the cone would continue to oscillate!!

Yes, that is exactly what I said above, but that's not what your earlier response was when taken/read literally. It would be helpful to throw in "the math", so there's no misunderstanding. You never answered my earlier question

OK............. let's call it a decrease in attenuation, which from a certain point of view is gain. As you go below 20Hz towards 0, the voltage across the resistor (in this case the speaker) will continue to decrease... this is basic "CR" circuit stuff, the transfer function is (sR1C1)/(1+sR1C1) and the corner frequency of the pole is f=1/(2*pi*R1*C1) (C1=coupling cap and R1=speaker "resistance"). You said: "The frequency response will drop -3dB at 20 Hz and be flat down to about 100Hz. -3dB is half-power." I don't understand!!!!! As frequency increases above 20Hz, the signal is not going to drop...... it's going to increase as it

I would move the LED and current limiting resistor to the collector and add a base resistor to the transistor. Also, while the output is at the minus rail, the output clamp diode is making the poor 741 work really hard!!!

I'm missing something here................... If a capacitive coupled signal is driving a resistive load, this is like a high pass filter!!!!! Yes, you do apply the Maximum Power Transfer Theorem and the corner frequency will be, f=1/(2* pi*R*C) (the -3dB point, a.k.a. half power point), but as the frequency increases from 20Hz up, the gain goes up, not down provided the resistance (inductive reactance of the voice coil) remains relatively constant over the bandwidth, because capacitive reactance is going down. What does the "equivalent circuit" for a speaker look like, and what are some "real world" values??

Yes you can. You need 2 log amplifiers feeding an "adder" and the output of the adder needs to be fed through an anti-log amp. But why "over think the plumbing"!!!!

The range isn't "really" what is you think.... for example the 200mV range is 199.9 mV and the 2 volt range is really 1.999V. It's the leading digit.

Not exactly.... a LED will draw whatever current you allow it to by means of the current limit resistor! As you allow more current flow, it will get brighter and as it gets brighter, the LED (diode) forward voltage increases as dictated by the diode equation. Therefore, the "exact" amount is a little harder to calculate with 2 variables that are changing, and are non-linear!! In the "old days" most LED's needed around 20mA to be "bright". Nowadays LED's that only need 2mA to be bright can be found if you dig a little.

Try.......... http://www.analog.com/en/prod/0%2C2877%2CAD633%2C00.html

Phase shift is the relative shift of current or voltage of a point in respect to some reference. As soon as you introduce a time varying signal and you have capacitive and/or inductive elements, you will have phase shift. What 3 voltages? How and of what, did you "set" the phase shift?