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#1
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OK so I'd like to just use the Mach PWM spindle output go thru a low pass R-C filter and run my VFD from 0 to 5 volts. Has or can this be done and what pulse freq. , resistor size and capacitor size should I try. I saw a post(72321) for a homemade H drive where AL sugested 10K and 0.005uf for the pwm going into it. And I found a simple R-C Filter cutoff calculator at http://www.muzique.com/schem/filter.htm but I'm not sure if I should shoot for the pulse freq or lower freq. I really would like to give the simple solution a try before buying a step or pwm to analog converter. Thanks Drew |
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#3
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| Drew, Think about this not based on the cutoff frequency but based on the charge/discharge time constant. A cap will charge to 63% of it's max voltage in 1 time constant. Conversely it will discharge to 37% of it's value in 1 time constant. In 5 time constants the cap will be fully charged or discharged. (check out http://www.kpsec.freeuk.com/capacit.htm) A time constant is defined as the product of the resistance in ohms and the capcitance in farads. So the 10K resistor and 0.005uF capacitor would have a time constant of 10E3 ohm * 0.005E-6 F = 0.00005 S or 0.05mS. Assuming the output current of the PWM will be limited limited enough by the 10K resistor the cap will be at 63% of the PWM voltage in 0.05mS. The output of the RC circuit could be calculated using a spreadsheet and the On/Off times of the PWM. But the easier approach is to get a potentiometer and several caps from radioshack. Start with the 10K / 0.005uF setup with a resistor to simulate the load of the VFD. Look at the voltage across the load resistor with a DC voltmeter. Vary the pot and see what happens. Then make small changes to the cap value and see what happens. Also if you are using this in a non temperature controlled environment forget the RC method, the swings in temperature would make the circuit useless. Good Luck, Gerard |
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#4
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| Hi, if I understand the problem correctly you are using a PWM output, converting it to a DC voltage and using this Voltage to control the Motor speed.. Gerard is quite correct that the answer to this is in the RC- time constant. However I think for what you are doing the values are not that critical. If you choose a RC time constant that is way too low, there will be significant ripple at the input to the VFD. and I guess there is no telling how it will react to it! Possibly this would cause a bit of speed jitter at the spindle. If you choose a RC time constant thats way too high, then, the reaction time of the circuit to changes in PWM frequency will be slow. in other words when Mach decides to change the spindle speed it might take the system a few seconds to settle at the new speed. If it were me I would try a time constant of 50 times or 100 times your PWM period. So for example Mach gives you a 2Khz PWM signal.. then the period is (1/2000)= .0005... so if you take a RC time constant of 100 times this ie 0.05 seconds you should be fine, even if the values double or half over temperature. so somewhere in the ball park of 10k with 5uF. If you find theat the system is reacting too slowly you can reduce the C or increase the R. one other thing to check.. try and use an 'R' value thats at least 10 time smaller than the input resistance of the VFD. So in your case, if you are using a 10K.. the input resistance (of the whatever analog input on the vfd) should be greater than 100K. Hope this helps, Bernard Bryce |
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#5
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| Oops.. re.." If you find theat the system is reacting too slowly you can reduce the C or increase the R." That should have read "If you find that the system is reacting too slowly you can reduce the C or the R. Bernard |
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#8
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Go to this link. http://ww1.microchip.com/downloads/e...tes/00538c.pdf It covers the theory and math you need to design you LPF. Remember that your max output voltage (at PWM = 100%) will only be as great as your port voltage unless you use an opamp to scale (amplify) the output. You will only get a max of 3V if your port has a 3V output. To complicate things more if your VFD doesn't have a high impedance input it will furture reduce the max output from your LPF without the opamp buffer by Vin * Rvfd/ (Rvfd + Rlpf) = Vout : Where Vout is the output voltage of your parallel port or breakout board, Rvfd in the input resistance of the VFD, Rlpf is the value of the resistor in your LPF and Vin is the voltage at the input to the VFD. Remember to use an opamp designed for a single supply voltage. |
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