hubble86
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Hello, I have got a small 4 wire 200 step 2-phase stepper motor which I wish to use a 10mF 63V capacitor. I am getting around 6Vrms (open circuit) at each of the two coils when rotated at around 350 rpm. How do I step up the voltage and charge the capacitor with good efficiency(low power loss as a result of I2R heat, if any)?
Thanks.
If you are getting AC from the motor, then you can use a voltage multiplier to boost the voltage, while turning AC into DC.
A boost converter accepts DC, and outputs DC.
Obviously it is important that you find out whether the motor produces AC or DC. Try hooking up two parallel led's to it, in opposite directions.
It's quite obvious that the output voltage must be AC, simply because the average voltage across an inductor (= a motor winding) must be zero. Only a DC motor/generator with commutator can output DC voltage.
I don't believe that the exact waveform matters much for your purpose, assuming a sine voltage can't be completely wrong, even if it's somehow distorted.
Besides open circuit voltage (e.m.f.) you would want to determine the short circuit current respectively the windings impedance. This gives you an idea of expectable output power. You might find windings R and L numbers in the motor datasheet (R can be easily measured of course).
I presume that the motor inductance forms a considerable part of motor impedance at your intended operation frequency. The L/R corner frequency can be expected in a few 10 Hz range, but you are operating the motor at about 300 Hz if I calculated right.
But why don't you check with an oscilloscope?
Because I am not having any. Is it okay to use pc based scope like given here
I only know the coil resistance and nothing else about the motor How to find the max current produced by motor?
Consider my comments about windings L/R in post #8. The current at 200 Hz will be limited by motor inductance, unless it's compensated by a C load.
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