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The cheap L298 modules don't have a chopper or current setting so you have to add a big power resistor in series with each winding if you want relatively high performance. You could also lower the supply voltage to about 5V from 12V which would work, but the motor would not be able to accelerate as fast. To get 2V with a 12V supply, you would need to throw ...

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It pays to read specs. Standard Voltage: 2V Phase current: 1.2A Phase resistance: 1.7 ohm The Hi+Lo side drivers have some specs which are roughly equivalent to 1 ohm per side, so you have... Pd= 5V^2/3.7=6.75W per phase x2=13.5W with 2/3.7=54% of the total power dumped into the heatsink which based on about 1sq"/W (my rule of thumb) makes this ...

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Your motor has a total step of 600. It means, for a complete revolution, you have 600 steps. Hence the resolution appears to be 360°/600 = 0.6° per step. This is the resolution in what is called FULL MODE. However, your motor can works in STEP MODE 1/8. That means that you can move the motor by 1/8 of a step. Hence, you can achieve 8 more steps and thus ...

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As I commented on the question, you're probably better off with conventional relays given that: their cost scales better as you go for larger/more powerful motors they are less complicated (since you stated that you've never designed a circuit before) Having said that, if you wish to pursue the solid state approach then you're probably aiming at (MOS)FETs ...

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Comparing the specs of the two motors side-by-side, there are some significant differences: Model 17HS13-0404S1 datasheet Model 1704HS168A-OB datasheet Both are standard NEMA-17 size mounting plate. Both are bipolar connection. Both are rated for 12VDC operation. Specification | Model 17HS13-0404S1 | Model 1704HS168A-OB | Amps/Phase | ...

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I found out that this Spike/swing is also present on the supply voltage... The swing on the supply voltage looks basically like the one over the shunt, but inverted. This is proof of current going through both upper and lower FETs rather than from lower FET Gate drive, but it is not 'shoot-through'. It is caused by Drain-Source capacitance, it's ...

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