The sense resistor is put in series with the coil. There are 2 of them, one for each coil.Quote
MeltManBob
I looked at the datasheet some more trying to figure out how the supply voltage effects the output voltage even though they are not directly linked and couldn't find much. The trip current is set by Vref/(8*Rsense) and Rsense appears to be a combination of 2 external resistors. Basically what I'm not getting is that I've read that if you increase the supply voltage then you can drive the motors faster to the trip current but I'm not seeing where the correlation lies since the supply voltage is not the voltage used to calculate the current. Otherwise there wouldn't be any benefit of using a 24v supply over a 12v but I know that it is better to use a higher voltage.
To see the correlation you need to do the math. Just look at one coil of the motor alone. You put a voltage over it. How much time does it take for current to grow from zero to the level set by the pot. The function is I(t)=V/R*(1-e^(-t/tau)), and we put condition that I(t)=0.5A (setpoint), and then we know all the rest just need to solve for t. And that t will say how much time it takes for current to go from zero to 0.5A. Thats all. Put the following in wxMaxima:
"--- rise time single ---";
kill(all)$ declare([t,R,L,tau],real,[Fs],integer)$ assume(t>=0,Fs>0,Ts>0,R>0,L>0,C>0)$ ratprint:false$
"****date****";
R:19$ L:0.032$ tau:L/R; setpoint:0.5; V:12;
"****solve ****";
Curent(timp):= (V/R) *(1- %e^(-timp/tau)) ; Fs:1/(Ts) $
Ts:solve(Curent(timp)=setpoint,timp),numer; Fs,numer;
"*** end ***";
And change the variables, then hit the enter from the keypad (not the usual enter) to recompute. If V=12V then ts=0.0026 seconds. If the V=24v, then ts=8.4..*10^−4 seconds. It means with 24v supply, the currrent reaches 0.5A mark much faster. Coils energize faster. We are interested to make this as fast as it can be, so higher voltage is better because of that purpose.
Then, put the specs of a normal reprap motor, that is low inductance and low resistance, try something like L:0.003 and R:2 and V:24 and see that difference ts=6.3*10^−5, now that is different, and suddenly that time period is equivalent to 16kHz instead of 378Hz and 1.2Khz like it was in first case.
Quote
MeltManBob
Anyway I'd like to reiterate that I would like to make sure the rest of my math logic outside of the drivers is good to go or not that way as we continue to hash out the details of the driver I can keep a big picture perspective. Ultimately I know that a faster time constant for a motor would appear to be best because the current/torque can build up faster BUT being able to actually calculate these things would let me figure out if a more powerful motor might be better even if it has a slower time constant but more than made up for it in torque. For example a smaller motor with a faster time constant that could reach it's rated current within some period of time 'X' compared to a larger motor where in that same time period 'X' it only reached 50% of it's rated current/torque but that torque being higher than the smaller motor. In that situation the more powerful motor would actually produce more torque. Thanks again!
The mass, acceleration, force, ratio of pulleys, and that part is pretty easy and clear. But the deal with the motor is something else. It does not mean you can take the number written as the holding torque, multiply it with step angle and have your output torque. First, the motor you mentioned, driven at 12v would need a time period of 0.84 miliseconds just to reach 0.5A. That time frame is much too long. As a result, the current wont ever have enough time to reach 0.5A. As a result of that, the average current will be very low. And because the flux is proportional to the current, then the flux is low too. Then the motor has actually no power. It does not matter what datasheet says about holding torque at this point, that motor will not output anything that resembles that torque number.