Note: Minibot current calculation vs battery voltage

I just want to review the equations for motor current in terms of spec motor constants when the voltage is other than the spec 12 volts for the Tetrix motor.


tq_nm = output torque

tq_stall = spec stall torque at 12 volts

w_free = spec free speed at 12 volts

i_free = spec current at w_free speed at 12 volts

i_stall = spec stall current at 12 volts

V = motor voltage

kt = motor torque constant

ke = motor speed constant

R = motor resistance

Lets start with the  pure resistive (no inductance assumed)  motor current formulation :

i = (V- ke*w)/R

Substituting known values at the spec 12 volts , and using R = 12/i_stall

ke =(12- i_free*R )/w_free = 12*(1 – i_free/i_stall)/w_free

plugging this back into equation for i gives

i = (V- 12(1-i_free/i_stall)*w/w_free)*i_stall/12

= i_stall*V/12  – (i_stall-i_free)*w/w_free  or

1) i = i_stall*V/12( 1  -(1-i_free/i_stall) * w/ (w_free*V/12))

Now with a little manipulation we can write this in terms of parameters that might have been spec’d at some other voltage V.


2) i = i_stall@V*( 1 – (1-i_free@V/i_stall@V)* w/w_free@V)

where  i_stall@V = i_stall*V/12

i_free@V = i_free*V/12

w_free@V = w_free*V/12

In other words if we just multiply all the spec values by a ratio V/12 and plug them into the  current normalized  formula  2) we will get the same results as using 1).  This is artificial for i_free and w_free .  i_free  doesn’t change with voltage… ie it is based upon the drag torque and since kt is independent of supply voltage then i_free must also be.

Also, w_free@V is not just w_free*V/12 it is slightly different.  It is ok if i_free is << i_stall.   The exact formula can be derived  from equating i at both free speeds.

So i_free@12 = i_free@V

12-ke*w_free = V – ke*w_free@V

so w_free@V = (V-12 +  ke*w_free)/ke  and plugging in for ke gives

w_free@V = w_free*(V/12 – *i_free/i_stall)/(1-i_free/i_stall)

One can see that as i_free/i_stall -> 0 then w_free@V -> w_free*V/12

For the minibot, i_free/i_stall =.5/7.5 = 1/15  so it is not a bad approximation.

At V = 14 volts, w_free@14 = w_free*(V/12)*1.01  so only a 1% error.


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