Electronic – Controlling the “amount” of regenerative braking (variable back torque?)

You seem to be confused about what you want. If you want to decrease the motor speed, but you still want maximum torque, then you must apply full rated electrical power to the motor, and put a mechanical brake on the motor until it slows to the speed you desire. Or, you must somehow make your motor less efficient. I don't think that's what you want.

Think of it this way: electrical power is the product of current \$I\$ and voltage \$E\$:

$$ P = I E $$

Mechanical power is the product of torque (\$\tau\$, in newton-meters) in and angular velocity (\$\omega\$, in radians per second):

$$ P = \tau \omega $$

A motor is an electrical to mechanical power converter. The mechanical power always equals the electrical power after losses.

Furthermore, current is proportional to torque, because the more current you apply, the stronger the magnetic field inside the motor, and the attraction between the motor's poles becomes greater.

If the mechanical and electrical powers are correlated, as are the current and torque, then voltage and speed must be, also. And they are, because the faster the rotor spins through the stator field, the greater back-emf it will generate. This is Faraday's law of induction.

So, if you want to decrease speed, decrease voltage. If you want to decrease torque, decrease current. If you increase torque (say by putting a brake on the motor), you are increasing motor torque. But if you don't change the supply of electrical power, then the mechanical power also won't change. If torque increased, the only way to keep mechanical power constant is to decrease speed, so the motor slows down.

There is one kink here: as torque goes up, current goes up. The resistive losses in the motor also go up, because the windings have some resistance, and those resistive losses are proportional to the square of the current:

$$ P = R I^2 $$

So, as current goes up, the resistive losses increase, making the motor a less efficient converter of electrical energy to mechanical energy, because some of that electrical energy is now creating heat. If you stall the motor, then the motor reaches 0% efficiency: speed is zero, so mechanical power must be zero, but the motor is drawing a ton of current, and there is a voltage drop over the winding resistance, so electrical power is very high.

Interesting fact: if you can make a motor with no winding resistance (or other losses), and you connect it to a perfect voltage source, then the speed regulation (how much speed changes with torque) is perfect. That is, the motor won't slow down if you try to stop it: it will just draw exactly enough more current from your battery to keep spinning at the same speed, no matter what.

PWM is all irrelevant to this. PWM motor control is just a way to efficiently apply less than the full battery voltage to the motor. It works because a PWM driven motor is equivalent to a buck converter. Changing your PWM duty cycle is equivalent to changing your supply voltage:

The maximum torque you could have (which you will get when the motor is stalled) is limited by the current your power supply can supply and the losses in the motor, just as it is without PWM. Your PWM driver might add a bit of resistance to the circuit, reducing the current and torque a bit, but usually this isn't significant compared to the resistance of the motor windings.

First, the torque constant \$k_t\$ for a DC machine is derived as follows. If you assume a constant speed and you neglect any losses or saturation, then the power into a motor equals the power out of the motor, or \$E*I = T*\omega\$, where \$E\$ is the line to line EMF, \$I\$ is the DC input current, and \$T\$ is the torque at speed \$\omega\$. From this we can say that \$k = \frac{E}{\omega} = \frac{T}{I}\$. Let me repeat the assumptions that were made for this equality to be true:

1. Constant speed
2. Lossless energy conversion
3. DC input current

If those 3 assumptions hold, then \$k\$ is a constant of proportionality. Traditionally for a brushed DC motor, we've given \$k\$ two different names, \$k_E = \frac{E}{\omega}\$ and \$k_T = \frac{T}{I}\$, where \$k = k_E = k_T\$ (assuming your units are \$\frac{Volts}{rad/sec}\$ and \$\frac{N*m}{Amp}\$, respectively).

Second, assumption #3 above poses a problem when we switch from DC motors to brushless motors because brushless motors are typically driven with either square (trapezoidal) currents or sinusoidal currents. Another issue that arises is that due to the commutator in an ideal DC machine, the EMF \$E\$ is a mean rectified EMF over all the coils in the machine. In a brushless machine, we aren't dealing with a mean EMF but rather an EMF with a waveform that depends on how the motor is built. The two ideal cases are a trapezoidal EMF and a sinusoidal EMF. Another issue that arises that in the ideal DC motor above the line to line EMF is just 1 phase while in a brushless motor, the line to line EMF be 2 phases in the case of a Wye connected brushless motor. In some cases (for example, a motor with trapezoidal EMF and driven by square wave currents), the \$k = \frac{E}{\omega} = \frac{T}{I}\$ equality still holds for a brushless motor. In other cases (for example, a motor with sinusoidal EMF and driven by sinusoidal currents), the equality does not hold.

Third, you can't calculate \$k_T\$ based on 1 data point. Typically a motor manufacturer would calculate \$k_T\$ by hooking the motor to a dynamometer and then measuring voltage, current, speed and torque while increasing the torque. They would then take a best fit line of the torque vs. current curve and the slope of that line would be \$k_T\$. This line will not go through the origin because of friction (that is, the motor requires a certain minimum amount of current to get the motor started) and many motors will not have a linear torque vs. current curve for high values of current (due to saturation). Also, generally this test is done at room temperature and it is done quickly to keep the temperature of the motor as close to room temperature as possible. The ratings in the chart above would have been performed with the motor windings at a high temperature (at the rated temperature of the insulation).

Fourth, why did I tell you all of that? Because most people when dealing with brushless motors assume things about \$k_T\$ that aren't true. Most often they treat it as the same as a DC brushed motor with a commutator. They also neglect losses due to friction and saturation. The other issue is that there really is no standard definition of \$k_T\$ in the industry. It could refer to the line to line or the line to neutral value. It could refer to the RMS or the peak value.

Fifth, if you forget about \$k_T\$ for a moment (since you aren't given enough information in your chart to determine \$k_T\$ anyway) your question boils down to "If 2 motors have the same rated torque but different input currents, what determines that input current?" Your intuition that it has to do with voltage is correct. If you increase the current from 48 V to 85 V, then in order to maintain the same input power, your current will decrease by the ratio \$\frac{85 V}{48 V} = 1.7\$. You'll see that that the current does indeed decrease by that amount \$\frac{2.4 A}{1.7} = 1.4 A\$. Motor designers have a rule of thumb where if have a motor design and you want to increase the voltage by a certain ratio, then all you need to do is increase the number of turns by that ratio and decrease the wire area by that ratio. Doing that changes the resistance by the correct amount but keeps the flux in the motor the same.