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This question largely covers the question I have, however the answer is unsatisfying, making the current output a function of an otherwise undescribed 'current control loop'.

I see by studying my own PWM controlled motor a similar phenomenon - a high-frequency PWM switching on all phases, where the current variation is on a much lower frequency and much nearer to an ideal current profile for the magnetic fields passing coils etc.

The linked question draws the distinction between going open-circuit and being always linked to one of the power lines either positive or neutral. As I write this question the answer begins to crystallise, so now I ask if my understanding is correct:

schematic

simulate this circuit – Schematic created using CircuitLab

In the pictured circuit, SW1 is in one of two positions:

  1. Attached to the positive terminal, increasing the current in the inductor
  2. Attached to the ground terminal, creating a circuit for the developed current to exist

It would be clear in this case that switching time is critical, as without some kind of buffer, the switching IC would have to accommodate the 'spark' and loss of energy the linked answer refers to. Further, any ambiguity in switching could feasibly see the battery terminals being shorted, wasting energy or damaging switching equipment. Is this simply handled with precise timing or are the switching ICs handling this in a much more graceful or intelligent way? At very least a well positioned capacitor might accommodate the current flow while the switch state is ambiguous.

I think if the proposed mode of operation is correct, that being connected to the battery ground is actually irrelevant, and that closing the inductor's own circuit is the critical factor. Is this an accurate statement?

Finally then, a circuit is sensing the developed current in the inductor, and making high-speed adjustments to the PWM output to keep current to that prescribed by the controller. Where in the control circuit would this current measurement take place and in practice what electronic device is making that measurement? (Example part numbers would be of interest)

Ultimately, I see a high-frequency voltage applied to the terminals, but no equivalent high-frequency current being absorbed by the motor phases. There appear to be good physical reasons why the current does not follow the somewhat violent voltage variations, and I would like to confirm here that the observed current vs voltage profile is as expected and I should not be looking for errors or issues in my sensing equipment.

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  • \$\begingroup\$ Is the switch make-before-break, or break-before-make? \$\endgroup\$ – user_1818839 Apr 1 at 12:48
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    \$\begingroup\$ @BrianDrummond that would be my question to the group. In the diagram a make-before-break would short the battery, a break-before-make would leave the inductor open-circuit. Neither is optimal. \$\endgroup\$ – J Collins Apr 1 at 12:49
  • \$\begingroup\$ You're thinking along the right lines. A practical answer would involve break before make, and some form of circuit protection (snubber, spark gap, efficiency diode for some examples) \$\endgroup\$ – user_1818839 Apr 1 at 12:52
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    \$\begingroup\$ The voltage~current relationship is all embedded in the simple formula \$V = L\dfrac{di}{dt}\$. \$\endgroup\$ – Andy aka Apr 1 at 12:52
  • \$\begingroup\$ Also imagine what happens with VI when stray capacitance or PN junction capacitance is added? Here's another variation with a simple SCR control switch . You can change any variable. tinyurl.com/yj8k3m75 \$\endgroup\$ – Tony Stewart EE75 Apr 1 at 13:47

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