1. VBUS is, I guess, three phase supply. This is useful for large loads.Ok, So if i understand correctly this is what it represents. Am i correct?
View attachment 167047
There will be voltage V1 from the motor terminals and V2 after the voltage drop in the resistor. The difference voltage is what op amp will amplify.
No, you are wrong. Make a detailed sketch and show us.Ok i understand, One additional question i forgot to ask and really i want to know is that the voltages V1 and hence V2 will be the back emf voltages from motor. Am i correct here?
No, you are wrong. Make a detailed sketch and show us.
R & B also drop voltage. They are pretty much resistive loads for the following reason.View attachment 167057
...complete VBUS will be dropped across the resistor R.
There will be voltage drops across various components in the circuit; but the total voltage across all of them added together (algebraic addition with sign and phase in case of AC) will be equal to the VBUS. As drawn, the potential at Pt1 is undefined. It is wrong to say that the complete VBUS potential will be dropped across the resistor R, as shown.View attachment 167057
I can see a VBUS voltage applied at the top of the resistor but the complete VBUS will be dropped across the resistor R. The voltage at Pt1 is "VBUS" and the voltage at Pt2 is "Gnd". Is it correct? I am still confused. I am assuming that switch A and C' is closed. The switches A',B, B',C are open.
Completely wrong. The circuit is a three phase inverter, the switches are continuosly switching, no DC current through the motor windings.I can see a VBUS voltage applied at the top of the resistor but the complete VBUS will be dropped across the resistor R. The voltage at Pt1 is "VBUS" and the voltage at Pt2 is "Gnd". Is it correct? I am still confused. I am assuming that switch A and C' is closed. The switches A',B, B',C are open.
I presume you mean RMS current. Average current will be zero for a sinusoidal waveform. RMS value (root mean square values taken over a complete period or cycle) is different from the average (arithmetic mean) value unless the value is a constant (DC quantity). But even the RMS current is NOT a true measurement of the "real power consumed by the motor" because the motor is rarely a resistive load. To get the real power, you need to know the voltage waveform too and the relationship with the current waveform (phase factor). This relationship can be messy but for sinusoidal waveforms we can make simplifications.The average current through the shunt resistor is a measurement of the real power consumed by the motor.
In post#1 the shunt shows the current of the DC bus.I presume you mean RMS current.