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How to identify voltage or current to Opamp

electronicsman

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I have the following high level diagram and an application note for measuring the currents, i am confused with the inputs to the op-amp, how do i know if it is currents or voltages? I have attached the application note. I assume it is the currents.

1611025562078.png
 

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BradtheRad

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The op amp inputs detect voltage difference across the low-ohm resistor. Current flowing through the resistor automatically creates a proportional voltage across it (per Ohm's law). It's a popular method to detect current levels.

The op amp inputs are high impedance and admit minuscule current.

There is the Norton op amp which does detect current at its inputs. It's used much less than everyday op amps.
 

betwixt

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The important word Brad used is 'across'.

Very few op-amps, and certainly not in that application, would sense current. The 'IBUS' voltage here is the difference in voltage between the the two op-amp input pins. It uses standard Ohms Law V=I*R to measure the current flowing through the resistor. If 'R' is known, V dropped across it is proportional to the current flowing through it.

Brian.
 

electronicsman

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Ok, So if i understand correctly this is what it represents. Am i correct?
1611067046486.png
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.
 

FvM

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The amplifier used in this circuit would be an instrumentation amplifier (differential amplifier with defined gain), not an OPAMP which has ideally infinite gain.

A practical circuit needs protection means for the amplifier in case of high inrush current or inverter short.
 

electronicsman

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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?
 

c_mitra

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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.
1. VBUS is, I guess, three phase supply. This is useful for large loads.

2. The three wires corresponding to three phases are not shown. They are bundled together. In reality they are three independent wires.

3. The fat black rods are indeed switches. To measure currents in three different phases, you close the corresponding switches (bottom rods).

4. The lower connection is made to the neutral of the three phases.

5. V1-V2 is the unbalanced current. If the motor is running happily, the unbalanced current will be zero over the three phases.

6. This is only a schematic, you MUST NOT take it as it is.
--- Updated ---

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.
 

KlausST

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Hi,

The easiest and reliable way is to use a dedicated "current sense amplifier".

But for sure Instrumentation amplifier or standard Opamps work as well...with the according circuit (mainly resistors).

Klaus
 

electronicsman

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No, you are wrong. Make a detailed sketch and show us.
1611112009315.png
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.
 

danadakk

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A useful ref handbook for IA work -





Regards, Dana.
 
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BradtheRad

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View attachment 167057
...complete VBUS will be dropped across the resistor R.
R & B also drop voltage. They are pretty much resistive loads for the following reason.

A motor is not solely an inductive load. It switches current through its coils rapidly one after another. The net effect is for the motor to resemble a train of brief resistive loads. In between are switching glitches. So for the most part a motor can be considered a resistive load.
 
Last edited:

c_mitra

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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.
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.

You are correct to say the the potential at the other end of the indicated R will be the ground potential. That is correct because this end (Pt2) is directly connected to ground.

I understand your confusion because this is very common.
 

FvM

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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.

The average current through the shunt resistor is a measurement of the real power consumed by the motor.
 

c_mitra

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The average current through the shunt resistor is a measurement of the real power consumed by the motor.
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.

In this particular case I guess that only the off balance current will be measured.
 

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