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Peak Current in AN1292 Microchip Application Note

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electronicsman

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This is the file available in the application note AN1292. My doubt is the Peak current input parameter 10A. How did he know that the peak current is 10A by design? Please help me. If i want to design on my own and design 20A how do i do it? Please advise.
 

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  • Tuning_parameters.PNG
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Hi,

The "10A peak current" is listed at the "input parameters".
Thus I assume it's his own decided design parameter. Not calculated and not measured.

Klaus

Btw: Maybe the AN tells more about it.
Can you give a link to the AN?
 

Thank you very much for the reply. The document i am referring to is attached and on page 8 i find something related to peak current. Could you please help to understand it.
 

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Hi,

It clearly says it's just an example value:
For example, having a phase current of 4.4A peak

Just follow the examples, but use your own values.
If you expect 20A, then do the calculations with 20A.

Klaus
 

Still searching for the solution i found the following link which is useful
https://ww1.microchip.com/downloads/en/AppNotes/70640A.pdf
Some of the information i need to share is i am using the same MCLV-2 board and i am trying to understand the parameters and calculations. Now referring to the document 70640A on page 2. 70640A.PNG. Now i want to understand the experimental methodology which is explained in the initial attachment from my post. The methodology is as shown but i don't understand it.DMCI.PNG. Do I need to have the current probe?
 

Another question in the development board MCLV-2 is there are 3 current test points as shown belowBusCurrentTestPts.PNG. The circuit as per schematic diagram is CurrentFeedback.png. Now my doubt is the normal scope in voltage measurement gives me these measurements or do i require separate measurements. If i check on these current test points on the scope i am not finding anything. Please advise. The link of the MCLV-2 development board is https://ww1.microchip.com/downloads/en/devicedoc/ds-52080a.pdf and i am using the internal op amp configuration.
 

Hi,

IMOTOR1 and IMOTOR2 show two phase currents (AC) of the motor. They are referenced to 1.65V DC to be able to measure positive as well as negative currents.

Klaus
 

TorqueCurrent.JPG I am confused with the statement "If it is too high, the motor can overheat if it runs in open loop for a long period of time". If there is no load connected as well i can decide how much current the motor can draw for example 1A, 1.5A? Or the condition is true (drawing 1A, 1.5A etc) only if a load is connected? My confuses arises because if you connect a resistor to a circuit then only current is drawn, in case of open load no current is drawn. This situation does not seem to apply to a motor? Am I correct? Please advise.
 

Hi,

a motor is a device that tranforms electrical power to mechanical power (and vice versa).

The mechanical power is: P = 2 * Pi * RPS * torque.
The higher the torque --> the higher the current --> the lower the resistance (impedance)

Klaus
 

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