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Forum: µC & Digital Electronics Turn off I2C on MSP430


Author: knpr (Guest)
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Hello,
i am developing a pcb for low power purposes and want to implement a i2c 
Bus for data transfer of several Ics. The µC is a CC430F5137 and is 
located on a diffrent pcb.
My main goal is to reduce energyconsumption of the i2c-bus while not 
sending data.
I just wondered if it is possible to just enable the internatl pull-up 
resistors of the µC to take the bus out of service to save energy.
I found this comment in the Ti forum:
"While this is obvious for normal I/O, this is important if you use the 
I2C controller where the output drives only to GND but not to VCC (open 
collector). If using I2C, you cannot rely on the internal pullups (which 
are too weak anyway) and need external pullups."
To me that means I cannot use the internal pull-ups for driving the bus. 
Am I right?
Otherwise I thought about driving the bus by a output of the µC and 
external pullups and just disable this output if necessary. Any thought?

Thanks in advance.
Best regards knpr

Author: Clemens L. (c_l)
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When the I²C bus is idle, no output driver is active, and the pull-up 
resistors pull the voltages to the high level. In this state, no current 
flows, so there is no power consumption.

You do not need to do anything to save energy.

That remark on the TI forum is about making rising signal edges faster 
by reducing the pull-up resistance. But in the idle state, there are no 
edges.

Author: knpr (Guest)
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thanks for the answer

Author: knpr (Guest)
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Hello,
I have one question left..
What happens when the µC goes to sleep or deep sleep mode. I think the 
Open-Drain Stage while not switch the output to high bevor going to 
sleep. Then there would be power consumption by the Pull- up resistors. 
wouldnt it?
best regtards

Author: Clemens L. (c_l)
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knpr wrote:
> I think the Open-Drain Stage while not switch the output to high bevor
> going to sleep.

Why should it? Open drain (high impedance) is perfectly fine.

> Then there would be power consumption by the Pull- up resistors.

How?  There is power consumption only if some output actively drives the 
line low.

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