Question on actual I2C pin wiring for an MC9S08QG8

[mykepredko]

Hey Folks,

Hopefully a quick one.  I am planning on using a MC9S08QG8 for power management implementing a boost supply from alkaline/NiMH batteries to 5V and then back down to 3.3V via a buck topology for a 32 bit MCU (CF or Kinetis) as well as allowing the 32 bit MCU to communicate with the '08GQ8 via I2C. 

Will there be problems if the '08QG8 is running on battery power (potentially as low as 2.2V) if the I2C pull ups are on 3.3V?  Really what I am asking is are there ESD protection diodes on the '08QG8 pins that will cause a current flow from the pull ups that are always enabled? 

I guess the simpler/best solution would be to power the I2C pull ups from the battery source (the same source as the '08QG8) correct? 

Thanx,

myke

[bigmac]

Hello Myke,

The IIC pins on the 'QG8 will have internal diodes present, and these must not conduct to a significant degree, under normal circumstances.  In your case, you also have to consider the battery life.

I presume that the battery would consist of two alkaline cells, so the maximum voltage, when they are fresh, will be less than the regulated 3.3 volt level.

I guess there are two possible solutions - connect the IIC pullup resistors to the battery supply, or supply the 'QG8 device from the regulated supply.

I would estimate that the boost-buck arrangement, assuming a non-switching buck regulator, would probably have an efficiency of 30-35 percent only.  Therefore the second connection method would require that the average current draw of the 'S8 be significantly less than that of the 32-bit MCU, so as not to degrade the battery life too much.

If the overall current levels are relatively low, and the regulation requirement is not too severe, you might also consider using a regulated charge pump device.  It may even be feasible to utilize a single boost regulator with 3.3 volt output, depending on what happens when the battery voltage is only slightly below the output level.  Either of these methods should give a much improved efficiency, resulting in longer battery life.

Regards,

Mac

[mykepredko]

Hi Mac,

Thank you for your reply - I think the right solution is to use the battery voltage for the I2C pull ups (as you suggest). 

I'm surprised at your estimation of the efficiency (I presume that this is the overall efficiency of boosting up and then back down).  In my calculations, assuming a 250kHz PWM for the Boost/Buck, I found I would be closer to 60% (75% or better, depending on component selection) - is there anything I should be looking at or (hopefully) haven't explained correctly? 

Unfortunately a charge pump device just isn't realistic with the current draw of the 32bit MCU and other circuitry. 

Thanx,

myke

[bigmac]

Hello Myke,

If you are using  a 3.3 volt linear regulator, as I had assumed, the ideal efficiency (with zero quiescent current) will be 66 percent, with one third the total power dissipated within the regulator.  Allowing for some regulator quiescent current might reduce this to, say 60 percent.  If the boost switching regulator had an efficiency of 70 percent, the overall power efficiency would be about 42 percent.

However, since the battery capacity is usually rated in mA-hour, rather than watt-hour, the battery life will be determined by "current efficiency".  This will be even less than the power efficiency, as the battery voltage is usually lower than the regulated voltage.  If the battery voltage were say 2.8 volts, at a given point in the battery discharge curve, the overall current efficiency would be about 36 percent, i.e. the battery current draw would be nearly three times the load on the 3.3 volt output.  This will further degrade for lower battery voltage.

You did not mention the actual load levels to be accommodated.  If the 32-bit MCU spends most of the time in a low power mode, with full operating current present only for short bursts, the boost regulator would need to handle a peak load that is many times the average load.  Under these circumstances, it is possible that the average efficiency of the boost regulator would be considerably less than the assumed 70 percent.

Regards,

Mac

[mykepredko]

Hey Mac,

No linear regulators are to be used in the design. 

Let's see how good ASCII art works here:

2x "C" Batteries -+-> 5.0V Boost -+-> 200mA draw @ 5.0V

                  |               |

                  |               +-> 3.3V Buck -> 170mA draw @ 3.3V (MCU)

                  |

                  +-> 'HC08 control Boost/Buck & MCU !Reset

Does that make things clearer as to how I am architecting the power?  I should point out that when the system is "off", I will be turning off the Boost and Buck power to the various devices (as well as pulling down the MCU !Reset pin even though there is no opportunity for parasitic power to drive the MCU).  I put "off" in quotation marks because the 'HC08 will still be powered (in Stop2 mode) waiting for a button press (on !IRQ) to resume

Do you have a reference of "current efficiency" for Boost supplies?  I've just done a quick Googe search and couldn't find anything useful.  To calculate the Boost/Buck efficiency, I'm using the standard equations and I get 75+% for a 250kHz PWM in both cases. 

In Stop2 mode ("off") mode, I am expecting to have a total draw of around 20uA in "off" mode which allows for a reasonable life for two "C" cell alkaline batteries. 

Thanx - I appreciate your comments. 

myke

[bigmac]

Hello Myke,

I did not previously realize that you had a high external current draw at 5 volts.  However, the overall efficiency would be somewhat higher if you were able to use two separate boost converters, one for 5 volt output, and the second to directly provide 3.3 volts output.

Is your boost converter capable of being completely turned off to zero volts?  Some that use a simple topology, including use of an inductor rather than a transformer, would actually output battery voltage, minus a forward diode drop, when not oscillating.

The term "current efficiency" was my own invention.  Battery capacity is normally specified in ampere-hours, rather than watt-hours.  So current draw, rather than power draw from the battery is more meaningful.

Regards,

Mac

[mykepredko]

Hi Mac,

Thanx for the reply. 

No, there is no opportunity to turn off the +5V boost - I was going to leave the controlling N-Channel MOSFET off during "Stop" mode.  The circuitry on +5V is controlled by the +3.3V circuitry and if the +3.3V circuitry is off the +5V circuitry will be held in reset. 

That's an interesting suggestion to have two separate boosts and I'll have to look to see what that does for the overall system efficiency - I could see a significant advantage there.  It would mean that I would have to add MOSFET switches to the two Boosts' inputs to prevent any leakage current, which shouldn't be an issue. 

I understand what you mean by "current efficiency" and I've been thinking through it to maximize the battery life. 

myke