Current profiles

ozzy 19 years ago

Hi folks,

I've just had a look at the AN606 application note from microchip

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I'm having problems in calculating the average current from the current profiles given in figures 3 and 4 (page 2) (The values of current are also shown on page 1, Table 1). This voltage is across a 99.4 ohm resistor (Rg).

For fig 3 I get:

Power-up region:

Vaverage = 40mv

Iaverage = 40mv/99.4 = 400 microAmps

Active region

Vaverage = 84mv

Iaverage = 84mv/99.4 = 840 microAmps

Sleep region

Vaverage ~ 6mv

Iaverage ~ 6mv/99.4 = 60 microAmps

The above does not give the correct answers (found on Table 1; 261KHz RC mode) - anyone know what i'm doing wrong?

regards,

Ozzy

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James Arthur 19 years ago

Ozzy, You're right, and Table 1 and Fig. 3 do not agree. The table gives an i(Active)-to-i(Power-up) ratio of about 8:1, but the Fig. 3 graph clearly does _not_.

The scope shots are mostly useless--we can't see enough detail to know the duty cycle of that pulsing current, or compute any reasonable estimate. I also suspect they used a 10x probe, so the scale would be

10x your assumptions.

James Arthur

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ozzy 19 years ago

James,

Thanks for checking my answers. One point I'd like to pick up on is the term average voltage. If you 'assume' the signals are a rectangular form (I did) then the average voltage is defined as:

Average voltage = area under curve/ length of base (ref:Electrical circuit theory by J.O Bird)

As this is a rectangular waveform the 'length of base' ( 1/2 of duty cycle) should'nt matter should it?

Anyway something is obviously wrong - maybe I should write to them.

Regards,

Ozzy

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