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Current profiles

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

Reply to
ozzy
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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

Reply to
James Arthur

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

Reply to
ozzy

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