AVR-Chat

are you sure you got your formulas right?!  
that's just a voltage divider and im pretty sure you can't generate 650kV from a 12V battery. im sorry but i can't remember the formula for the voltage divider circuit. 
maybe someone might confirm that the multisim was right with its output. as far as i know, current can be computed by dividing Vbatt over the total resistance, which is 1,300,010 ohms. pretty low current there! 

Bruce Parham <obparham@jpl.nasa.gov> wrote:


jay marante wrote:
> 
> here's what i wanted to do:
> 
> first, i wanted to determine if the battery back-up supply is low on voltage output.
> i had the circuit this way:
>     main ---------------- >| ------------
>                                              \__________________  dc-dc converter
>                                              /
>    battery -------------- >| ------------                                         * >|     - diode
> 
> max battery voltage : 12V
> 
> the max639 dc-dc converter has a low-battery output(LBO)  pin that goes low when the battery input(LBI) is less than
> the reference voltage of about 1.23V. the LBI pin is connected to the main power supply. im planning to connect the
> LBO pin to ICP of mega16 and the battery output to ADC input 7 of the chip. to have the battery voltage lowered to 5V
> before it goes in the ADC pin, i have this circuit:
> 
>       battery
>          |
>         <
>         > 1.3 Mega ohm
>         <
>          |
>          |------------------- ADC7   ~ 5V as simulated in multisim
>          |
>         <
>         > 10ohm
>         <
>          |
>      GND
> 
> now, when circuit switches to the back-up battery, the LBI goes low; which in turn, switches the LBO low.
> my question is:
> 1. will the ICP work even if timer1 is off? i won't enable the timer, only the ICP interrupt bit. i cannot use the
> external interrupts since i have these pins connected to other devices. i'll use this interrupt to start an ADC
> conversion.
> 2. i don't know much of the ADC of mega16. i understand the datasheet says that it can be configured to use the AVCC
> as the reference voltage. does this mean that the AVCC pin can be connected to VCC?
> 3. the datasheet also says that the there should be an external capacitor at the aref pin. what should be the
> capacitor value?
> 
> thanks for the help.
> -jay
> 

Let's see now, I = E/R so I[10 ohm res] = 5/10 = 0.5 Amps.  
And E = I*R so E[1.3 Meg res] = 0.5 * 1.3e+6 = 6.5E+5 Volts or 650 KV. Wow, some battery!

(Typical SPICE result...)

Bruce

P.S.: I do wish thay would go back to teaching real EE.

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