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re: electric car batteries
2 aug 2002
don kelly  wrote:

>i see the diode- however, as shown, assuming "top" of battery is + and
>current (conventional)  direction with switch closed is to the right, the
>inductor current will tend to flow the same way- in the direction blocked by
>the diode.  do you really want backflow in the forward direction of the
>diode?- the object is to charge the capacitor, not discharge it.

got a better suggestion?

>in any case, returning the inductor energy to the battery doesn't help
>the charging. it would be better to dump it into the capacitor- a free
>wheel diode across the l-c would allow this.

can you draw that?

>possibly i am mis-interpreting your diagram due to font problems...

try viewing it with a fixed font.

>>        diode
>>     ----|<----
>>    |     /    |
>>    |----/  -------uuuu-----
>>    |  switch    inductor   |
>>   ---                     ---
>>    -  battery        cap  ---
>>    |                       |
>>    -                       -

>> in the first pulse, a peak current ip (roughly vdt/l) flows during dt.
>> the cap charges to v = ipdt/(2c), approximately, storing ec = 1/2cv^2
>> = ip^2dt^2/(8c) j.

>no problem so far

>> if the switch, esr, wires, coil, and so on have total resistance r, the
>> charging circuit dissipates ed = ip^2r/2, approximately, so efficiency
>> e = ec/(ec+ed) = dt^2/(dt^2+4rc), if i did that right, approaching 100%
>> as dt increases.

>your dissipation figure is the peak power dissipation (assuming the current
>rises linearly from 0 to ip)-not energy. the energy dissipated will be the
>integral of the power -assuming a linear rise from 0 to ip this gives
>ed=ip^2rdt/3

>the efficiency then becomes 1/(1+8rc/3dt)... which leads to the same
>conclusion -efficiency increases as(not as quickly) as dt increases.

i agree with you. i made an integration mistake. this makes dt
= 8rce/(3(1-e)) = 8(0.006)10kx10^-6(0.9)/(3(1-0.9)) = 0.00144
seconds for 90% charging efficiency in the first pulse, with
that 10k uf cap with the 0.006 ohm esr. 
 
>as dt increases, the approximation of a current ramp becomes less realistic

you might redo this with exponentials.

>what you appear to be  saying, is that, within the limits of the ramp
>approximation, the longer the ramp, the slower the average charge rate
>ip/2dt.

i was surprised to see the peak current disappear as an efficiency term.

nick




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