[andyr]

I am planning on using a pair of 1.3AH batteries, in series (delivering 24v), on each channel of a JFET phono stage.  I could use larger batteries ... but I am wondering if these would be sufficient? 

Each channel pulls 35ma - so after 12 hours straight listening, I will have sucked out 0.42AH.

This is only 1/3rd of the total energy charge - can you advise:
1. is 1/3rd of the total storage a large amount of charge to have consumed?
2. what is the DC+ rail voltage likely to be?  IE. with 1/3rd of the stored charge gone, by how many volts will the DC rail be reduced?


Thanks,

Andy

[srb]

You said that each channel uses 35mA, so two channels would use 70mA and 12 hours of use would consume .84AH or 65% of the capacity.
 
Because the voltage drop with discharge varies with the ratio of total battery capacity to current drawn, your total battery capacity of 1.3AH divided by your constant draw of 70mA would be equivalent to C/18.5 relative to this 12V lead acid battery discharge chart.
 
It looks like the battery voltage should only decrease by ~ .5V per battery or ~ 1V for the two in series down to 30% capacity.  You can see that even if you only needed 4 hours of runtime, going to a battery with 1/3 of the capacity would result in a larger voltage drop.
 

 
Steve

[andyr]

You said that each channel uses 35mA, so two channels would use 70mA and 12 hours of use would consume .84AH or 65% of the capacity.
 
Because the voltage drop with discharge varies with the ratio of total battery capacity to current drawn, your total battery capacity of 1.3AH divided by your constant draw of 70mA would be equivalent to C/18.5 relative to this 12V lead acid battery discharge chart.
 
It looks like the battery voltage should only decrease by ~ .5V per battery or ~ 1V for the two in series down to 30% capacity.  You can see that even if you only needed 4 hours of runtime, going to a battery with 1/3 of the capacity would result in a larger voltage drop.
 

 
Steve

Thanks, Steve ... but can you redo your calcs.

I said I have 2 x 12v SLAs per channel!  So there is only 35ma coming out of each battery bank.


Regards,

Andy

[srb]

  Sorry, missed that with a late night posting.
 
Your calculations are correct, only ~ a third of the AH capacity would be depleted.  The discharge current (ratio of capacity/current draw) would be 1.3AH/35mA = C/37 (.03C), and the resulting voltage drop even less, maybe only ~ .2V.
 
More importantly, the number of discharge cycles and battery life would be greatly increased with a discharge depth of only ~ 30% versus ~ 70%.  Any of these charts are only rough estimates as there are so many variables including discharge current, ambient temperature, charging current and charging profile.
 

 
Steve

[andyr]

Thanks very much, Steve. 

Following a C/37 curve, there would appear to be only a very small DC+ voltage drop when 1/3rd of the AH has been consumed.  And this results in a satisfactory battery life, in terms of # of charging cycles.

So I think I will choose 1.3AH batteries, instead of 2.2AH (so I can use a smaller case). 

Regards,

Andy

[srb]

So I think I will choose 1.3AH batteries, instead of 2.2AH (so I can use a smaller case). 

Yes, in the end, your original thought of the 1.3AH seems to be the near perfect choice for your current draw and anticipated runtime with respect to voltage drop and battery life and size.  Well done!
 
Steve