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what data can be stored on the 800 Kb memory of the PM850MG store?

 To answer your question of logging in PM800 series, first of all, please allow me to explain the structure of the storage, which can make you easily to understand as all depends on the number of parameters you want to log, time intervals for the logging and specification of each data log. 

PM820, PM850 & PM870: 
Data Log 1 
Max. Records Stored - 5000 
Max. Register Values Recorded - 96 + 3 D/T* 
Storage (Bytes) - 14,808 

PM850 & PM870: 
Data Log 2 
Max. Records Stored - 5000 
Max. Register Values Recorded - 96 + 3 D/T* 
Storage (Bytes) - 393,216 

PM850 & PM870: 
Data Log 3 
Max. Records Stored - 5000 
Max. Register Values Recorded - 96 + 3 D/T* 
Storage (Bytes) - 393,216 

* - 3 D/T means Start Date & Time - is simply a timestamp for each set of records being logged and occupy 3 register but it is excluded in the "96 max. register value recorded", so you can simple say max. register value is 96 as date & time is on top of this. 

Let me put it in an example for better understanding - 
To log current A only with time interval 15 mins, number of register is 1 (according to table 8-5 & 8-6, current A = 1 register) and so it is lower than the limitation of "max. register values recorded - 96". 
Then, you can obtain the total number of recordings (i.e. 1250hrs [5000/4 as interval = 15mins] = 52.083 days) and have the result of 52 days in total. 
But at the same time, you also need to consider the storage of Data Log as either the max. records or storage is full, FIFO or Fill/Hold would begin. 

For 5,000 records with current A only, there would be (1+3)*2*5,000 {(register of parameter to log + 3 D/T)*(2 bytes/register)*(total records stored)} = 40,000 Bytes. 
So, it is within the range for Data Log 2 & 3 (393,216 bytes) and you can log 52 days but over the range for Data Log 1 (14,808 bytes). 

Therefore, max. records is 1851{14,808 bytes/[(1+3 registers)*2 bytes/register]}, i.e. 19.28 days, if you are using Data Log 1. 

If you would like to log current A & current B with time interval 15 mins, you can also log for 52 days as the total number of register is 2 (still smaller than the limitation of 96) for data log 2 & 3 but 15.4 days for Data Log 1. 

If you are logging Real & Reactive Energy with 15 mins time interval, there are 8 registers in total and it is less that max. register value recorded - 96. 
Then, you have to calculate the number of bytes and obtain 7 days as 14,808 bytes in data log 1.
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