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手册里只能看到这样的表述,但不清楚具体的计算方法,只能知道与设计容量有关,现在的情况是同一批次按相同的老化步骤后,有几组电芯老化过后的SOH掉的比较厉害,甚至从99%掉到93%这是什么情况呢
您好,SOH is calculated as (Full Charge Capacity) / Design Capacity. So as the Full Charge Capacity is updated as the battery ages, the SOH will decrease.
我后来又在手册中看到这样的方式,到底用的是哪一个呢,是直接SOH=FCC/DC,还是与SOH Load Rate有关的等式?单就从手册中的描述可以得知SOH Load Rate是用来设置工作电流的,最后还是基于这个电流仿真出来的FCC得出的SOH。
那是不是可以理解成SOH Load Rate设的越小,那么仿真的工作电流就越大,那么如果我设SOH Load Rate=1,那么typical current=1C,这时候如果我以0.5C放电到保护,因为相对来讲,0.5C放比1C放的时间要长,那么现在以0.5C放电到保护的时间,而使用基于1C来仿真的FCC算得的SOH应该会偏大,反之则偏小,是不是这样呢,因为之前通过试验降低设计容量的值来提高SOH的,但是FCC在都大于DC的情况下,按照FCC/DC,应该都100%才对,但现在情况不是这样,所以猜测是不是上述关于SOH Load Rate的逻辑导致这样的问题
我的SOH Load Rate设置为5,即typical current=0.2C,但是实际放电电流为0.5C,那么在相同的时间放出的容量,0.2C放的要少与0.5C,所以导致计算SOH的FCC要少于BQstudio上位机看到的FCC,而我为了实验,特地把设计容量改小,使得上位机看到的FCC/DC>1,也就是理论上至少100%的程度,而现在偏低,就是因为前面说的计算SOH的FCC少于BQstudio上位机看到的FCC的原因吗
您好,
SOH is calculated as QMax*IR@25C. You can see the gauge calcualation at [SOHFC@25C Q] register. In this equation I = DesignCapacity/SOH Load rate (default C/5). R is from the Ra table of the gauge. This value is then divided by Design capacity to report a %.
The only thing that could really impact you would be
1) Writing Design Capacity wrong
2) QMax update very small
3) Ra update very large.