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【求助】紫外分光光度法的定量限和检出限的确定

作者 Billd_G@126.com
来源: 小木虫 350 7 举报帖子
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药品检测的清洗验证中需要用到紫外分光光度法来测定物质的吸收值,需要对此方法进行验证,请问用紫外分光光度法来做分析方法验证时,如何确定其检出限和定量限。确定的标准是否有出处。
我有同事提出配制一定浓度的溶液,反复检测6次,计算6次检测结果的RSD值,当RSD值为10%时为定量限,当RSD值为30%时为检出限,但是我查找了一些指南和通论,并没有找到相应的依据。请问哪位高人指点一下。 返回小木虫查看更多

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  • zhuwei2053

    配制一定数量的梯度浓度,进液相,一般检测限峰高是基线噪音峰高的3倍,定量限是基线噪音峰高的10倍

  • huoshixin

    可以参考《GB/T 5009.1-2003 食品卫生检验方法 理化部分 总则》 的附录A部分。

  • shufang5072

    计算出不同浓度标样作出的标准曲线的标准偏差Sy/x,并将其假定为空白标准偏差,其3倍及10倍值被假定为估计的LOD和LOQ

  • shufang5072

    这里的LOD和LOQ也可认为是仪器的LOD和LOQ,即IDL和IQL

  • Billd_G@126.com

    引用回帖:
    Originally posted by shufang5072 at 2011-03-30 12:40:28:
    计算出不同浓度标样作出的标准曲线的标准偏差Sy/x,并将其假定为空白标准偏差,其3倍及10倍值被假定为估计的LOD和LOQ

    请问此方法的依据是什么?药典、专论或者指南?

  • yugijiang

    ICH:
    QL = 10 σ  /S
        DL=3.3σ  /S
    where  σ  = the standard deviation of the response
              S = the slope of the calibration curve
    The slope S may be estimated from the calibration curve of the analyte. The estimate of σ  may be carried out in a variety of ways for example

    LOQetermination of the signal-to-noise ratio is performed by comparing measured signals from samples with known low concentrations of analyte with those of blank samples and by establishing the minimum concentration at which the analyte can be reliably quantified. A typical signal-to-noise ratio is 10:1.
    LODetermination of the signal-to-noise ratio is performed by comparing measured signals from samples with known low concentrations of analyte with those of blank samples and establishing the minimum concentration at which the analyte can be eliably detected. A signal-to-noise ratio between 3 or 2:1 is generally considered acceptable for estimating the detection limit,

  • w?x?l

    引用回帖:
    4楼: Originally posted by shufang5072 at 2011-03-30 12:40:28
    计算出不同浓度标样作出的标准曲线的标准偏差Sy/x,并将其假定为空白标准偏差,其3倍及10倍值被假定为估计的LOD和LOQ

    请问如何计算标准差

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