Question
Asked 9th Mar, 2016

How to create certificate of analysis (CoA)?

How to create certificate of analysis (CoA)? or is CoA the same as certificate that you got when you identify any sample in some laboratories?

Most recent answer

17th Mar, 2016
Sergio Sbrenni
Istituto Superiore di Sanità
For the issue of a certificate of analysis you can usefully refer to the requirements of paragraph 5.10 of the EN ISO/IEC 17025:2005 standard. The most important requirements are shown below:
  1. a title (e.g. "Test Report" );
  2. the name and address of the laboratory, and the location where the tests were carried out, if different from the address of the laboratory;
  3. the name and address of the customer;
  4. identification of the method used;
  5. a description of, the condition of, and unambiguous identification of the item(s) tested;
  6. the date of receipt of the test item(s) where this is critical to the validity and application of the results, and the date(s) of performance of the test;
  7. reference to the sampling plan and procedures used by the laboratory or other bodies where these are relevant to the validity or application of the results;
  8. the test results with, where appropriate, the units of measurement;
  9. where applicable, a statement on the estimated uncertainty of measurement; 
  10. where relevant, a statement of compliance/non-compliance with requirements and/or specifications;
  11. where appropriate and needed, opinions and interpretations.
  12. the name(s), function(s) and signature(s) or equivalent identification of person(s) authorizing the test report certificate;
Hoping to be helpful
2 Recommendations

All Answers (5)

9th Mar, 2016
Bachir Achour
Université de Biskra
Dear Didy,
This is a document signed by a competent analyst that includes the product name, list of ingredients, product batch number, the analysis, the method used, results, conclusion of the analysis ( satisfactory or unsatisfactory), the name and position of the analyst and the date of issuance of the document. Generally, this certificate is provided by the laboratories by giving them the lot number.
With my best regards
Prof. Bachir ACHOUR
10th Mar, 2016
Didy Yoga Lucky Prayoga
Airlangga University
Dear Prof. Bachir Achour
I see, thanks for the answer..
Best regards,
Didy Yoga
10th Mar, 2016
Rishabh Srivastava
Free lance
A signed COA is a regulatory document and it may be used as a judicial proof. It shall provide complete information relating to quality and performance of products. 
1 Recommendation
11th Mar, 2016
Didy Yoga Lucky Prayoga
Airlangga University
Let's say, my product (spirulina) use a continuous culture. How often should I make CoA? since CoA is made per batch.
Can you help by adding an answer?

Similar questions and discussions

How to Solve second order differential equation with Runge Kutta method?
Question
8 answers
  • Pramod KumarPramod Kumar
Hello Everyone!!
I am trying to solve non-linear differential equation with runge kutta method but i am facing issue with the results. i have attached my matlab code also with this.
The problem is that, as with increasing iteration , the deflection (y) increases so fast( which not supposed to happen).
Results
y(1)= -1
y(2)=-1.4 e+10
y(3)= -1.6 e+96
I am not able to find out the reason behind this exponential increase.
i would appreciate if anyone can help me to see, what i am not doing !!
thank you for your time.
% differential equation
d^2(y)/d^2(t) = -2*e*(u/m)*dy/dt - (1/4)*e*(NF)^2*y^3 + e*k*cos(w*t)- ((NF)^2/m)*y;
%%MATLAB CODE%%
clc;
clear all;
n = 1; % number
e = 0.2; % dimensionless parameter
u = 0.01; % damping
m = 0.333; % number series
k = 100; % Excitation amplitude
NF= 374.82 ; % Natural FGrequency
w = NF - e*50; % excitaion frequency in radian per second
w_hertz= w/(2*pi()); % excitaion frequency in hertz
l = 1; % length
E = 114*10^9; % stiffness
H= 1.5*10^(-3); % thickness
h = 0.1;
t = 0:h:100; % time
y = zeros(1,length(t)); % displacement
z = zeros(1,length(t)); % z= dy/dt
d = zeros(1,length(t)); % deflection
% Boundary condition
y(1)= -1; % y at time t=0
z(1)= 0; % z = dy/dt
% all first order differential equation
c_typ = @(t,y,z) z;
d_typ = @(t,y,z) -2*e*(u/m)*z - (1/4)*e*(NF)^2*y^3 + e*k*cos(w*t)-
((NF)^2/m)*y;
for i=1: (length(t)-1)
k_1 = c_typ(t(i),y(i),z(i));
l_1 = d_typ(t(i),y(i),z(i));
k_2 = c_typ(t(i)+0.5*h, y(i)+0.5*h*k_1, z(i)+0.5*h*l_1);
l_2 = d_typ(t(i)+0.5*h, y(i)+0.5*h*k_1, z(i)+0.5*h*l_1);
k_3 = c_typ(t(i)+0.5*h, y(i)+0.5*h*k_2, z(i)+0.5*h*l_2);
l_3 = d_typ(t(i)+0.5*h, y(i)+0.5*h*k_2, z(i)+0.5*h*l_2);
k_4 = c_typ(t(i)+h, y(i)+h*k_3, z(i)+h*l_3 );
l_4 = d_typ(t(i)+h, y(i)+h*k_3, z(i)+h*l_3);
y(i+1) = y(i) + (1/6) * (k_1 + 2*k_2 +2*k_3 + k_4)*h;
z(i+1) = z(i) + (1/6) * (l_1 + 2*l_2 +2*l_3 + l_4)*h;
d(i) = y(i);
d(i+1) = y(i+1);
end

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