HRRL Pos Test 27Jul2012 Data Ana 3 MeV

From New IAC Wiki
Jump to navigation Jump to search

3 MeV

2 ways to calculate electron charge

    NE(Int_t ri, Int_t Reprate, Int_t calc_meth){
    printf("%d %d \n",ri, Reprate);
    if(ri==1) TFile *f = new TFile("C:/Users/sadiq/Documents/phd/Runs_Jul_2012/rootfiles/7-27-12/r3735.root");
    else if(ri==2) TFile *f = new TFile("C:/Users/sadiq/Documents/phd/Runs_Jul_2012/rootfiles/7-27-12/r3737.root");
    TTree *ntuple = (TTree*)f->Get("ntuple");
    TH1F *f9=new TH1F("f9","ADC9",4000,0,4000);	 	 
       TCanvas * c2 = new TCanvas("c2", "c2",600 ,500);
       c2->cd();
       Long64_t ne;
       printf("%d \n", ne);
      //method 1 is reprate times single pulse charge

method 1

       Double_t Qp1=0,Qp2=0;
       if(calc_meth == 1){	
       //ntuple->Draw("ADC9*(1.22*1e-9)/50>>f9","1"); 
       ntuple->Draw("ADC9>>f9"); 
       //Qp2 = (f9->Integral())*(1.22*1e-9)/50; //charge calculate with method 2.
       //f9->Fit("gaus");
       ADC9_fit_mean = f9->GetMean(1); 
       //ADC9_fit_mean = ADC9->GetMean(1); 
       printf("ADC9_fit_mean = %d \n",ADC9_fit_mean);
       Qp1 = ADC9_fit_mean*(1.22*1e-9)/50; //Qp: charge per pulse. The calibration: 1.22 (9nV s)/(ADC channel), 50 ohm Terminal.
       printf("QP1=%g\n\n",Qp1);
       printf("e- charge per pulse with method 1: %g\n",Qp1);
       //printf("Reprate: %d\n",Reprate);
       Qs=Reprate*Qp1; //Qs: Charge per second.
       //printf("e- charge per second: %g\n",Qs);
       ne=Qs/Q_e; //Ne: number of electrons per second.
               }


method 2

       if(calc_meth == 2){	 
       //ntuple->Draw("ADC9*(1.22*1e-9)/50>>f9","1"); 
       ntuple->Draw("ADC9>>f9"); 
       printf("%bin# Bin Content%d\n");
       for(i=1; i<=4000; i++){
       //for(i=0; i<=600; i++){
       bin_content[i] = f9->GetBinContent(i);
       //Qp2+=(i*(1.22*1e-9)/50)*bin_content[i];
       Qp2+=i*bin_content[i]*(1.22*1e-9)/50;
       //printf("%d  %d  %g \n",i,bin_content[i],Qp2);
       }
       //printf("Integral = %d \n",f9->Integral());
       //Qp2 = (f9->Integral())*(1.22*1e-9)/50; //charge calculate with method 2.    
       printf("QP2=%g\n\n",Qp2);
       printf("total e- charge in this run with method 2: %g\n",Qp2);
       //printf("Reprate: %d\n",Reprate);
       Qs=Qp2; //Qs: Charge per second.
       printf("e- charge per second: %g\n",Qs);
       printf("e- charge: %g\n",Q_e);	 
       ne=Qs/Q_e; //Ne: number of electrons per second.
       printf("1. total number of electrons in this run: %g\n",ne);
               }		 
       printf("2. total number of electrons in this run: %g\n",ne);
       c2->Close();
       f->Close();
       return(ne);
       }

results in 3 MeV

results on 3 MeV with Method 2

run in run out NaI Left: in sub out NaI Right: in sub out NaI Left: in sub out NaI Right: in sub out NaI Left: overlap in and out NaI Right: overlap in and out e- rate (Hz) e+ rate (Hz) e+/e+ ratop e- current (A)
3735 3736 Hrrl pos 27jul2012 data ana with Cuts r3735 sub r3736 DL.png Hrrl pos 27jul2012 data ana with Cuts r3735 sub r3736 DR.png Hrrl pos 27jul2012 data ana with Cuts r3735 sub r3736 DL 1.png Hrrl pos 27jul2012 data ana with Cuts r3735 sub r3736 DR 1.png Hrrl pos 27jul2012 data ana with Cuts r3735 over r3736 DL 1.png Hrrl pos 27jul2012 data ana with Cuts r3735 over r3736 DR 1.png


1.18636e+012 0.25656 2.16258e-013 1.90076e-007
3737 3736 Hrrl pos 27jul2012 data ana with Cuts r3737 sub r3736 DL.png Hrrl pos 27jul2012 data ana with Cuts r3737 sub r3736 DR.png Hrrl pos 27jul2012 data ana with Cuts r3737 sub r3736 DL 1.png Hrrl pos 27jul2012 data ana with Cuts r3737 sub r3736 DR 1.png Hrrl pos 27jul2012 data ana with Cuts r3737 over r3736 DL 1.png Hrrl pos 27jul2012 data ana with Cuts r3737 over r3736 DR 1.png


1.97625e+012 0.251457 1.27239e-013 3.1663e-007

Hand calculation on electron rate

At full current on Faraday cup 1 we have 1201 nVs/ADC channel.

[math]\frac{1201~nVs}{50 \Omega} = 24~nAs = 24~nC[/math]

charge rate under 300 Reprate: [math]300 \times 24~nAs = 7200~nC = 7.2 \frac{\mu C }{ s}[/math]


[math] \frac{7.2 \times 10^{-6}\frac{ C }{ s}} {1.6 \times 10^{-19} C} = 4.5 \times 10^{13}[/math]


[go back]