Why is my Bit Error Rate (BER) become constant, is my code incorrect?

I am working to simulate the BER between 2 users in CDMA with convolutional codes and modulation scheme of QAM-16. From the graph i've attached, the BER of user 1 and user 2 are the same and constant. It seems that the SNR does not affect the transmission. Is there anyway i can improve the graph? Here is my code:

M = 16;                 % Modulation order
k = log2(M);            % Bits per symbol
EbNoVec = (0:20)';      % Eb/No values (dB)
numSymPerFrame = 1;   % Number of QAM symbols per frame
users=2;            % Number of Users
trellis = poly2trellis(7,[171 133]);
tbl = 32;
rate = 1/2;
%------------------Generation of Walsh code--------------------------------
noOfSubCarrier =20;                               %Number of  Data Sub-Carriers
walsh=hadamard(noOfSubCarrier);
code1=walsh(10,:);                   
code2=walsh(20,:); 

berEst1 = zeros(size(EbNoVec));
berEst2 = zeros(size(EbNoVec));%Initialize the results vector

% The main processing loop executes the following steps:
% Generate binary data and convert to 64-ary symbols
% QAM modulate the data symbols
% Pass the modulated signal through an AWGN channel
% Demodulate the received signal
% Convert the demoduated symbols into binary data
% Calculate the number of bit errors

for n = 1:length(EbNoVec)
    % Convert Eb/No to SNR
    snrdB = EbNoVec(n) + 10*log10(k*rate);
    % Reset the error and bit counters
    numErrs1 = 0;
    numErrs2 = 0;
    numBits = 0;

      % Generate binary data and convert to symbols
      B=10;
      dataIn1= rand(1,B);
      dataIn2=rand(1,B);
      symbols1= unique(dataIn1);
      symbols2= unique(dataIn2);
      probs1 = histc(dataIn1,symbols1)./numel(dataIn1);
      probs2 = histc(dataIn2,symbols2)./numel(dataIn2);
      [dict1, avglen1] = huffmandict(symbols1, probs1);
      [dict2, avglen2] = huffmandict(symbols2, probs2);
      comp1 = huffmanenco(dataIn1,dict1);
      comp2 = huffmanenco(dataIn2,dict2);
       % Convolutionally encode the data
      dataEnc1 = convenc(comp1,trellis);
      dataEnc2 = convenc(comp2,trellis);

      % QAM modulate
      txSig1 = qammod(dataEnc1,M,0);
      txSig2 = qammod(dataEnc2,M,0);
      %------------------Spreading & IFFT for User1------------------------------
      Tx_data1=txSig1';
      Spread_User1=Tx_data1*code1;    % Spreading 
      Spread1=(Spread_User1)';
      ifftdata_user1=ifft(Spread1);      % Taking the IFFT
      ifftdata1=ifftdata_user1';
      %------------------Spreading & IFFT for User2------------------------------
      Tx_data2=txSig2';
      Spread_User2=Tx_data2*code2;    % Spreading 
      Spread2=(Spread_User2)';
      ifftdata_user2=ifft(Spread2);      % Taking the IFFT
      ifftdata2=ifftdata_user2';
      %----------------------Addition of all signal------------------------------
      TotSignal = Spread1+Spread2;
      % Pass through AWGN channel
      AwTotSignal = awgn(TotSignal,snrdB,'measured');
      %-----------------------Removing the FFT & De-Spreading--------------------
      fft_data_received =fft(AwTotSignal);
      Rec_Data1=(AwTotSignal'*code1');
      Rec_Data2=(AwTotSignal'*code2');
      % Demodulate the noisy signal
      rxSym1 = qamdemod(Rec_Data1,M,0);
      rxSym2 = qamdemod(Rec_Data2,M,0);
      data1=vitdec(rxSym1,trellis,5,'cont','unquant');
      data2=vitdec(rxSym2,trellis,5,'cont','unquant');
      % Convert received symbols to bits
      %dataOut1 = de2bi(data1,k);
      %dataOut2 = de2bi(data2,k);
      % Calculate the number of bit errors
      nErrors1 = biterr(comp1',data1);
      nErrors2 = biterr(comp2',data2);
      % Increment the error and bit counters
      numErrs1 = numErrs1 + nErrors1;
      numErrs2 = numErrs2 + nErrors2;
      numBits= numBits + numSymPerFrame*k;

      % Estimate the BER
      berEst1(n) = numErrs1/numBits;
      berEst2(n) = numErrs2/numBits;
  end

Answer 1

[Sujay Mondal]

Hello, psychecrew have got the solution. Actually I am also facing the same problem during QPSK transmission I also getting a flat line with increasing SNR. Is there any solution to this particular problem.

Thanking You,

Sujay Mondal.