Hi Markus,
  The tester cannot define the transistor with interal diode protection:
  so that when we have changed the position of test-leads , the tester gives 
  different data NPN or PNP and the pinouts also changed for the same device!!!
     After test , here is an algorithm to distinguish among: 
     NPN+diode(case1);
     PNP+diode(case2);
     3diode(case3);
     3diode(case4);
     2dioden in serie(case5).
     
          C(Y)                                          C(X)
            o                                            o
            |--+                                         |--+
            | _|_            +--|>|--+o C(Y)             | _|_              +--|<|-+-o C(X)
          |/  /_\            |      _|_                |/ _\_/_             |     _|_                       Z
 B(X) o---|    |   = B(X)o---+      /_\       B(Y) o---|    |    = B(Y) o---+    _\_/_        A(X) o--|>|---o---|>|---o K(Y)
          |> --+             |       |                 |< --+               |      |
            |                +--|>|--+o E(Z)             |                  +--|<|-+-o E(Z)
            o                                            o
          E(Z)                                         E(Z)

   NPN+diode(case1)    =     3diode(case3)     ;      PNP+diode(case2)    =    3diode(case4)   ;  2dioden in serie(case5) .
   
  1)  We see that in the 5th cases there are 3 diodes, so the tester will 
detect them by the first test: 
  CheckPins(TP1, TP2, TP3);
	CheckPins(TP1, TP3, TP2);
	CheckPins(TP2, TP1, TP3);
	CheckPins(TP2, TP3, TP1);
	CheckPins(TP3, TP2, TP1);
	CheckPins(TP3, TP1, TP2); 
under : (NumOfDiodes == 3) .
  2) Now it is possible to define one COMMUN ANODE=X and one COMMUN CATHODE=Y  
(so if it is a transistor: the Base is X for NPN ,or Y for PNP , 
 Z is the Emmiter of course ) .
  3) there is a second test:
  N°  X   Y   Z    NPN+D       PNP+D      3diode(case3)    3diode(case4)   2dioden in serie(case5)
  1-  5   5   0   vX=0.7;vY=0 vX=0.7;vY=5  vX=0.7;vY=5     vX=0.7;vY=5      vX=0.7;vY=5
  2-  5   0   5
  3-  5   0   0
  4-  0   5   5
  5-  0   5   5
  6-  0   0   5  
  So the test n°1 can help us :
   if: vY= 0 volt ==> NPN+diode with B=X and C=Y and E=Z
   if: vY= 5 volt ==> PNP+diode with B=Y and C=X and E=Z
   
   Thanks Markus.