GO - NO GO TRANSISTOR:
This circuit consists of comparing a fixed beta to that one of the transistor that the user chooses. This circuit consists of two current sources that one of them is on the base of the transistor and the other one is with the collector of the DUT transistor. The collector voltage is then connected to a verification circuit which switch either a red Led (No Go transistor) or a green Led (Go transistor). A choice for the user is made to choose, whether to make a test of beta = 100 or beta = 200. When the user selects the switch to 100 beta, automatically an arrangement to the circuit is made and vice versa for the beta = 200.Transistor Tester Circuit
Description of circuit :
This is the power input of the circuit. A 10v is the voltage input and then a 7805 voltage regulator will give a 5V to the circuit. The 10V will be used to power up the amplifiers and the 5V will be used for the amplifier reference.
This part will be the current source that will be inputted into the collector of the DUT. This current source will generate 5mA since the positive terminal of the op-amp it has 5V, and so it will be reflected to on the negative terminal (assumed as same point). On the 1K resistor, R1, it will have 10V – 5V = 5V on it. So the current will be:
Isource = (10V-5V) / 1K = 5mA.
This part will be the second current source in the circuit that will be inputted into the base of the DUT. This current source will generate two current which will depend upon the choice of the user. Since the user will select whether beta = 100 or beta = 200, 2 different current (2 different resistors) will be selected from a switch. By selecting the switch on the beta = 100 then the 100k resistor will be selected. This will generate a 50uA current source. (The same concept as current source of part (2)):
Isource = (10V-5V) / 100K = 50uA.
First of all when it is beta 100, collector current will be of 5mA and the base current will be of 50uA as seen on the following formula:
Beta = Ic / Ib
Ib = Ic / Beta => Ib = 5mA / 100 = 50uA
By selecting the switch on the beta = 200 then the 200k resistor will be selected. This will generate a 25uA current source. (The same concept as current source of part (2)):
Isource = (10V-5V) / 200K = 25uA.
First of all when it is beta 200, collector current will be of 5mA and the base current will be of 25uA as seen on the following formula:
Beta = Ic / Ib
Ib = Ic / Beta => Ib = 5mA / 200 = 25uA
This part is where the Device Under Test (DUT) will be placed. On the collector point, there will be the 5mA current source (part 2), on the base the uA current source (part 3) and the emitter connected to ground.
The last section of the circuit is the logic part where it shows the user whether the transistor is good or not (according to the choice of the beta made from switch). If the choice of the user is beta = 100, then the transistor must be greater than or equal to 100 to lit the green Led. If less than 100 beta, then the red Led will lit.
For the beta = 200, it will be the same thing. If the transistor is have an actual beta less than 200 then the red Led will light else the green Led will light on.
After the testing part that is the two current sources, there will be the comparator circuit. If the beta of the device under test lower than the reference value than a red LED will be lit. But if the beta of the DUT is greater than the reference beta, than a green LED is lit. The circuit is composed of two transistors with a common emitter resistor. One of thetransistors has a fixed base voltage while the other base is connected to the output of the collector of the DUT.
If the DUT collector voltage is high, then Q4 will turn on and lit the Red Led. Else Q5 willturn on and the green led will turn on. This two transistors work opposite each other (since the combination that they are connected). The resistor R5 and R7 are as a potential divider for the transistor Q5.