Electronic Circuits/Detailed Answers to Quiz 1
Vc = 5v and Vb = 0v. 0v is lower then 5v which means that the diode is not conducting. Which means that it is Reversed biased. So the CB junction is Reversed. Ve = -1v and Vb = 0v. Which means that the base is at a higher potential (or read voltage) then the emitter. Because of this the diode is conducting. So the BE junction is forward biased. Based on our chart this means that the npn BJT transistor is in active mode.
Vc = 3v and Vb = 0v based on this we know that the diode is not conducting. 0v is lower then 3v so the diode can not conduct. Beacuse it can not conduct the diode is reverse bias. So, the CBJ is reverse. Since, Ve = -1v and that value is lower then the base voltage of 0v the diode is conducting. Therefore, the EBJ is forward baised. Because this is true and from the chart we can conclude that the npn BJT transistor is in active mode.
C 2v 0v 2v
Here Vc = 2v and Vb = 0v which means that the CB diode is not conducting. Hence, CBJ is reverse baised. The Ve = 2v and the Vb = 0v. Which mean that the EB diode is not conducting, making the EBJ junction reverse biased. So we end up with reverse-reverse case. From our chart we know that the reverse-reverse case means our BJT transistor is in cut-off mode.
Vc = 5v and Vb = 0v. So the collector has a higher potential then the base. Since diodes will not conduct in this condition the CB diode is reverse-biased. Ve = 4v which is at a higher potential then the ground connected base, the EB diode will not conduct. Making this diode reversed as well. This puts the BJT in Cut-off mode.
Ve equals 2 volts and the base voltage is 1 volt. The EBJ junction is reversed. The collector voltage is 5 volts 4 volts higher the base making CBJ reversed. Once, again we have cut-off mode
This one is tricky. Since Ve has the same voltage as Vb you cant really say the diode conducts. Diodes conduct when the anode is at a higher potential then the cathode. Not at equal. So the diode is reversed. Same holds true for the CBJ. Therefore this diode is in cut-off mode
Both the emitter and the collector are at higher voltages then the base. Making both of them Reverse biased. The reverse-reverse case correspounds to Cut-off mode.
10 volts found on the collector compared to the grounded base shows us that the CBJ diode has to be reversed. The negative emitter voltage compared to the 0v base bais shows us that the Emitter-base junction diode will have to be forward. Making this BJT transistor an active mode one.
The voltage found at the collector is -3v, the voltage found at the base is 0v. The bas is at a higher potential then the collector. The BC diode conducts, the CBJ is forward. The emitter's -3v makes the EBJ conduct, making EBJ forward. This transistor is saturated.
Doing a quick check. We find that Vc < VB and VB > VE. Since Vc < VB the EBJ diode is forward, Since VB > VE the EBJ diode is forward. This is the forward-forward case making the BJT transistor saturatied.
The voltage at the bas is higher then both the Collector and the emitter voltage. So right away we know that both diodes are conducting. Resulting in the forward-forward condition again. Forward-forward condition gives us saturation mode.
Finding a higher voltage at the base then the emitter tells us that the EBJ diode is conducting. Finding a Higher voltage at the collector then the base tells us that CBJ diode is not-conducting. This makes the BJT transistor an active mode device.
Remember that in this problem we have a NPN with the Vbe(on) and Vbc(on) condition present.Vbe(on) = .5 v and Vbc(on) = .4v
The 5v found at Vc compared to the 3v found at Vb makes CBJ reverse. The grounded emitter compared to the positive base makes EBJ forward. Giving us an Active mode BJT transistor.
Ve is lower then the grounded base voltage. So, EBJ is forward. CBJ is reverse because of the 3 volts found at the collector. This is an active mode transistor.
Both Ve and Vc are higher then the 0 volts found at Vb. So both of them are reversed. Refereing to the table we see this makes the transistor in cut-off mode.
Clearly an active mode BJT. The Collector is higher then the base voltage. The emitter is lower then the base voltage.
E 5 10 2
With the emmiter voltage less then the base we have conduction. Causeing EBJ to be forward. So we know that the transisotr has to be in either active mode or saturatied mode. With the Low collector voltage compared to the high base voltage, we find the transistor is saturatied.
Have seen this case before. Everything is grounded the BJT is in cut-off mode. All junction diodes are reversed.
G -3v 0v 3v H 10v 0v -2v I -3v 0v -3v J -5 11 -2 K 2 5 -3 L 7v 10v -10v M 0v 10v -.1v N 2.2v 1v -1v O 7v 0v -3v