Hi Guys,

I have this schematic and I want to know the output at Vo. I am confuse on how to compute for the output since the emitter of the BJT is connected to -5V with a current of 12A. I am not sure if Vo will be -0.7V or -5V. Can anyone tell me what the output will be.

Attached is the schematic of the circuit.

 

I'd like to see how long a 2N2222 would survive at 12A current. There's nothing limiting the base current. It's a strange circuit. Maybe the 200Ω was meant to be in the emitter leg ...?

If you had time to measure before something died, the voltage at Vo would be dependent upon the voltage drop across each device (diode & transistor) at 12A - if that indeed is the actual current that would flow. How did you arrive at 12A anyway?

 

12A is the maximum current of the supply which is -5V. With this circuit, I do not expect that all the 12A will flow through the circuit, I am expecting that the current flowing in the circuit is less than 12A since it will be limited by the diode or the transistor. When this circuit is energized, I am confuse if the voltage at Vo is -5V (since Vce is small) or it will have a voltage equal to the diode drop which is 0.7V.

 

12A is the maximum current of the supply which is -5V. With this circuit, I do not expect that all the 12A will flow through the circuit, I am expecting that the current flowing in the circuit is less than 12A since it will be limited by the diode or the transistor. When this circuit is energized, I am confuse if the voltage at Vo is -5V (since Vce is small) or it will have a voltage equal to the diode drop which is 0.7V.

OK I understand where the 12A comes from.

But you haven't understood the very important point about the physical realities of your proposed circuit.

The base drive is from +5V to -5V without any limiting resistance other than the base-emitter forward voltage drop. You need to include a limiting resistor to have a sensible base current. Also there needs to be a limiting resistance in the diode - transistor series path.

Suppose as I suggested, you put the 200Ω in the emitter leg of the transistor. Also you would have to insert a limiting resistor in the base leg. Then you would have some sort of limit on both the base drive and the main diode - transistor paths. Alternatively you could drive the base at ground potential (rather than +5V) and you wouldn't need a base current limiting resistor.

In that case, if the transistor is in saturation the collector current would be approximately 4.3/200=21.5mA (a sensible value). The output voltage would then be at -0.7V or thereabouts. That is Vo would be at one diode forward bias voltage drop below ground. The bulk of the voltage drop would be across the 200Ω emitter resistor with a small Vce_sat drop across the transistor itself.

 

Like this ....

 

How about if the circuit looks like this attached file. This is an existing circuit and there is no emitter resistor. How will the circuit respond and what will be the Vo?