npn bjt connections

[SatyamBhujan]

Dear all,

I am having great troubles with the circuit shown below. According to me, this should work fine but for reasons unknown to me, the bulb is not lightning when I am closing the switch. Please Help me undertstand where and what I did wrong.

The specificatiosn are:

BJT NPN Transistor

12 V power supply

Vcc = 5V

Lamp: 12 V / 10 W

[johnsold]

You do not show any ground connection. As a result is is likely that no base current flows in Q2.

Also, there is no place in that circuit where you can connect a ground that will result in close to 12 V on the lamp.

BJT devices operate on current. Consider the current loops in your circuit. The ONLY loop which exists is V4, Q2 (collector to emitter), S4 (when closed) and X1.  When you add a gound somwhere, you can get base current in Q2: Vcc -> Q2 base -> Q2 emitter -> ???? -> Ground.  If you connect the ground on either side of S4 you will get very large base currents - currents which would destroy a real transistor instantly! If you connect the ground at the negative end of V4, the circuit will have about 10 mA of base current and the lamp will have about 4.3 V across it (with S4 closed). A real lamp would probably glow dimly at that level.  If you put the ground at the collector of Q2, you will get very large base to collector currents and about 16 V across the lamp.  Fortunately for your lamp, the transitor base will probably serve as a fuse and burn out before you destroy the lamp.  However, I do not want to be in your lab when you test this.  I do not like the flying pieces and bad odors which result.

Lynn, MSEE

[SatyamBhujan]

Dear Lynn,

Thank you very much for your enlightening comments. Sincerely, I never knew that there was so much dangers in that.

However, I tried a great deal of combinations and I've got one which worked ( meaning that the lamp lights up ).

I would like to have your views on that and I would highly appreciate it if you could tell me the good and bad things on the circuit.

[johnsold]

As I mentioned earlier, BJTs are current operated devices.  Check the currents in the base in your simulations.

The circuit on the left probably has thosands of amperes of base current, to get 1 ampere of collector current!  That is not a very good amplifier.

The circuit on the right will deliver about 4.3 V to the lamp. Most of the current will flow through the base and very little through the collector - again, not a very good amplifier.

Here is how I would build such a circuit.

Even these values are not entirely realistic.  They result in a base current of ~100 mA and a base voltage of about 4 V. Normal base voltage is about 0.7 V. They also result in over 10 W of dissipation in the transistor - far more than its rating.

Lynn

[SatyamBhujan]

Thank you once again Lynn for replying in such a short notice.

Your circuit reflects the confidence of working properly if the resistor values are correctly calculated. 

I am surely going to try it out.

Coming back to the cicrcuit that I posted, I am getting a current of approximately 4.11 A. for the left circuit. I am very interested in rectifying this circuit because I

want to know where and what went wrong... ( I guess this is how we grow up in any particular endeavour )

Please find attached the circuit.

[johnsold]

I do not have Multisim.  Can you post an image of your schematic as you did with the other circuits?

Lynn

[SatyamBhujan]

I have tried to collect as many readings as I deemed necessary. I hope this helps.

[johnsold]

I am surprised that the base current is only 4 A.  The model used for the transistor must have a significant base resistance.

Note that the current gain in that circuit is Ai = Ic/Ib = 0.75/4.117 = 0.18. You need to put in five times more current than you get out!

The power dissipation in the base is (5 V)*(4.117 A) = 20.6 W.  For a transistor with a 500 mA absolute maximum collector current and 0.8 W total power dissipation rating, this is assured instant destruction.

The problem is that you have nothing in the base circuit to limit the current.  See the circuit I posted earlier.

For the Q2 circuit the load in the emitter serves as feedback to reduce the base current to an acceptable value.

Lynn

[SatyamBhujan]

Hmm.... 

In fact I was usinng the Vcc source as a representation for a sensor which will actuate the circuit if ever the output is 1.

My aim is to open/close a gate whenever the motion sensor detects the presence of someone. And thus I have used the Vcc source to represent the motion sensor and the lamp to show actuation of the motor ( I  know it should be bidirectional but this is only a trial circuit to understand whether transistors would be a good measn to implement it).

In short, the procedures are like that:

- When the motion sensor detects the presence of someone, the motor should open/close the gate ( which is controlled by a stepper motor ).

If the circuit works fine for a lamp, then I'll adjust it for a motor ( and of course I'll think somehting for the bididrectional motion as well ).

Well the above were the things that I had thought for the project. 

can you enlighten me on how can I implement something more efficient with less power loss?

[johnsold]

Is the output of the motion sensor a continuous voltage representing the distance or velocity of the object it senses or is it a binary device which switches between zero and 5 volts when motion it detected?

Lynn