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Power Dissipation Across A Transistor In Cutoff

Akshatha Venkatesh

Jan 14, 2017
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Several things...
  1. The red dots indicate where you didn't join nodes correctly (you did that several places)
  2. The input is best as a sawtooth. This is a bit tricky.
I've fixed both of these here.

You can see the voltages on the circuit and on the scope traces. If you stop the simulation you can move your mouse cursor over the scopes to see how the voltages compare. It may be a little easier if you reduce the frequency of the sawtooth (try 3Hz). To do this, right click on the sawtooth, pick "edit", change the frequency, and click on OK. Then reset or restart the simulation.

The behaviour you see is not exactly as predicted by the datasheet, but this is because the datasheet doesn't describe what happens at intermediate voltages and it assumes a much higher collector current.

However the behaviour is, exactly what you would expect, given sufficient experience.
hey please tell me how do I calculate the Vce of the first transistor ?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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The Vce is the voltage across it. This will depend on how it is biased. If it is in cutoff, what do you think the voltage across it will be?
 

Akshatha Venkatesh

Jan 14, 2017
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The Vce is the voltage across it. This will depend on how it is biased. If it is in cutoff, what do you think the voltage across it will be?
If it is in cutoff , i think the voltage across it will be the voltage at the emitter . The entire emitter voltage drops across the transistor. Am I right ?
 

(*steve*)

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So, apply that to a transistor.
 

(*steve*)

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If you look at the datasheet you'll note that voltages are typically measured with reference to the emitter. In that respect, you are not applying a voltage to the emitter.

There are good technical readings for this, one of which is that the base emitter junction is forward biased, so you need to vary both the base and emitter voltages out the collector voltage. The latter is simpler.

It is better to say that the voltage across the transistor rises to [some definition of of the maximum voltage]. In your case the collector rises to it's Vcc (which is ground), compared to the reference voltage on the emitter. So if the voltage on the emitter is 8V with respect to ground, the voltage across the transistor would be defined in the datasheet as -8V. If you use ground as a reference (to allow voltage to be compared) this would be measured as 8V.

The next question you need to answer is "what is the voltage across any collector or emitter load?"
 

Akshatha Venkatesh

Jan 14, 2017
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So if the voltage on the emitter is 8V with respect to ground, the voltage across the transistor would be defined in the datasheet as -8V. If you use ground as a reference (to allow voltage to be compared) this would be measured as 8V."
This is exactly what I meant when I said the voltage at the emitter will be dropped across the transistor, but I forgot about the negative sign.
 

(*steve*)

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Sorry I don't understand this , please explain.

The transistor and it's load form a voltage divider. To determine the voltage across the transistor you need to consider the voltage that would be dropped across the load.

In this case the impedance of the transistor in cutoff will be vastly higher than the impedance of the load. That will allow you to determine the voltage across each and the current through both.

In this case it's reasonable to assume very small values are zero. This will simplify your calculation. The practical difference is typically inconsequential.

The important part is to ensure that the voltages are defined. For example, a voltage divider formed by two open circuits or two short circuits will have an undefined voltage at the junction. In contrast, a voltage divider formed by an open circuit (or a really high resistance) and a short circuit (or a relatively tiny resistance) can be assumed in most cases to have a trivially simply defined voltage at the junction.
 
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