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transistor collector to emitter voltage

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electronicslearner77

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Sorry for asking a very basic question. I had this doubt for a very long time. Assuming a common emitter configuration and an npn transistor we know emitter is connected physically with base similarly base is connected to collector as well. when we take transistor equation I can understand Vbe and Vcb voltage drops since they are physically connected but from where the Vce voltage drop is coming into picture as they are not physically connected. What am I missing?
 

Vce is not physically connected . it is drop across collector to emitter.
if you apply , Kirchoff's laws , then it is that volage.
 

When the base is "physically" connected to the emitter as well as to the collector node, there is also a connection between C and E.
 

OK, little bit understanding. Even there is no physical connection there is electrical connection. Now from the standard text book I took the sign convention of transistor here the assumption is made that all the currents flowing into the transistor are considered positive that is ok for me but regarding the voltage drops VEB, VCB, VCE. From what i understand Emitter voltage is higher than Base, but in reality if you see Base is at higher voltage than emitter by around 0.6V and sometimes 0 in cutoff which are fixed. Then why the book considers Emitter at higher voltage than base. I am not sure i am clear in my question but somehow i want to know the sign notation of voltage drops in transistors that is which terminal voltage is at which level compared to others.
 

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. I am not sure i am clear in my question but somehow i want to know the sign notation of voltage drops in transistors that is which terminal voltage is at which level compared to others.

Yes - you are not clear in your question. Are you talking about an npn or a pnp transistor?
This important as far as the B-E voltage (positive/negative ) is concerned.
 

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