Simplifing my Circuit

hgmjr

Joined Jan 28, 2005
9,029
Let Rb= 1k , Rc=100 and Vcc= 5v

Can anybody simplfy this circuit and explain why is Vcc-V(Rb)-V(Be)=0

Thanks
The above equation states that if you begin with the voltage applied to the circuit, 5V in this case and then you subtract the voltage drop across the resistor Rb and then you subtract the voltage across the base-emitter junction, you should end up with 0 volts.

hgmjr
 

wmodavis

Joined Oct 23, 2010
739
It is also called Kirchoff's Voltage Law. The sum of voltage drops around a loop is equal to zero.

[SIZE=+1]Kirchhoff's Voltage Law - Introduction[/SIZE]


Kirchhoff's Voltage Law - KVL - is one of two fundamental laws in electrical engineering, the other being Kirchhoff's Current Law (KCL).
  • KVL is a fundamental law, as fundamental as Conservation of Energy in mechanics, for example, because KVL is really conservation of electrical energy.
  • KVL and KCL are the starting point for analysis of any circuit.
  • KCL and KVL always hold and are usually the most useful piece of information you will have about a circuit after the circuit itself.
 

Thread Starter

ahmedzica

Joined Jan 9, 2012
20
The above equation states that if you begin with the voltage applied to the circuit, 5V in this case and then you subtract the voltage drop across the resistor Rb and then you subtract the voltage across the base-emitter junction, you should end up with 0 volts.

hgmjr
I already know that. But I'm wondering if we used KVL why ignored V(Re) as in the left loop

Thanks for your help
 

Thread Starter

ahmedzica

Joined Jan 9, 2012
20
I really appreciate your efforts. and I know how to use KVL. But I can't imagine how this circuit would be if we replaced this Vcc (1 Terminal ) with 2 Terminal and Removed that ground over there
 

Adjuster

Joined Dec 26, 2010
2,148
If Vcc were replaced by a connection to another network, you would have another more complex circuit. If there is a particular circuit you are thinking about, perhaps you might want to explain what you have in mind

What are you actually trying to analyse - the simple transistor circuit considered as a load on Vcc?
 

Adjuster

Joined Dec 26, 2010
2,148
At low input voltages the current may be estimated approximately by supply current I of (Vcc) = (β+1)(Vcc-Vbe)/Rb.

Since here Rb is only 10 times Rc it is likely that, with a reasonably high gain transistor, above a certain voltage the transistor will saturate. This is because (β+1)(Vcc-Vbe)/Rb will exceed Vcc/Rc. At higher input voltages therefore, I(Vcc) ≈ (Vcc-Vbe)/Rb + Vcc/Rc
 
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