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what will be the value of Vb.

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the_falcon

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Hi all,
I am getting a bit confused with the voltage division rule in this problem.the picture is attached below.when Vcc is 15 V and Ra and Rb are 100 Kohms and Ib is 0.1 mA, what will be the value of Vb.

I actually thought the procedure would be like (Vcc * Rb)/(Ra+Rb).But it is wrong.

Can anyone explain me this as where I am thinking wrong.

 

Use the quation for the voltage divider with load assuming that the load is represented by the input impedance (resistance) R[in] and that:
[*] I[in]= 0.1mA = U[in]/R[in]

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html

U[in] = Vcc * {R ll R[in]} / {R[a]+R ll R[in]}

Solve the equation for R[in]
I[in]*R[in] = Vcc * {R ll R[in]} / {R[a]+R ll R[in]}
then go back to [*] to calculate U[in] ..

IanP
:D
 

Hi.

Voltage divider is circuit that consists of two impedances connected in series. In your case base of CE transistor is connected to the node between resistors and it sinks a non-zero current, so these resistors don't form divider (note that if base current were zero, Ra and Rb could be considered as divider).

You can derive expression for Vb in the next way:
1) Apply KCL to the node between Ra and Rb: Ira = Irb + Ib.
2) Apply Ohm's law to Ra and Rb: (Vcc-Vb)/Ra = Vb/Rb + Ib <=> Vb = Vcc*Rb/(Ra+Rb) - Ib*Ra*Rb/(Ra+Rb) => Vb = 2.5 V.
 

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    the_falcon

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so good..that was a very nice explanation.

thanks so much

falcon
 

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