Transistor potential divider query

Hi all,

The attached circuit doesn't do anything, its just my attempt to brush up on transistor operation while evaluating some software I downloaded on my android tablet.

The divider is really straightforward, with switch A open there is 100uA .

What I need explaining is the extra current that is drawn in the divider when switch A is closed.

116uA with 33uA going into the base of the transistor and remaining 83uA going through the bottom resistor of the divider.

The divider has not changed at all 50k/50k.

I have already established a relationship between the base current and the collector current (10 to 1) and the emitter current consists of the collector current plus base current.

Any one care to jog my memory.

Regards.

Mark

The current must change because you connect the load. The base current loaded the voltage divider.

And the base current is equal

Ib = (Eth - Vbe)/( Rth + (β+1)*Re )

And

Eth = Vcc * R2/(R1+R2) = 5V

Rth = R1||R2 = 25KΩ

So

Ib = (5V - 0.75V)/( 25k + 101k) = 0.03373mA = 33.73μA

And the base voltage

Vb = (β+1)*Re + Vbe = 101K*33.73uA + 0.75V = 4.156V

Wow, thanks ill get my calculator out later and try some different values later.

Since a certain transistor part number has a wide range of current gain, usually we make the current in the divider 10 times the typical base current. Then the base voltage is almost the same if the transistor has a low current gain or a high current gain.