# transistor voltage divider bias

Discussion in 'General Electronics Chat' started by alexmath, Jun 12, 2014.

1. ### alexmath Thread Starter New Member

May 2, 2014
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Hey everyone, i have a little problem here. When we use a voltage divider bias configuration and i don't understand why the base voltage of the transistor is exactly the same as the voltage on the second voltage divider resistor. Isn't the base of the transistor in paralel with the second resistor and comporting like some kind of load? Please check this short little video. Thank you for your answering!

http://www.youtube.com/watch?v=cbvUMOmUgaM

2. ### MrChips Moderator

Oct 2, 2009
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4,278
You want to make the current through the voltage divider at least ten times the current required by the load on the divider.

3. ### Veracohr Well-Known Member

Jan 3, 2011
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There's your answer. The point between the divider resistors and the transistor base are the same point, how could they not be at the same voltage relative to ground?

I suppose your question is about why the transistor isn't loading down the voltage divider and causing the voltage to drop. Transistors typically have very small base currents, so like MrChips said, a much higher divider current means the loading and the thus voltage drop is negligible.

4. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
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And if the voltage on that point is <0.6v the transistor base pin may not be drawing any current at all.

5. ### BobTPH Active Member

Jun 5, 2013
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Try taking out the emitter resistor and see what happens.

Bob