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Let R1 be upper resistor and R2 be the lower resistor of a voltage divider bias bjt circuit. Then can I visualize the DC bias of the base emitter junction as follows: I am trying to look at DC biasing in a different way. Is it correct that if I consider the input into the base emitter junction as a D.C. resistance that is chosen by my choice of Re (not r'e) and an arbitrary beta. Then can that portion of circuit appear as a simple current divider with R2. If this true, then it would seem R2 has to chosen so that adequate current drive is present for the base emitter junction to conduct property. I.e. R2 is not loading the base emitter junction down.

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  • \$\begingroup\$ "R2 is not loading the base emitter junction down"...May I ask: What do you want to say with this sentence? \$\endgroup\$ – LvW Mar 7 '17 at 19:23
  • \$\begingroup\$ i am trying to say that R2 would draw a smaller amount of current in relation to the base current \$\endgroup\$ – Creative Mar 7 '17 at 19:52
  • \$\begingroup\$ No - that is contraproductive; read Andy aka`s answer (factor 10). \$\endgroup\$ – LvW Mar 7 '17 at 20:11
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it would seem R2 has to chosen so that adequate current drive is present for the base emitter junction to conduct property

Yes this is very true and the rule of thumb is that the current that flows though R1 and R2 should be about ten times higher than that current that flows into the base (determined by beta and Re). This rule of thumb means you can estimate R1 and R2 without considering base current and this estimation yields the volage at the base.

That base voltage then determines emitter voltage (about 0.7 volt lower than the base) and that voltage divided by Re tells you the current that largely flows through Rc. This of course is chosen (in most circumstances) to position the quiescent collector voltage at about 50% of Vcc.

Having said all that, there are exceptions to the rule of thumb especially when you want a more accurate positioning of the quiescent colelctor voltage and value tweaks will be made; you can consider the loading effect of the base current on the potential divider formed by R1 and R2 and make an adjustment to one of those resistors in order to reposition the quiescent collector voltage.

Or you bite the bullet and learn how to use a sim tool like LTSpice (free of course). But it always helps to understand the underlying trends.

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  • \$\begingroup\$ Thank you very much for your help. I will download LT Spice and use to compare my mathematical circuit analysis to what is simulated \$\endgroup\$ – Creative Mar 7 '17 at 21:50

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