enter image description here

Sorry for the noob question. From my understanding, R1s acts as pull down resistors for the base and R2 will limit the current flow to the base. I have seen both configurations, but couldn't work out which one is actually appropriate, if there's any. Given the same resistance for R1 and R2, wouldn't the circuit on the left draw a lot more power? But if we choose appropriate resistor values so that the current draw are similar, will there be any difference between the two circuits?

How about in the case of MOSFETs?

Thank you!

  • \$\begingroup\$ Yes, there is a difference. Use mesh current analysis to see what the difference is. \$\endgroup\$
    – user103380
    Sep 18 '18 at 20:21
  • 4
    \$\begingroup\$ In your 1st circuit, R1 has no effect on the operation of the transistor. \$\endgroup\$
    – brhans
    Sep 18 '18 at 20:21
  • \$\begingroup\$ Hint: if Vcc=1V, the left xtor will be on full but the right won't. \$\endgroup\$ Sep 18 '18 at 20:32

Your two circuits break down into the following:


simulate this circuit – Schematic created using CircuitLab

On the left side, you can see that \$R_1\$ is isolated and irrelevant to the BJT circuit. On the right side, you can see that there is a difference that I'll get to, shortly.

Note that you are correct about the possibility of setting \$R_2\$ on the left side schematic so that similar base current occurs in \$Q_1\$ in both schematics.

Okay. So what's better about the right side schematic? On the left side, you are stuck using your voltage rail, \$V_\text{CC}\$. But on the right side you can fabricate any source voltage: \$0\:\text{V}\le V_\text{TH}\le V_\text{CC}\$. As well as arranging for whatever you need as \$R_\text{TH}\$. This is a huge advantage and provides another degree of design freedom.

By the way, few people would ever consider hooking up an NPN BJT so that the collector is tied directly to the positive voltage rail and the emitter is tied to ground. That is just asking for trouble.

  • \$\begingroup\$ I didn't downvote. - If OP needs to ask this kind of question then I doubt s/he understands how to calculate the equivalent Thévenin circuit, which you have done and not mentioned that you've done. \$\endgroup\$ Sep 19 '18 at 4:42
  • \$\begingroup\$ @HarrySvensson I was just directly answering the question about differences. If the OP doesn't understand and points out a question, then I'll try and expand what I wrote to help more. You might be right and thanks for the comment. As far as downvoting, my problem when they don't take time to explain themselves is: (1) I may not be able to repair the problem if they won't tell me what they see; (2) Others may not understand what the difficulty might be if they don't tell others what they see; (3) They are cowardly and also disrespectful of other readers by failing to explain problems they see. \$\endgroup\$
    – jonk
    Sep 19 '18 at 4:57

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