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I have the following circuit and I am trying to find VB, VE, and VC, and consequently IB & IC. The "hints" I am given are (1) to assume that vs has a DC value of 0V, (2) to use the large-signal BJT model, and (3) I can neglect CE. enter image description here


From what I understand, the equivalent large-signal model for an npn BJT in the forward active region (that is assuming the setup above is in forward active region) is shown below: enter image description here
Now given that the DC value of vs is 0, and applying KVL to the base-emitter loop (also assuming that VBE = 0.7V), I'm getting VB = -5V and VE = -5.7V. Now before I go further, I have a gut feeling that this is wrong, but I'm having a hard time understanding why. My thought is that since vs has 0VDC, VB should be 0 as well?


The theory behind BJT operation is a major point of weakness for me so any help is appreciated!


schematic

simulate this circuit – Schematic created using CircuitLab


Here is the circuit I performed the KVL on based on the hints given to me and the assumption I made that the BJT is in forward active region.

blackmcgraw
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  • Maybe you should spell out exactly how you applied KVL to get those numbers. I see two unknown voltages (without making simplifying assumptions). – Spehro Pefhany May 12 '21 at 10:42
  • Why do you think that the voltage at the base will be equal to VB = -5V if vs is 0V? – G36 May 12 '21 at 12:24
  • @SpehroPefhany I added the schematic I performed KVL on, which as I'm looking at it again I can see that it gets me nowhere. I also forgot to add that I was given another hint that 0V = V_R2 + V_BE + V_Rbias – blackmcgraw May 12 '21 at 17:39
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    Can you apply KCL and KVL at the same time? Because your equivalent circuit looks correct. – G36 May 12 '21 at 17:44
  • @G36 Thank you for your input and sorry for not clarifying, but yes I'm not constrained to just performing KVL. I haven't applied KCL to the full model yet, though. I'll try that now thank you – blackmcgraw May 12 '21 at 17:49
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    I meant something like this: Vee = V_R2 + Vbe + R_E thus Vee = IbR2 + Vbe + IeR_E – G36 May 12 '21 at 17:56
  • Hmm, I'm not sure how to proceed with that sorry. I would think that V_B is needed to get I_B? And I_E = I_B + I_C, which according to the large_signal model: I_C is dependent on I_B. – blackmcgraw May 12 '21 at 18:11
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    For the BJT we have \$I_E = I_B + I_C = I_B + β I_B = (β + 1)I_B\$ Thus, we can solve for Ib or Ie. Try to analyze this example https://electronics.stackexchange.com/questions/471906/calculation-of-base-current-and-what-decides-the-current-through-collector-emitt/471923#471923 and maybe you should try to read this as well https://electronics.stackexchange.com/questions/355899/how-is-possible-that-with-same-ibase-there-is-more-than-one-vce/355955#355955 – G36 May 12 '21 at 18:45

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