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Hello today we had a class in electronics and we discussed the BJT operation.Here are my notes:

Suppose we have this circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Ie = (Vb-Vbe)/(Re+Rb/β)

Ib = Ie/β

Ιc = Ie

VR2 = IcxR2 = 4.3/101x100 = 4.25V

VCE = V1-VR1-VR2 = 10-4.25-4.3 = 10-8.55 = 1.45V

Isnt that correct?

Miss Mulan
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    No, I don't think you wrote everything down correctly. You should go through your textbook and double-check everything. – Elliot Alderson Apr 01 '21 at 19:00
  • Where am I wrong? – Miss Mulan Apr 01 '21 at 19:04
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    Your professor said things. You heard, perhaps, different things. You wrote, perhaps, still different things -- not capturing what you heard precisely. Beyond that, we've no way of looking into your head to see what you understand of (a) what you wrote, (b) what you heard, or (c) what the professor said. But it would be okay for you to post up a schematic, analyze it with what you understand, and then ask us if you got that right. – jonk Apr 01 '21 at 19:05
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    I'm not going to proofread all of your notes and do your work for you. Your labels for currents and their directions seem odd. Your symbols seem odd. You have random comments and mathematical expressions and the relationship of all of these things is unclear. – Elliot Alderson Apr 01 '21 at 19:07
  • @ElliotAlderson I use the direction of the flow of electrons bad me:( – Miss Mulan Apr 01 '21 at 19:13
  • @MissMulan I'd like to add two links you might look at. This [first one](https://electronics.stackexchange.com/a/429995/38098) justifies your schematic by showing how a textbook CE BJT amplifier stage's biasing pair can be converted to its Thevenin equivalent at the base. At that point, the schematic looks like yours. So that is a segue. This [second link](https://electronics.stackexchange.com/a/438186/38098) goes into more detail in analyzing it. – jonk Apr 01 '21 at 19:16
  • @MissMulan Take a look at your first equation. Do you think it might require any parentheses? – jonk Apr 01 '21 at 19:20
  • Yes sorry I will edit. – Miss Mulan Apr 01 '21 at 19:20
  • @MissMulan Perhaps one more set of them?? – jonk Apr 01 '21 at 19:22
  • @jonk What more set? – Miss Mulan Apr 01 '21 at 19:22
  • @MissMulan Do you take Vb, which is a voltage, and then subtract from it Vbe/(Re+Rb/beta), which is a current? Or do you first take (Vb - Vbe) and then divide? – jonk Apr 01 '21 at 19:24
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/122548/discussion-between-miss-mulan-and-jonk). – Miss Mulan Apr 01 '21 at 19:24
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    @MissMulan I get numbers similar to yours. Assuming \$\beta=100\$, I get \$V_E\approx 4.2578\:\text{V}\$ and \$V_C\approx 5.7843\:\text{V}\$ so \$V_{CE}\approx 1.5265\:\text{V}\$. But I didn't use the exact same equations you did, as I used \$\beta+1\$ for the emitter-related bits. Close enough, though. – jonk Apr 01 '21 at 19:58
  • IE = IB * (beta+1) for more accurate analysis. – Meenie Leis Apr 01 '21 at 20:02

1 Answers1

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Bias point first order calculations (we neglect Ib):

Ie = (Vb - Vbe) / Re; /* Neglect Rb/β */

Ie = (5 - 0.7) / Re = 43 mA

Ιc = Ie = 43 mA

VR2 = R2 * Ic = 100 * 43 mA = 4.3 V

VR1 = R1 * Ie = 100 * 43 mA = 4.3 V

Vce = V1 - VR2 - VR1 = 10 - 4.3 - 4.3 = 1.4 V

Enrico Migliore
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