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I am trying to understand how BJT high side circuit works. I ran it in LTSPICE, but I fail to see how certain voltages get computed. enter image description here

Can someone, please, explain how to arrive at Vbq2 = 14.2V? Also, why Ieq2 = 5mA?

Thanks

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    \$\begingroup\$ See here for a discussion that should address all questions and more. \$\endgroup\$
    – jonk
    Jul 4, 2020 at 23:30
  • \$\begingroup\$ Welcome. Both Q1 and Q2 are saturated ON, so Q2 collector can only be 15 volts -Vbe of Q2. R4 is not enough of a load to drag the voltage down. Vbq2 is cause by the 600 mV Vbe drop of Q2. Please read all you can about bjt transistors. \$\endgroup\$
    – user105652
    Jul 4, 2020 at 23:34
  • \$\begingroup\$ Dear jonk, thanks for Bob Pease link. \$\endgroup\$
    – vgeng
    Jul 5, 2020 at 4:46

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When Q1 is saturated Vbq2 is 15V-Vbe of Q2. Vbe of a saturated transistor is around 0.7V for reasonable base current, so Vbq2 will be about 14.3V.

Ieq2 is the emitter current of Q2 and is equal to the sum of base current and collector current.

Note that Ic of Q2 is less than half the base current. You can infer something about how reasonable the base current is from that.

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  • \$\begingroup\$ Thanks all. Having been away from analog design since college, just getting back to real engineering and filling the gaps i made in my studies (shame on me). Vbe in Spice is 0.7V What throws me off is that SPICE shows VBq2 = 14.2V , not 14.3V. Any idea why? So, as i understand now, to get Ieq2 i need to get the base current first. Then, already having Icq2, i can figure out Ieq2 = Ibq2 + Icq2. Got it. Since Q1 is in saturation, Vcq1 is almost 0 or virtual ground - correct? Then i can compute Ir1 by just dividing 14.2 (or 14.3) by R1. Then Ibq2 = Ir1 + Ir2. Is that the right way to analyze? \$\endgroup\$
    – vgeng
    Jul 5, 2020 at 4:35
  • \$\begingroup\$ Vbe is larger with higher current and lower temperature, it is not fixed. See my last sentence which preemptively hints at your question. Yes, that is correct. \$\endgroup\$ Jul 5, 2020 at 5:13
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    \$\begingroup\$ yep, got it. I also checked jonk's link about Vbe and it explained what you just mentioned about Vbe . Thanks much all. \$\endgroup\$
    – vgeng
    Jul 5, 2020 at 15:03
  • \$\begingroup\$ One caveat now. If R2 =100R, Q2 becomes cutoff and Ibq2 = 0. I can see it in Spice, but not sure my explanation below is correct. My goal is to be able to explain this w/o using Spice. If i assume VBEq2 = 0.7, then IR2 for 100R is 7mA, and IR1 remains 4.3mA. IR1 - IR2 = Ibq2. Now it means IBq2 = -4.3mA - (-7mA) = +2.7mA, which does not make sense. I think max of Ibq2 should be zero. Then max IR2 should be -4.3mA. So,VBeq2 must be < 0.7V to achieve that IR2. If VBEq2 < 0.7,then Q2 is cutoff. This sounds a bit fishy to me.Please, suggest the correct explanation of why Q2 cuts off is R2 = 100R. \$\endgroup\$
    – vgeng
    Jul 7, 2020 at 4:04
  • \$\begingroup\$ If you ignore the transistor junction, the Vbe voltage would be about (0.1/3.4)*15 = 440mV. That is not enough voltage to cause much base current to flow, so the transistor remains off. The base-emitter junction behaves like a diode. If you apply a voltage much less than 0.7V (say 0.3V) then very little current flows, if you apply a voltage much greater than 0.7V (say 0.9V) then a great deal of current flows. For some purposes it's reasonable to say that when the transistor is 'on' Vbe ~= 0.7V. When the transistor is off, Vbe could be 0.5V or -5V, not much current flows. \$\endgroup\$ Jul 7, 2020 at 4:37

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