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Say we have this circuit below: enter image description here

And we were tasked to:

  • a.) Get Ve and Vc when Vb=0v.
  • b.) To find what Vb will make the transistor be in the cutoff and saturation region.

The following are given:

  • α is approximately 1; VBE=0.5V at the edge of conduction.
  • I1=8mA; I2=2mA; R1 = 2k; R2=2k

I already know what to do with B, just having trouble with the circuit analysis itself lol. I'm not sure how I1 would contribute to Ic and I2 with Ie. Could somebody point me in the right direction? I am thinking that the transistor might be in active mode because the emitter is connected to -Vcc and much more negative potential than Vb. So if i do a KVL eq around the base-emitter loop, ill get Vb-Vbe-Ve-(-Vcc)=0 so then Ve = 0-0.7+Vcc. Now at this point, i'm not really sure how to get Vc to prove that CB junction is reverse-biased and to prove that the transistor is in active mode. Would I1 be Ic and Vc be Vcc-I1R1?

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    *"I'm not sure how I1 would contribute to Ic and I2 with Ie."* The sum of the currents in the node should equal zero. So \$I_1=Ic+I_{R1}\$ etc. – Aaron Oct 20 '20 at 16:05
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    Notice that you know Vb = 0V thus Ve = -Vbe = 0.7V therefore I_R2 = -0.7V/2k = 0.35mA and Ie = Ic = 2mA - 0.35mA = 1.65mA and finally we have I_R1 = 8mA - 1.65mA = 6.35mA and Vc = 6.25mA * 2k = 12.7V And here you can read about saturation https://electronics.stackexchange.com/questions/276146/a-question-about-vce-of-an-npn-bjt-in-saturation-region/276266#276266 – G36 Oct 20 '20 at 18:22
  • @G36 May i ask why did you subtract IR2 from 2mA when getting Ie? Wouldnt the KCL equation be Ie= 2mA + 0.35mA? Since I2 and IR2 are going out of the node? –  Oct 21 '20 at 02:15
  • @Aaron Thanks. May I ask how would -Vcc in the emitter affect Ve? I'm thinking if I do a KVL around the base emitter loop, it would go something like Vb-Vbe-(-Vcc)=Ve? –  Oct 21 '20 at 03:33
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    @Batt The voltage at VE is negative and this means the the current will flow from into the VE node and the current path is GND--->R2--->I2--->-Vcc--->GND – G36 Oct 21 '20 at 12:27
  • @G36 VE is negative? Isn't VE=0.7 by KVL at the B-E junction? –  Oct 21 '20 at 12:36
  • Vb is at GND (0V) and -Vcc has a much more negative value than Vb. And when the BJT is in the active region the Vbe will be around 0.6...0.7V. But since VB = 0V thus VE must be at the -Vbe = -0.7V – G36 Oct 21 '20 at 13:27
  • @G36 You're right. I got confused with my KVL. Thanks! –  Oct 21 '20 at 14:14
  • @G36 For cutoff, Ib=Ie=Ic=0 so Vc would just be I1*R1 and Ve would be -IR2*R2? –  Oct 22 '20 at 04:44
  • Exactly as you said. And in saturation? – G36 Oct 22 '20 at 12:48
  • at saturation, would Ve still be Ir2/R2 and Vc still Ir1/R1? I'm having trouble doing the saturation analysis. –  Oct 22 '20 at 13:25

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