Timeline for Finding the resistors value in a transistor circuit that uses a constant current source
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| S Apr 30, 2017 at 16:55 | history | suggested | try-catch-finally | CC BY-SA 3.0 |
Clarified title (added textual description of the schematic); mathjaxed smybols and values; fixed typos; added [current-source]
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| Apr 30, 2017 at 16:29 | review | Suggested edits | |||
| S Apr 30, 2017 at 16:55 | |||||
| Apr 30, 2017 at 12:49 | vote | accept | elecV1 | ||
| Apr 30, 2017 at 12:09 | comment | added | elecV1 | Thanks again G36.You are right for VE1=0.7V I was confused.For your second thought I believe it's a good explanation but maybe there are another solution without assumptions. | |
| Apr 30, 2017 at 11:58 | comment | added | G36 | The Q3 emitter voltage is: Ve = Vb3 - Vbe3 = Vbe1+Vbe2 -Vbe3 = 0.7V + 0.7V - 0.7V = 0.7V. As for why Ie2 = Ib3 + 50Ib3 the Ie2 current should be much larger then Ib3 so I assumed 50 times larger. | |
| Apr 30, 2017 at 11:50 | comment | added | elecV1 | G36 thank you for your answer.Indeed BJT's are identical.However, your equation R4 = 0.7V/404uA I think it's wrong because emmiter voltage of Q3 it isn't O.7V but 1.4V.Also I don't understand why Ie2 =50Ib3 as you mentioned above.Thank you | |
| Apr 30, 2017 at 11:33 | comment | added | G36 | Ie3 = Ic3+Ib3 = (beta +1)/beta * Ic3 = 404uA so R4 = 0.7V/404uA = 17.326kΩ. IR5 = Ib3 + (Ic2+Ib2) = Ib3 + Ie2 = Ib3 + 50Ib3 = 204uA--->R5 = (8V - 1.4V)/204uA = 32.352kΩ | |
| Apr 30, 2017 at 11:19 | answer | added | Neil_UK | timeline score: 1 | |
| Apr 30, 2017 at 11:07 | comment | added | G36 | No Shockley equation ? And the BJT's are identical ? | |
| Apr 30, 2017 at 10:28 | history | asked | elecV1 | CC BY-SA 3.0 |