Timeline for BJT Emitter Degeneration Rπ
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2017 at 18:07 | vote | accept | lucenzo97 | ||
| Aug 9, 2017 at 13:35 | comment | added | G36 | @Keno In general, you should never neglect r_pi resistance in your calculations. Maybe when you have a large RE resistance ( larger than 1k) without Ce capacitor. | |
| S Aug 9, 2017 at 6:02 | history | rollback | lucenzo97 |
Rollback to Revision 2 - Edit approval overridden by post owner or moderator
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| Aug 9, 2017 at 5:21 | history | suggested | Mayank | CC BY-SA 3.0 |
corrected grammer
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| Aug 9, 2017 at 5:12 | review | Suggested edits | |||
| S Aug 9, 2017 at 6:02 | |||||
| Aug 8, 2017 at 7:38 | vote | accept | lucenzo97 | ||
| Aug 8, 2017 at 7:39 | |||||
| Aug 8, 2017 at 5:00 | answer | added | analogsystemsrf | timeline score: 1 | |
| Aug 7, 2017 at 13:05 | comment | added | G36 | No, CC output resistance = (r_pi/(beta+1))||RE if we ignore R1,R2 | |
| Aug 7, 2017 at 12:59 | comment | added | lucenzo97 | @G36 But that last equation (r_pi + (beta+1)*RE) applies for CC output resistance, right? | |
| Aug 7, 2017 at 12:35 | comment | added | G36 | Rin = R1||R2||r_pi (with CE across RE or without RE at all) The input resistance with RE and without CE capacitor is Rin = R1||R2||(r_pi + (beta+1)*RE) | |
| Aug 7, 2017 at 9:54 | comment | added | LvW | When you have a sum of two quantities C=A+B and B is much smaller than A, the you can perhaps live with the error by letting C=A. In the world of electronics nothing is correct by 100% because we ALWAYS neglect some effects. But depending on the specific application and the allwowed simplification error you have to decide if you may neglect the influence of on quantity - or not. | |
| Aug 7, 2017 at 9:49 | comment | added | lucenzo97 | @LvW I don't quite get what you mean.. | |
| Aug 7, 2017 at 9:46 | comment | added | LvW | The answer is simple: Write down the input resistance for all cases you are intersted in (with/without RE or CE). And then - as a second step - use real values and see if the base-emitter resistance plays a major role (lets say > 10%) or not. | |
| Aug 7, 2017 at 9:10 | comment | added | lucenzo97 | @Andyaka Edited. | |
| Aug 7, 2017 at 9:09 | history | edited | lucenzo97 | CC BY-SA 3.0 |
deleted 147 characters in body
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| Aug 7, 2017 at 9:04 | comment | added | Andy aka | You say that "Emitter resistor will be most probably bypassed by capacitor" and this is just not true. Then you say "so emitter can't really have much influence on small signal input impedance" and this is also not true. Your claims and your questions are not obviously separated enough to make an answer. Clean up your question and don't make false assertions. | |
| Aug 7, 2017 at 8:25 | comment | added | lucenzo97 | @Andyaka But how? Capacitor to ground from emitter lead equals as grounded emitter for small signal analysis, right? Why else would emitter resistor be bypassed. And I didn't muddled up anything... | |
| Aug 7, 2017 at 7:16 | comment | added | Andy aka | The emitter resistor is NOT "most probably bypassed by capacitor" at all and this claim is putting me off answering because it seems you have things muddled up. I'm not sure what your question is based on your false claim. | |
| Aug 7, 2017 at 6:25 | history | asked | lucenzo97 | CC BY-SA 3.0 |