Timeline for How to increase CE amplifier gain to 12?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 7, 2021 at 14:59 | comment | added | Tom | Thank you very much. I understand that now. Again, much appreciated! | |
| Dec 7, 2021 at 14:58 | comment | added | jonk | @Tom It comes from the 190 mV minimum plus the 1 V peak signal value. The emitter resistor (1) will go from 190 mV across it to 2.190 V across it. The quiescent point is halfway between the two. | |
| Dec 7, 2021 at 14:53 | comment | added | Tom | Guessing about 700mV for the base-emitter voltage, this means the base voltage for 𝑄1 is −15V+460mV+1.19V+700mV=−12.65V. - Please can you clarify where the 1.19V has come from? Maybe it is very simple, but I should ask rather than guess. | |
| Dec 7, 2021 at 14:30 | history | edited | jonk | CC BY-SA 4.0 |
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| Dec 7, 2021 at 14:27 | comment | added | jonk | @Tom You interacted well, here, which suggests you want to know more. That's a good and worthy trait. And thanks very much for letting me know it helped. Makes it all worth the time! Best wishes and be sure to help others when you find a moment and someone else interested. Pass it on. | |
| Dec 7, 2021 at 11:21 | vote | accept | Tom | ||
| Dec 7, 2021 at 11:21 | comment | added | Tom | Thank you so much for the help! It’s really appreciated that you’ve done so much. This has really given me some extra understanding of the circuit, which I have not been as clear previously | |
| Dec 7, 2021 at 7:05 | history | edited | jonk | CC BY-SA 4.0 |
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| Dec 7, 2021 at 6:37 | history | edited | jonk | CC BY-SA 4.0 |
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| Dec 7, 2021 at 5:18 | history | edited | jonk | CC BY-SA 4.0 |
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| Dec 7, 2021 at 5:10 | history | edited | jonk | CC BY-SA 4.0 |
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| Dec 7, 2021 at 5:03 | history | answered | jonk | CC BY-SA 4.0 |