Timeline for common emitter amplifier output distortion
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Oct 21, 2017 at 23:37 | history | edited | user156489 | CC BY-SA 3.0 |
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| Sep 8, 2017 at 20:19 | vote | accept | CommunityBot | ||
| Sep 8, 2017 at 20:18 | comment | added | user156489 | I GET IT my second question was when the base voltage equal the collector voltage the junction is still reverse bias ,but i remember that when the input signal is apply the base voltage will rise and the collector will decrease and it will saturate the transistor , Thank you @DaveTweed you save me from headace | |
| Sep 8, 2017 at 20:15 | answer | added | Dave Tweed | timeline score: 5 | |
| Sep 8, 2017 at 20:07 | comment | added | Dave Tweed | Yes, the B-C junction must remain reverse-biased, which is the same thing as saying that you don't want the transistor to saturate. I don't understand the second part of your question. | |
| Sep 8, 2017 at 20:03 | comment | added | user156489 | @DaveTweed i think you just get the answer as i remember in a book that i read if you want to use the transistor as amplifier base collector junction must be reverse bias but if the voltages is equal it is still reverse bias am i wrong ? | |
| Sep 8, 2017 at 20:01 | comment | added | user156489 | No i just feel lazy and i use low value resistor to make a stiff voltage divider @Keno | |
| Sep 8, 2017 at 19:58 | history | edited | user156489 | CC BY-SA 3.0 |
added 66 characters in body
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| Sep 8, 2017 at 19:53 | comment | added | lucenzo97 | @isam I suppose you wanted to achieve impedance matching with R1||R2=50 ohms = output impedance of function generator? | |
| Sep 8, 2017 at 19:50 | comment | added | user156489 | it is not distorted it looks clean same as the input @Trevor | |
| Sep 8, 2017 at 19:50 | comment | added | Vladimir Cravero | Run a DC simulation and add the results to the question. Also, move the windows around so that your full schematic is visible. | |
| Sep 8, 2017 at 19:50 | comment | added | Dave Tweed | Note that the Thevenin voltage for your base bias is ALSO 4.5V, which is a very bad idea. Your transistor is already close to saturation with no input signal. When the base is driven more positive, the transistor saturates, and you in fact end of driving the input signal itself directly through to the output via the B-C junction of the transistor! | |
| Sep 8, 2017 at 19:49 | comment | added | Trevor_G | what does the signal on the base look like.... | |
| Sep 8, 2017 at 19:46 | comment | added | user156489 | no its connected i check it again it just does not show in the picture i dont know why @VladimirCravero | |
| Sep 8, 2017 at 19:43 | history | edited | Dave Tweed | CC BY-SA 3.0 |
fix formatting
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| Sep 8, 2017 at 19:43 | comment | added | Vladimir Cravero | There is no DC path to bias the base, as the tiny dot between base and R1R2 is missing. | |
| Sep 8, 2017 at 19:40 | comment | added | user156489 | what is the effect of the low input impedance here ? Q point is Vc=4.5V and Ic=5ma | |
| Sep 8, 2017 at 19:39 | history | edited | user156489 | CC BY-SA 3.0 |
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| Sep 8, 2017 at 19:36 | comment | added | JIm Dearden | Because the design of your circuit is wrong. R1,R2 too low, Vb too high, RE too high, Q point ? | |
| Sep 8, 2017 at 19:29 | history | asked | user156489 | CC BY-SA 3.0 |