Timeline for Class A amplifier - Transistor Q-Point calculation
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2019 at 15:12 | comment | added | Jaywalk | Yep, take note of the B+1 term to compensate for the current ratio between the base and emitter. This allows you to use a single current value Ie to mathematically equate the two terminals in your KVL, enabling you to solve the current. | |
| Jun 12, 2019 at 15:10 | comment | added | user3728501 | Ok that makes a lot of sense, so it is a KVL with a loop around GND, RE, the transistor BE junction, and RB2, back to ground? | |
| Jun 12, 2019 at 15:01 | comment | added | Jaywalk | Oh, Rb is required to calculating the KVL around the base to determine base current. You can see the KVL in the first link you posted. | |
| Jun 12, 2019 at 14:55 | comment | added | user3728501 | Perhaps I should try and rephrase the question: "Why are we calculating RB and what does it correspond to in the real world?" Is RB the input impedance of this entire amplifier circuit? If so I don't understand where that impedance is seen from, or why we are calculating it, since this is a DC analysis, and the input impedance is only relevant when doing the AC analysis part. (Perhaps I'm wrong about that?) | |
| Jun 12, 2019 at 14:51 | history | edited | Jaywalk | CC BY-SA 4.0 |
added 47 characters in body
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| Jun 12, 2019 at 14:50 | comment | added | Jaywalk | The equivalent of "turning the device off" is shorting independent voltage sources and opening independent current sources. | |
| Jun 12, 2019 at 14:49 | comment | added | Jaywalk | Oh yeah sorry, you also short Vcc during resistance calculations. | |
| Jun 12, 2019 at 14:49 | comment | added | user3728501 | This is the DC analysis part | |
| Jun 12, 2019 at 14:46 | history | answered | Jaywalk | CC BY-SA 4.0 |