Timeline for Transistor Current Source Biasing
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2017 at 21:04 | vote | accept | VanGo | ||
| Oct 31, 2017 at 14:46 | answer | added | analogsystemsrf | timeline score: 0 | |
| Oct 31, 2017 at 10:16 | comment | added | JIm Dearden | A general 'rule of thumb' is to have at least 10x the required base current through the biasing resistors (R1,R2). (source impedance < load) This allows for a simple voltage divider calculation by ignoring the base current. | |
| Oct 31, 2017 at 9:16 | comment | added | dirac16 | If you're reading AoE also read page 10 and 11 (Edition 3). It explains why the load impedance should be way larger than the source impedance. | |
| Oct 31, 2017 at 8:55 | answer | added | jonk | timeline score: 6 | |
| Oct 31, 2017 at 8:25 | comment | added | Bimpelrekkie | To make more current flow through R1, R2, they need to have a lower value right? What is the output impedance of the voltage divider R1, R2? And the input impedance at the base of Q1, is \$\beta*R_{emitter}\$ right? Since we do not want to disturb the voltage coming out of the voltage divider, the impedance from the transistor loading it should be much higher than the output impedance of the voltage divider. | |
| Oct 31, 2017 at 8:18 | comment | added | Bimpelrekkie | I would use an (in my view) easier description: The current flowing through R1, R2 should be significantly larger than the current flowing into the base of Q1. Imagine if Q1 needs 10 uA base current to make the desired Ic (=current through R_load). If the current through R2 is normally only 20uA, that extra 10 uA to the base makes a huge difference. Now we make 200 uA flow through R2 and R1, then an extra 10 uA for the base makes much less difference. | |
| Oct 31, 2017 at 8:01 | history | asked | VanGo | CC BY-SA 3.0 |