Timeline for Why is a BJT considered "current-controlled"?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2016 at 12:54 | comment | added | LvW | @Analog Arsonist, as you have mentioned the definition of a voltage divider: Are you sure that the principle of voltage division applies to LINEAR parts only? Where did you find such a restriction? May I remind you on the task of non-linear resistors in stabilizing of oscillator circuits? | |
| Aug 22, 2016 at 3:46 | comment | added | Analog Arsonist | @alfred centauri He's already explained that it's a simplification but that's what designers do. And you're posting another model that is also an approximation, even though it is more accurate than assuming 0.7V. I use datasheets to get worst case vbes though, not an equation. Yes it changes but not as much as your base current changes! | |
| Aug 22, 2016 at 3:35 | comment | added | Analog Arsonist | @lvw it's called a current source, which is what mmize described. The current is fixed. It's not a voltage divider because vbe doesn't really change based on a change in vin, which is the definition of a voltage divider. | |
| Aug 22, 2016 at 3:18 | comment | added | efox29 | I am not sure why this answer is the top most rated. It's a good estimate but doesn't answer OPs question about why (or why not) its current controlled. | |
| Aug 18, 2016 at 13:55 | comment | added | Alfred Centauri | The first sentence is false and the equation given at the end is an approximation that ignores the logarithmic dependence of \$V_{BE}\$ on \$I_B\$. $$V_{BE} = V_T \ln \frac{\beta I_B}{I_S}$$ So, while it is true that \$V_{BE}\$ doesn't change by much, it isn't true that \$V_{BE}\$ doesn't change at all. | |
| Aug 17, 2016 at 18:36 | comment | added | LvW | What really happens is the following: R1 realizes - together with the base-emitter path - a voltage divider. And the signal voltage Vin causes a corresponding voltage drop across the B-E path which controls the collector current. Hence, it is NOT the base current Ib which determines Ic. Just the opposite is true: Ib and Ic are both caused by Vbe. | |
| Aug 17, 2016 at 18:05 | history | edited | mmize | CC BY-SA 3.0 |
added 2 characters in body
|
| Aug 17, 2016 at 17:10 | comment | added | Dmitry Grigoryev | Sorry. I was trying to keep it simple by ignoring this voltage entirely in my answer, but comments got me started ;) | |
| Aug 17, 2016 at 17:02 | comment | added | mmize | @DmitryGrigoryev You are correct. I was trying to keep things simple to not add any confusion. | |
| Aug 17, 2016 at 16:59 | comment | added | Dmitry Grigoryev | Nitpick: the voltage drop across Vbe will always be around whatever the datasheet says, which could be as low as 0.3V for some BJTs. | |
| Aug 17, 2016 at 16:44 | history | answered | mmize | CC BY-SA 3.0 |