Timeline for bjt transistor circuit - finding the resistances
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2012 at 0:12 | comment | added | Kaz | By the way, you'd never build a real circuit this way. The amount of current that will flow through the BE diode is very sensitive to small voltage changes, so you would never be able to reliably control the current to a 2mA trickle with this. The slightest variations in the resistors or the transistor (including dynamic ones: temperature) will foil this. The circuit is susceptible to thermal runaway. With rising temperature, the BE junction's voltage lowers: it conducts more at a given voltage. That means more collector current, which means more temperature ... | |
| May 16, 2012 at 23:28 | vote | accept | Neyo | ||
| May 16, 2012 at 23:26 | comment | added | Neyo | I was about to say, it was 0.177 after all. So overall I have Rc=5ΚΩ as you said, and R2=24.3KΩ | |
| May 16, 2012 at 23:18 | comment | added | Kaz | Yes, the drop across R2 is 4.3V. If we divide this by the current (R = V/I) we get the resistance. Are you sure about that 0.159? That looks like you're adding .157 and .002. But base current is .02 mA (Ic/beta). Am I wrong? Check twice. | |
| May 16, 2012 at 23:11 | comment | added | Neyo | So in the end its 4.3V? | |
| May 16, 2012 at 23:08 | comment | added | Kaz | R2 is connected between the base (whose voltage we are given: +0.7) and the collector, whose voltage we also know (+5V, from the VCE we are given). | |
| May 16, 2012 at 23:04 | comment | added | Neyo | I have calculated I2 as 0.159mA, I'm just not sure what the voltage accross R2 is. I must be missing something obvious. | |
| May 16, 2012 at 22:59 | comment | added | Kaz | First, calculate the current going through R2. This current is the sum of two currents, because it forks into two destinations: the transistor base and R1. The base current is obtained from dividing the collector current by the transistor's beta. The R1 current is obtained from knowing the voltage across R1, and it resistance. Once we add these two currents, we know the current across R2, and since we know the voltage across R2, Ohm's Law gives us its resistance. | |
| May 16, 2012 at 22:56 | comment | added | Neyo | Also, how do i find the voltage surrounding R2 again? | |
| May 16, 2012 at 22:48 | comment | added | Neyo | Fair enough. Concerning Ib, is it correct to say its direction is left to right? Meaning overall its I2=Ib+I1 (I know you already said "sum" but didn't know if you meant it like this or didn't take +/- into account) | |
| May 16, 2012 at 21:59 | comment | added | Kaz | Don't worry about that. These Vbb, Vcc labels are largely just traditional nonsense. Why Vcc in the first place? Because Vc is the collector voltage, so Vcc is the voltage "beyond" Vc. See here: en.wikipedia.org/wiki/IC_power_supply_pin#History . What happens if you have a larger circuit in which some of the transistor emitters go to Vee, and some go to ground? I avoid these silly names when designing circuits and use descriptive network names or numeric ones like +V12 and -V12. | |
| May 16, 2012 at 21:47 | comment | added | Neyo | Thanks, I think I'm begining to understand the circuit with what you said. I'm still puzzled though as to why there's Vee voltage in what I would guess should be a Vbb. | |
| May 16, 2012 at 21:33 | history | edited | Kaz | CC BY-SA 3.0 |
added 45 characters in body
|
| May 16, 2012 at 21:26 | history | edited | Kaz | CC BY-SA 3.0 |
added 144 characters in body
|
| May 16, 2012 at 21:21 | history | answered | Kaz | CC BY-SA 3.0 |