One number that pops up a lot in BJT calculations is the 26mV used to calculate the base-emitter resistance. Where does that value come from? • 29. Common Emitter Cir...
ACTUALLY r'e INVOLVES THE APPLICATION OF DIFFERENTIAL CALCULUS TO SHOCKLEYS EQUATION BY TAKING ITS DERIVATIVE AND,USING ALGEBRA SUBSITUTION TO COME UP WITH THE 26mV.
I have a few points to make: 1 - the conversion is K = C + 273.15 ( not 273.16, that is the triple point of water ) 2 - My text book uses 25mV but more important is that this is based on a specific ambient temperature and perfect PN junction, in practice this is only an estimate 3 - My experiments on general purpose transistors generally give 30-35mV as a more accurate estimate for practical circuits 4 - You can calculate Vt from the 2N3904 datasheets as approximately constant at 30mV for the range Ic = 100uA to 10mA but it varies with Ic in general. It's not given directly in the datasheet but you can calculate it from: Vt = Hie * Ic / Hfe or R'e = Hie/Hfe
Hmm. What if the emitter resistor was replaced with a temperature probe, a microcontroller, a DAC, and another transistor? Sure, it's clearly more academic than practical. Who would want to buy something that makes a typically expensive stereo amplifier system even more so?
ACTUALLY r'e INVOLVES THE APPLICATION OF DIFFERENTIAL CALCULUS TO SHOCKLEYS EQUATION BY TAKING ITS DERIVATIVE AND,USING ALGEBRA SUBSITUTION TO COME UP WITH THE 26mV.
great video! Just one thing, the unit is not called degrees Kelvin (°K), it's just Kelvin (K)
I have a few points to make:
1 - the conversion is K = C + 273.15 ( not 273.16, that is the triple point of water )
2 - My text book uses 25mV but more important is that this is based on a specific ambient temperature and perfect PN junction, in practice this is only an estimate
3 - My experiments on general purpose transistors generally give 30-35mV as a more accurate estimate for practical circuits
4 - You can calculate Vt from the 2N3904 datasheets as approximately constant at 30mV for the range Ic = 100uA to 10mA but it varies with Ic in general. It's not given directly in the datasheet but you can calculate it from: Vt = Hie * Ic / Hfe or R'e = Hie/Hfe
That was actually helpful. I could not think that this was the thermal voltage on my own. Also now I know that transconductance is (r'e)^(-1)
I was doing a bjt problem literally today, looking at old notes and asking myself WTH is Vt. Then this video pops up just amazing lol
Excellent topic thank you!
Hmm. What if the emitter resistor was replaced with a temperature probe, a microcontroller, a DAC, and another transistor? Sure, it's clearly more academic than practical. Who would want to buy something that makes a typically expensive stereo amplifier system even more so?
Hello Friend.
Excellent explanation. Congratulations.
Thank you! Cheers!
wow nicely explained
Thank you so much 🙂