Calculate resistance for bootstrap resistor in common emitter amp - Page 1
Forum section: electronics / beginners

author: Amoebius
author_rank: Contributor
timestamp: October 22, 2022, 02:33:35 pm
content: I have problems with a calculation.The formula for Rbeq is shown in the picture i included.I read the formula as follows:Rbeq = Rb/(1-((1/R1)+(1/R2)+(1/Re)))/((re+((1/r1)+(1/R2)+(1/Re)))R1 = 1,1K OhmR2 = 13,9k OhmRe = 1k Ohmre = 50 OhmRb = 5k OhmDoing this I end up with Rbeq = 250.506,16 Ohm, 5o times the value of Rb. On the website they get a value approximatly 20 times Rb.What am I doing wrong?https://electronics.stackexchange.com/questions/481707/bootstrapping-common-emitter-amplifierLink to original

author: magic
author_rank: Super Contributor
timestamp: October 22, 2022, 03:01:51 pm
content: Quote from: Amoebius on October 22, 2022, 02:33:35 pmThe literal meaning of the formula as printed is as follows:Rbeq = Rb/(1-((1/((1/R1)+(1/R2)+(1/Re)))/(re+(1/((1/R1)+(1/R2)+(1/Re))))))Which is to say:1. R1||R2 means parallel combination, and you forgot to 1/(...) everything in each case.2. I think you are supposed to subtract the ratio of the two X||Y||Z thingies from 1, rather than include 1- in one of them.3. You forgot another pair of parentheses around everything after Rb/, not sure if you calculated correctly regardless of this error or not.4. Writing complex formulas like that is pure masochismWe could at least make some use of spaces to group related items and make it a little more readable:Rbeq = Rb / (1- ( (1/ ((1/R1)+(1/R2)+(1/Re)) ) / (re + (1/ ((1/R1)+(1/R2)+(1/Re)) ) ) ) )

author: Amoebius
author_rank: Contributor
timestamp: October 22, 2022, 04:05:40 pm
content: Ok, thank you.But even with the corrected formula i dont't get the 20 times Rb...

author: magic
author_rank: Super Contributor
timestamp: October 22, 2022, 06:31:59 pm
content: This makes things harder because we now need to start thinking about the circuit.I see that there is reappearing in the formula. If the author didn't screw it up, this lower-case r should refer not to the Re resistor, but the intrinsic emitter resistance of the transistor under given bias condition. Perhaps this is your mistake? Intrinsic emitter resistance is given by the formulare= VT/ IEwhereVTis the thermal voltage, about 26mV at room temperature, up to 30~40mV when hot or really hot.IEis the emitter current, of course, and you will need to calculate it.There are many ways to explain what it is, one way that may be relevant here is that reis the output resistance of the emitter, assuming that base voltage is fixed and considering the transistor as an emitter follower.I would have to think harder about the circuit to explain what the math means and how reis involved here. I think what the author meant and what would make sense, is that Rbeq is Rb multiplied by the ratio of input voltage swing to the corresponding base-emitter voltage variation, which is due to the non-negligible and non-zero output impedance of the emitter driving a bunch of resistors, acting like a voltage divider between the "ideal" output voltage of the emitter follower (zero emitter output resistance, constant Vbe) and ground, giving the voltage you actually get at the emitter terminal.Or something very similar. I may have got details wrong

author: Amoebius
author_rank: Contributor
timestamp: October 22, 2022, 07:31:27 pm
content: It is called bootstrapping, and i think it is the positive feedback of the ac signal through Rb to base, Rb being in series with the biasing output of R1 and R2. It is increasing the input impedance somehow, i dont't quite understand it yet.

author: Grandchuck
author_rank: Super Contributor
timestamp: October 22, 2022, 08:22:02 pm
content: The transistor emitter signal is almost the same amplitude as the input signal and is in phase. Thus, the signal voltage across the base resistor is almost zero and so is its signal current. The resistor acts as a high impedance.

author: Amoebius
author_rank: Contributor
timestamp: October 23, 2022, 05:14:44 pm
content: Ok, thanks.

author: magic
author_rank: Super Contributor
timestamp: October 23, 2022, 08:31:31 pm
content: Yep, this is how it works. You can imagine that for AC signals the capacitor is a short. Also, imagine that the transistor consists of an ideal voltage follower with constant offset voltage of about -650mV (Vbe) and then 50Ω output resistance re. The author says at the bottom of the text that he assumes re=50Ω. IMHO bias current will be closer to 0.45mA and resistance about 60Ω, but whatever.The capacitors shorts R1 and R2 in parallel with Re, giving 505Ω. The 50Ω/505Ω voltage divider causes a 50/555=9% fraction of input AC voltage to appear as AC voltage across Rb. This means that AC voltage across Rb is 9% of input AC voltage and AC current into Rb is 11 times less than it would be if Rb were connected to ground, so "equivalent" Rb resistance appears 11x higher.Not sure where 20x came fromThe formula for Rbeq given by the author is equivalent to my calculation. So it looks like author's calculation was wrong and 11 is the correct answer.

author: Amoebius
author_rank: Contributor
timestamp: November 05, 2022, 12:33:44 pm
content: Thank you, just found your answer