R2 function on BJT Base - Page 1
Forum section: electronics / projects

author: abdulbadii
author_rank: Frequent Contributor
timestamp: August 09, 2022, 04:14:04 am
content: Anyone outright understand what actually R2function when R1completely rules to control the B of BJT, why still need R2 ?Thanks before

author: Marco
author_rank: Super Contributor
timestamp: August 09, 2022, 04:20:09 am
content: It's always part of a resistive divider to lower the base voltage. It can be a pull down resistor for a driver which can only source current.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 09, 2022, 05:21:48 am
content: If all you want to do is turn the transistor on you don't need R2. If you want to set the operating point then you do.R1 and R2 form a voltage divider that sets the base voltage at half the supply rail. The emitter will be 0.6 V lower than the base voltage. The collector will draw whatever current is needed to supply the emitter, assuming the base current is negligible.

author: Whales
author_rank: Super Contributor
timestamp: August 09, 2022, 06:04:54 am
content: You are correct that R1 alone could work (with some tuning), but it's fiddler.I'm assuming the voltage at node 3 is the output of this amplifier.If you only have R1You have to tune R1's resistance so that the output of this amplifier (when there is no AC signal on the input) is approximately halfway between the voltage rails. This gives your output voltage lots of space to move before it clips.If R1 is too small then the transistor will turn on too much and output a low DC voltage, giving it little room before it clips. Similarly if R1 is too big then the transistor will not turn on enough, so the output voltage will be near the positive rail and give your output less room to move before it clips.Now consider that the beta/HFE of an NPN changes between each transistor, ambient temperature and the value of the resistors on the output. There is no one value of R1 that will always work in this circuit, you would have to connect a pot to tune it for every one you make (and maybe attach a heatsink to it).If you only have both R1 and R2R1 and R2 divide the input voltage to be half of the supply voltage. Your transistor's Vbc will be perhaps a few volts, so this puts the output voltage somewhere near the middle too. Changing the beta/HFE or temperature of the transistor won't change this too much. You could even swap it out for different models of transistor and it will still probably be tuned well enough to work.(In reality I'd choose slightly different values for R1 and R2 and I wouldn't exactly centre the output voltage because that's not ideal for distortion in this style of common-emitter amp, but hey it's probably close enough).

author: Whales
author_rank: Super Contributor
timestamp: August 09, 2022, 06:08:41 am
content: Hmm, this might not really be a common emitter amp. The emitter resistor is bigger than the collector resistor. It's a sort of hybrid between common emitter and common collector -- I suspect its gain might be less than 1?

author: Zero999
author_rank: Super Contributor
timestamp: August 09, 2022, 11:20:18 am
content: Quote from: Whales on August 09, 2022, 06:08:41 amIt will also have a relatively small voltage swing.Quote from: abdulbadii on August 09, 2022, 04:14:04 amIt's difficult to tell from looking at that fragment. Please post the entire schematic.

author: abdulbadii
author_rank: Frequent Contributor
timestamp: August 09, 2022, 01:00:26 pm
content: a two BJTs oscillating DC wave. ~4.1V peak

author: David Hess
author_rank: Super Contributor
timestamp: August 09, 2022, 03:07:35 pm
content: That is my favorite headache inducing circuit.R1 and R2 in this case form a voltage divider which hold the base, and by extension the emitter, at a constant voltage. R2 could be replaced with a zener diode but there is no need for it.

author: TimFox
author_rank: Super Contributor
timestamp: August 09, 2022, 03:59:28 pm
content: R1 through R4 are the classic method for biasing a BJT to a given quiescent current.The rest of the circuit is the interesting part.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: August 09, 2022, 11:42:50 pm
content: Hi,I recognize this circuit. It is the same as the oscillator that was used in the Jay_Diddy_B 5-transistor ESR meter.Link:https://www.eevblog.com/forum/projects/5-transistor-esr-meter-design/Forum member W2AEW made a nice video about this meter:ModelThese are all the waveforms in a mess ;-)Separated:The oscillator is an emitter-coupled oscillator.I have attached the model.Regards,Jay_Diddy_Bemitter coupled oscillator.asc(2.17 kB - downloaded 34 times.)

author: David Hess
author_rank: Super Contributor
timestamp: August 10, 2022, 10:23:37 pm
content: Quote from: Jay_Diddy_B on August 09, 2022, 11:42:50 pmTektronix used it in a few of their probe compensation output circuits. It has the virtue of providing a 50/50 duty cycle while using only a single capacitor.