Way back to the basics. Transitor question

[wannaBinventor]

Three years since I really tinkered and I've forgotten most everything. I never really understood things like this well in the first place.
Okay, lets say I have the above circuit.
If R2 and Rc are not there, the behavior is very straightforward. Ib = (Vcc-Vbe)/R1 and Ie = Ib(hfe+1).
I simulated all that and made sure it made sense to me again. If we add R2 back, I was thinking that it should be Ie = (Vb - Vbe)*hfe, but I'm looking to define Vb. I know the voltage divider equation, but I think I've gotten confused on where my starting point should be because Vb in this case will change based on Re. I know that dVb/dRe > 0, but I can't think of how it is actually changing. I'm pretty sure that Vb = Ie*[(Rb/hfe)+Re]+Vbe, but I'm thinking Ie is itself dependent on Vb.... I think I'm just confused on the starting point.

Then, matters are further complicated with the addition of Rc. How does that impact everything?

Thanks!

 

[crutschow]

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[JimB]

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[JimB]

Oh Crutters!
You just beat me to it!

JimB

 

[wannaBinventor]

The image should be showing up in the first post now. If not, it should be below. Sorry about that.

 

[audioguru]

I design it like this:

 

[wannaBinventor]

Thanks very much for that audioguru. I have some follow-ups to that.

I think one of the things that I'm getting confused as I focus on the theory aspect is which variables are what I'd call "choice" variables and which are dependent upon them. I take it that you wanted 10x voltage gain, and 0.4mA current through Rc and then selected other resistor values from there. Is that right?

You mention that "if the collector with no input is 5v then it can swing.....then Rc = 10k and Re = 1k"
That's where I'm getting kind of hung up, I'm not fully understanding how the presence of the transistor is affecting the voltages across the collector and emitter resistors. I seem to remember that voltage gain at the collector is the ratio of Rc to Re, so I can see why you've selected that if you are going that

....but let me make sure I understand the rest of it. You knew you wanted .4mA through the collector, so you divided that by hfe to get the required base current. Then you found what total resistance through the divider would generate 20 times base current, which was about 259k ohms. Then, knowing you had a .65V drop base to emitter, you allocated the resistance between r1 and r2 such that you would achieve the 1.74uA base current you knew you needed for your desired 4mAmps through the collector. For what it's worth, I get (1.15-.65)/226k = 2.12 uAmps. Is this just rounding error, or am I not following it right?

 

[audioguru]

For a transistor to use an AC signal you usually want its circuit designed for the maximum possible output level. Then you bias the transistor so that its output can swing as high as possible and as low as possible.
If the collector idles with no signal at 4.5V which is half the supply voltage then the collector voltage can swing 4.5V up to 9V when the transistor is completely turned off, but it can swing down only 3.7V to 0.8V because when the transistor is completely turned on then the 1k emitter resistor has 0.8V across it. If the collector voltage idles at 5V then it can swing up 4.0V to 9V and it can swing down close to 4V to 0.8V.

R1 calculates to be 226k ohms which is not a standard value so 220k should be used.
R1 has a voltage of 9V - 1.15V= 7.85V across it, not 1.15V - 0.65V the emitter resistor has 1.15V - 0.65V across it because its current is (1.15V - 0.65V)/1k= 0.5V. Then Ohm's Law calculates its current to be 7.85V/226k= 34.7uA which is 20 times the base current of 1.74uA.
The current in R2 is 34.7uA minus the base current of 1.74uA= 32.96uA. Then its value is 1.15V/32.96uA= 34.9k ohms which is not a standard value so 33k was used.

Sorry, I made a mistake when I calculated the emitter voltage. With a collector current of 0.4mA then the emitter current is almost the same so the idling emitter voltage is 0.4V, not 0.5V. So the idling base voltage should be 0.4V = 0.65v= 1.05V, not 1.15V.
Then the values for R1 and R2 will be a little different but you can calculate them.

 

[Jony130]

wannaBinventor do you know why audioguru chosen the divide current 20 times larger than base current ?

 

[wannaBinventor]

Thanks audioguru . I appreciate your time. I'm not one hundred percent sure why that was chosen, but I have heard of a ten time a rule before. I believe it to be chosen to keep the reference voltage stable as the transistor is turned on. Is that correct?

 

[audioguru]

The current in the voltage divider for the base of the transistor is 10 to 20 times the typical base current because the hFE of a transistor is a range of numbers.
For example the hFE of a BC547B at a collector current of 2mA is from 200 to 450 but is typically 290. With the voltage divider current much higher than the base current then a low base current or a high base current will not make much difference.

 

[wannaBinventor]

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