I have been studying and researching this for a while now and I have come across this a couple or so days ago. I have come across the formula for input capacitance for the common emitter amplifier. Xc< or=0.1×Zin or Rin. The formula is: C=1/(( 2π×f(low)×(0.1×Zin)). A formula for the input impedance of a common emitter amplifier with voltage divider bias is: Zin=R1||R2||Zin(base). Where Zin(base) is: without bypass capacitor hfe(r'e + Re). With bypass capacitor, is: (hfe×r'e). A formula for the input impedance of a common emitter amplifier with voltage divider bias is: C=1/((2π×fclow×(Zin×0.1)).

 

I have been studying and researching this for a while now and I have come across this a couple or so days ago. I have come across the formula for input capacitance for the common emitter amplifier. Xc< or=0.1×Zin or Rin. The formula is: C=1/ 2π×f(low)×(0.1×Zin). Zin(base) is r'e + Re without a bypass capacitor. With a bypass capacitor to bypass the emitter resistor RE is hfe×r'e.

a formula for the input coupling capacitor for the common emitter amplifier with voltage divider bias: C=1/((2π×fclow×(Zin×0.1))

 

And you qestion is?

 

There are so many formulas in the field of electronics for designing and analyzing circuits.
When you are "studying and researching"such objects it is really necessary to understand the physical background of the formulas. Otherwise, you cannot know under which conditions the may apply (or not).
Simply to state "yes - correct" is not enough.

 

There are so many formulas in the field of electronics for designing and analyzing circuits.
When you are "studying and researching"such objects it is really necessary to understand the physical background of the formulas. Otherwise, you cannot know under which conditions the may apply (or not).
Simply to state "yes - correct" is not enough.

Yeah I understand where you're coming from. For we don't want to just memorize it, but understand it is more important. However, by understanding it, memory sticks. This is what I'm aiming for. Thank you for your reply and knowledgeable advice for this is what I want to understand and learn. I appreciate it...

 

Ohms Law is the ONE to memorize, and also to understand. It also applies with impedance, just substitute "Z" for R, and then add all of the vectors. The vector math gets a bit bigger, but it is all quite logical.

 

Ohms Law is the ONE to memorize, and also to understand. It also applies with impedance, just substitute "Z" for R, and then add all of the vectors. The vector math gets a bit bigger, but it is all quite logical.

Ohms Law is the ONE to memorize, and also to understand. It also applies with impedance, just substitute "Z" for R, and then add all of the vectors. The vector math gets a bit bigger, but it is all quite logical.

Okay I have learned how to add track vectors as far as phase angles and amplitude. I haven't applied them yet so this is something I might have to practice with. Thank you for the heads up I really appreciate it man.

 

For the bipolar CE amplifier Zin, the challenge of a rigerous evaluation is the feedback toward the base from the emitter voltage variation. I had to do it in school but never in my actual job. Of course, I never designed Top-End HIFI systems for a living, either. That was a bigger challenge back with germanium transistors, which had larger changes with temperature.

 

For the bipolar CE amplifier Zin, the challenge of a rigerous evaluation is the feedback toward the base from the emitter voltage variation. I had to do it in school but never in my actual job. Of course, I never designed Top-End HIFI systems for a living, either. That was a bigger challenge back with germanium transistors, which had larger changes with temperature.

Did that have something to do with the ac dynamic resistance between the base and the emitter? r'e=25mV/Ie. Does the number 25mV have something to do with the temperature? Bear with me please this is still pretty new to me... I just need to find out if I'm on the right track or not... Thank you for the reply though that helps and because of it I think I am on the right track. When it comes to the equation of the Zin, r'e is a factor in it when it comes to Zin(base).

 

Did that have something to do with the ac dynamic resistance between the base and the emitter? r'e=25mV/Ie. Does the number 25mV have something to do with the temperature? Bear with me please this is still pretty new to me... I just need to find out if I'm on the right track or not... Thank you for the reply though that helps and because of it I think I am on the right track. When it comes to the equation of the Zin, r'e is a factor in it when it comes to Zin(base).

YES!

 

Did that have something to do with the ac dynamic resistance between the base and the emitter? r'e=25mV/Ie. Does the number 25mV have something to do with the temperature?

No - the dynamic resistance between the base and the emitter is not identical to r'e.
This resistance is called "rbe" or in terms of h-parameters "h11=hie".

The quantity you have mentioned (r'e) is not a resistance (although it has the unit V/A=Ohm).
It is nothing else than the inverse "transconductance gm".

This quantity gm appears in the voltage gain formula - and some authors simply use the inverse of this value (r'e=1/gm).
The transconductance gm is identical to the slope of the voltage-control function Ic=f(Vbe/Vt) and can be calculated to be
d(Ic)/d(Vbe) = gm = Ic/Vt
(with the "temperature voltage" Vt~25mV for ambient temperature).

Using the relation ic/ib=beta we can find that the base-emitter resistance is rbe=hie=beta/gm=beta*r'e .

One further remark:
When there is an external resistor Re in the emitter path, the resulting input resistance at the base will be drastically increased due to signal feedback. In this case. the transistors input resistance is r_in=h11 + (beta *Re) with h11=hie=rbe.
The total input resistance of the cmplete stage must include also (in parallel) the resistors for base biasing.

 

And you qestion is?

Referring to #2 I just need some confirmation if this is correct or not and if there's any alternatives. Please keep it simple if you can..

 

Yeah I understand where you're coming from. For we don't want to just memorize it, but understand it is more important. However, by understanding it, memory sticks. This is what I'm aiming for. Thank you for your reply and knowledgeable advice for this is what I want to understand and learn. I appreciate it...

'Yes'

 

Referring to #2 I just need some confirmation if this is correct or not and if there's any alternatives. Please keep it simple if you can..

Personally, I prefer to use a simpler equation for the coupling capacitor :
C = 1/(2 * pi * F * R) = 0.16/(F * R)
Where:
R - Resistance seen by the capacitor (high-pass filter is formed by a combination of a resistance "seen" from the capacitor's point of view and the capacitor itself)
F - The lowest desired frequency you want to amplify.
In short, choose the coupling capacitor C so that all frequencies of interest are passed by the RC high-pass filter.

Of course, we must choose F wisely because, in the typical case, we have more than one high-pass filter in the circuit ( more than one coupling capacitor).
Thus, effectively two or more high-pass filters are connected in series, and this will effectively increase the cut-off frequency.

 

...................
Thus, effectively two or more high-pass filters are connected in series, and this will effectively increase the cut-off frequency.

Yes - for example the time constant formed by the capacitor Ce in parallel to the emitter resistor Re.

 

Personally, I prefer to use a simpler equation for the coupling capacitor :
C = 1/(2 * pi * F * R) = 0.16/(F * R)
Where:
R - Resistance seen by the capacitor (high-pass filter is formed by a combination of a resistance "seen" from the capacitor's point of view and the capacitor itself)
F - The lowest desired frequency you want to amplify.
In short, choose the coupling capacitor C so that all frequencies of interest are passed by the RC high-pass filter.

Of course, we must choose F wisely because, in the typical case, we have more than one high-pass filter in the circuit ( more than one coupling capacitor).
Thus, effectively two or more high-pass filters are connected in series, and this will effectively increase the cut-off frequency.

Thank you... I noticed you used 0.16 instead of 0.1 solving the equation for the capacitance: C=1/2π(f×R) as I see you used: 1/2π =0.15915 rounding it to the nearest hundred=0.16. This is accurate. Thank you. You hit the nail on the head and made confirmation letting me know I'm on the right track.

 

Thank you... I noticed you used 0.16 instead of 0.1 solving the equation for the capacitance: C=1/2π(f×R) as I see you used: 1/2π =0.15915 rounding it to the nearest hundred=0.16. This is accurate. Thank you. You hit the nail on the head and made confirmation letting me know I'm on the right track.

I would like to better understand formulas so that I can understand the behavior of these components and circuits.

 

I noticed you used 0.16 instead of 0.1 solving the equation for the capacitance: C=1/2π(f×R) as I see you used: 1/2π =0.15915 rounding it to the nearest hundred=0.16. This is accurate.

Yes, exactly the rounding 1/(2 pi) ≈ 0.16
In "your" formula
C = 1/( 2π × f_low× (0.1×Zin) )
0.1 * Zin factor will choose Xc = 0.1*Zin, so the actual cutoff frequency of a filter will be lower by a factor of 10.
Fcutoff = 0.1*F_low.
This ensures that your target frequency ( f_low) will pass through the circuit completely unharmed by the HP filter.

 

Consider that for top-line components the tolerance is +-5%, although some "audiophiles" can clearly sense less deviation than that, how much does being that precise actually matter??