Emitter Bypass Capacitor Class A Amp

[Amoebius]

Hi,

I have another question regarding the amp circuit.

As said in the other post, i calculated the circuit as follows:

Vcc = 12V
Hfe = 200
For Rc I chose 200 Ohm,
for Vre 1V,
Ic max = (Vcc-Vre)/Rc = 55mA
Ic qpoint =((Vcc-Vre)/2)/Rc = 27,5mA
Ib = Ic qpoint/Hfe = 137,5 uA
Ire = Ic qpoint+Ib = 27,6375 uA
Re = Vre/Ire = 36
R1 = (Vcc-(Vre+0,7))/(11*Ib) = 6744 Ohm
R2 = (Vre+0,7)/(10*Ib) = 1309 Ohm

Ce = 1/(2Pi*f*Xc) = 2199 uF
re = 25mV/Ie
Cin = 1/(2Pi*f*Zin) = 52uF
Cout = 1/(2Pi*f*Zout) = 71uF

My question is, when i split the emitter, can I just divide the 36 Ohm of the Emitter Resistor in two? Like 16 and 20 and leave the other values as they are?
And do i have to choose another emitter capacitor value or can I leave it at 2200uF?

Re1 would be 16 ohms and Re2 20 ohms with the bypass c.

Greetings,

Dirk

[MrAl]

Hi,

I have another question regarding the amp circuit.

As said in the other post, i calculated the circuit as follows:

Vcc = 12V
Hfe = 200
For Rc I chose 200 Ohm,
for Vre 1V,
Ic max = (Vcc-Vre)/Rc = 55mA
Ic qpoint =((Vcc-Vre)/2)/Rc = 27,5mA
Ib = Ic qpoint/Hfe = 137,5 uA
Ire = Ic qpoint+Ib = 27,6375 uA
Re = Vre/Ire = 36
R1 = (Vcc-(Vre+0,7))/(11*Ib) = 6744 Ohm
R2 = (Vre+0,7)/(10*Ib) = 1309 Ohm

Ce = 1/(2Pi*f*Xc) = 2199 uF
re = 25mV/Ie
Cin = 1/(2Pi*f*Zin) = 52uF
Cout = 1/(2Pi*f*Zout) = 71uF

My question is, when i split the emitter, can I just divide the 36 Ohm of the Emitter Resistor in two? Like 16 and 20 and leave the other values as they are?
And do i have to choose another emitter capacitor value or can I leave it at 2200uF?

Re1 would be 16 ohms and Re2 20 ohms with the bypass c.

Greetings,

Dirk

Hello,

Are you talking about putting 16 Ohms in series with 20 Ohms?
You do have to keep the resistance equal to 36 Ohms if you want to keep the same bias point, but what is your design goal here?

[Amoebius]

I want to improve the distortion and therefore split the emitter resistor.

I made a drawing of the section i am talking about.

[todorp]

Try The Art of Electronics, the BJT chapter. There are some examples and advice on how to determine the values of the emitter resistors and the capacitor for exactly your case.

[Terry Bites]

Splitting the reistor does help you separate the DC conditions from the AC current gain settting. Some details here:

https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/Book%3A_Semiconductor_Devices_-_Theory_and_Application_(Fiore)/07%3A_BJT_Small_Signal_Amplifiers/7.3%3A_Common_Emitter_Amplifier

[Amoebius]

Thanks for the answers. In general i belevieve i understand the concept, the ac gets shorted by the capacitor and with splitting the resistor you make a compromise between gain and distortion.

But i can't find the formula for calculating the Capacitor.

For a non splitted resistor bypass capacitor i believe it ist 1/(2Pi*f*Xc), where Xc is a tenth of the emitter resistor.

When i split the Resistor, is the formula still the same? And for Xc, which resistor do i choose, Re1 or Re2, when Re2 is bypassed?

[MrAl]

I want to improve the distortion and therefore split the emitter resistor.

I made a drawing of the section i am talking about.

This way is a bit easier because you can adjust the AC gain independently from the DC bias  by adjusting Re2.

We could work up a formula but for now the total impedance is:
Z=(Re1*(Re2*s*C+1))/(Re2*s*C+Re1*s*C+1)

and let s=j*w and w=2*pi*frequency.

[Amoebius]

OK, thank you.

Just one question, what is s and what is w? I am a total beginner, sorry

[iMo]

That formula works with complex numbers, where "j" is the "i" (imaginary unit), and "w" is the "omega" - see
https://en.wikipedia.org/wiki/Electrical_impedance

PS: in EE "j" is used instead of "i", such we do not mix the "i" - imaginary number with current..
Complex number in Mathematics A+iB
Complex number in Electronics   A+jB

[Amoebius]

OK, thanks. With total impedance you mean input impedance, C being the value of the emitter cap?

[MrAl]

OK, thanks. With total impedance you mean input impedance, C being the value of the emitter cap?

It's the impedance of the network so you can get an idea what the gain could be.
Yes C is the emitter cap value.
w is the angular frequency but it's just 2*pi times the regular frequency f in Hertz.

Since the cap is so large,  you can estimate the total impedance to be the total resistance which would be the 36 Ohm in parallel with Re2:
R=Re1*Re2/(Re1+Re2)

and that should be good enough for frequencies that are not super low.  It depends on the resistance of Re2 but probably ok for frequencies above about 50 Hertz.

For an example, if Re2 is 10 Ohms then the total resistance emitter to ground is around 8 Ohms.

Knowing that you can estimate the gain.

The 's' and 'j*w' forms are complex so you have to calculate the real and imaginary parts then square both, then add them together, then take the square root that gives you the magnitude of the impedance.  The above estimate should be good enough, but if you want to know more about this complex form i can show you a few examples or you can just look up "complex numbers" on the web.  They are called complex numbers but they are not that much more complicated than regular numbers you just add, subtract, multiply and divide a little differently that's all.

[Amoebius]

OK, many thanks, i am learning.

[Amoebius]

I have one more question. If i do it the way, MrAI suggests, how do i have to calculate the base imedance?

Is it hFE(r'e+Re2)?

r'e being 25mV/Ie?

[MrAl]

I have one more question. If i do it the way, MrAI suggests, how do i have to calculate the base imedance?

Is it hFE(r'e+Re2)?

r'e being 25mV/Ie?

Hi,

I would say it would be closer to
 hFE(r'e+Rp)
with Rp being the parallel combo of Re1 and Re2:
Rp=Re1*Re2/(Re1+Re2)
and remember that's only when the cap is large like your 2200uf, and for frequencies not too low.

that's why i suggested using the parallel combo.
That's great that you know about r'e.  Most 'beginners' dont know about that :-)

[Amoebius]

Ok, thanks.

[Amoebius]

But isn't Re1 in this case getting ignored because of the bypass cap?

[MrAl]

But isn't Re1 in this case getting ignored because of the bypass cap?

No, if you short the cap (again for frequencies not too low) Re1 is in parallel with Re2.  You can ignore Re1 if Re2 is very low value because then Re1 is insignificant.  So maybe if Re2 is one tenth the value of Re1 then Re1 can be ignored but they are still in parallel.
If you open the cap (low frequency) you have only Re1.

[Amoebius]

Ah, ok. Thank you very much.

[Amoebius]

One question remains.

Xc of the Emitter bypas cap in the non splitted emitter resistance circuit was chosen to be 1/10 RE.

What should it be in the split emitter circuit? 1/10 Re1?

Or do i have to consider the resistance of Re2 also?

[MrAl]

One question remains.

Xc of the Emitter bypas cap in the non splitted emitter resistance circuit was chosen to be 1/10 RE.

What should it be in the split emitter circuit? 1/10 Re1?

Or do i have to consider the resistance of Re2 also?

In the series resistance circuit if you have 26 Ohms on top and 10 Ohms on the bottom then if you short the cap you then have only 26 Ohms (again frequency not too low).  That means that if you use the parallel circuit with the two resistors separate, you have one 36 Ohm resistor (Re1) and one unknown resistor Re2. 
Since in the series arrangement you had 26 Ohms with the cap shorted, that would mean that the total resistance with the cap shorted in the parallel arrangement would also have to be 26 Ohms.  So to calculate the unknown Re2 now you would use:
26=Re1*Re2/(Re1+Re2)
You can solve this for Re2 and get:
Re2=(26*Re1)/(Re1-26)
and since Re1 was 36 Ohms that means Re2 would be 93.6 Ohms.

The more general formula is:
Re2=(Rcs*Re1)/(Re1-Rcs)
where Rcs is the upper resistor value (26 Ohms in your case).

Checking, 93.6 in parallel with 36 we get:
Rcs=36*93.6/(36+93.6)
and so Rcs does come out to 26 ohms again which is what the original series combination had when the cap was shorted.

We probably want to keep the reactance of the cap at the lowest operating frequency to be 5 percent or less than the new resistor value of Re2. If Re2 is 93.6 Ohms then that means we want it to be 4.68 Ohms or less.  Since the reactance of the cap is:
RC=1/(2*pi*f*C)

we can form the equation:
4.68=1/(2*pi*f*C)

and since C is 2200uf that means we can solve that for f to see what low frequency this would happen at.  Solving that for f we get:
f=15.5 Hz

so this should work down to around 15 Hertz or thereabouts.
You can test it in spice.

[Amoebius]

Ok,  i can't thank you enough, you helped me a lot.

[MrAl]

Ok,  i can't thank you enough, you helped me a lot.

You're welcome glad i could help.
We could do a more full analysis too at some point.