# CE - Full Emitter Resistor Bypass - Computing Capacitor Values

#### elec_eng_55

Joined May 13, 2018
214
I have some more questions in my ongoing saga re CE amplifiers.

(1) My Heathkit course manual calculates C1 And C2 using the following:

C1 = 3.18 / (f1 * Rin) = 3.18 / (100 Hz * 3.94K) = 8.07 uF
C2 = 3.18 / (f1 * RL) = 3.18 / (100 Hz * 5.6K) = 5.68 uF

Why wouldn't they use 0.159 rather than 3.18 for C1 and C2?

C3 = 0.159 / (f1 * RNF) = 0.159 / (100 Hz * 5 Ohms) = 318.47 uF

(2) I came up with a formula for computing the value of RNF:

RE = 500 ohms
RL = 5.6K
RC = 2.5K
re' = 37mV / 3mA = 12.3 ohms
Av = 100

RNF = (RE * ((RC * RL) - (Av * RC * re') - (Av * RL * re'))) / ((Av * RC * re') + (Av * RC * RE) + (Av * RL * re') + (Av * RL * RE) - (RC * RL)) = 5 ohms

I hope that is correct.

(3) How do I compute the ac current flowing in RNF as I want to calculate the wattage
for this resistor.

I think that I am now in the home stretch!

David

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#### Jony130

Joined Feb 17, 2009
5,230
Why wouldn't they use 0.159 rather than 3.18 for C1 and C2?
Safety margin due to the fact that the electrolytic capacitor has a very large tolerance +/- 20% or more and lose much of his initial capacitance when aging (30% loss in capacitance or more). So this is why they decide to use to this safety factor 3.18/0.159 = 20.

So, using this equation C = 3.18/(f*R) you get 20 times larger capacitor then you need.

Also, can you explain to me why didn't to use the standard capacitor values (1; 2.2; 3.3; 4.7; 6.8 )?

I came up with a formula for computing the value of RNF:
What can I say OMG? What a equation?

I will do it this way (in steps)

Desired voltage gain is 100V/V

Rc'' = 2.5kΩ||5.6kΩ = 1.72kΩ

Re'' = 1.72kΩ/100V/V = 17.28Ω

RNF = 17.28 - 26mV/3mA = 8.618Ω

And becouse RE >> RNF
I pick RNF = 8.2Ω as a standard E24 value. And the gain in simulation is 98.57V/V, not so bad.

And why your thermal voltage is 37mV?

(3) How do I compute the ac current flowing in RNF as I want to calculate the wattage
for this resistor.
This current will be very small so you do not have to worry about the wattage

V_RNF_peak ≈ Vin_peak * RNF/(RNF + re) = 1mVp * 8.2Ω/(8.2Ω + 8.67Ω) = 0.48mVp

But in terms of a power, we are interested in RMS value not in peak

#### elec_eng_55

Joined May 13, 2018
214
Safety margin due to the fact that the electrolytic capacitor has a very large tolerance +/- 20% or more and lose much of his initial capacitance when aging (30% loss in capacitance or more). So this is why they decide to use to this safety factor 3.18/0.159 = 20.

So, using this equation C = 3.18/(f*R) you get 20 times larger capacitor then you need.

Also, can you explain to me why didn't to use the standard capacitor values (1; 2.2; 3.3; 4.7; 6.8 )?

What can I say OMG? What a equation?

I will do it this way (in steps)

Desired voltage gain is 100V/V

Rc'' = 2.5kΩ||5.6kΩ = 1.72kΩ

Re'' = 1.72kΩ/100V/V = 17.28Ω

RNF = 17.28 - 26mV/3mA = 8.618Ω

And becouse RE >> RNF
I pick RNF = 8.2Ω as a standard E24 value. And the gain in simulation is 98.57V/V, not so bad.

And why your thermal voltage is 37mV?

This current will be very small so you do not have to worry about the wattage

V_RNF_peak ≈ Vin_peak * RNF/(RNF + re) = 1mVp * 8.2Ω/(8.2Ω + 8.67Ω) = 0.48mVp

But in terms of a power, we are interested in RMS value not in peak
Hi Jony:

I was just showing the theoretical values first and then I will convert to the standard values once
I have a good design. At least that was the thought behind it.

The 37mV came from my old Heathkit course manual but I will rework using 26 mV.

That crazy formula comes from reworking Av = (Rc//RL) / (re' + RE//RNF) and I must admit that
I couldn't isolate RNF by hand so I used a website to do the manipulation for me.

David

#### elec_eng_55

Joined May 13, 2018
214
I guess I have one last question (just call me Columbo). I am trying to understand how
one arrives at the maximum unclipped, lowest THD output voltage that such an amplifier
can produce. My course material says that this works.

The lesser of

vO_Plus = ICQ * rL = 3mA * 1.73K = 5.19 V peak

or

vO_Minus = VCEQ = 6 V peak

whichever is less.

Therefore the maximum unclipped low THD output should be 5.19 V peak.

When running a simulation, a Vin of 4mV yields a Vout of 378 mV and the THD
has just exceeded 1%. There is, however, no apparent clipping.

Is there a mathematical approach to find the "real" maximum ac output voltage?

It seems that having a VCEQ as high as 6 volts is not necessary!

That's me just thinking out loud.

Thoughts?

Sincerely,

David

#### Jony130

Joined Feb 17, 2009
5,230
Because you still forget that the BJT is a very nonlinear device. And that the voltage gain is not constant but will change with the input signal exponentially.

Av = Vout/Vin = - (Rc * Is)/Vt * exp^(Vin/Vt) (with CE capacitor across RE ressitor)

And this is why single stage BJT amplifier produces alot of distortion. We get about 1%THD per 1mV at the base (10% for Vin =10mV) .
https://www.researchgate.net/public...C_DISTORTION_IN_LOW_FREQUENCY_POWER_AMPLIFIER

http://www.kevinaylward.co.uk/ee/bipolardesign2/bipolardesign2.xht

You can improve the distortion by adding the emitter resistor (without CE capacitor). But to get "low THD" you need to pick RE >> re.
But it is impossible to get high voltage gain and low THD at low load resistance in such a simple circuit.
As you can see you're expecting too much from this simple circuit. And this is why no one uses this circuit as "low THD" amplifier.

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#### elec_eng_55

Joined May 13, 2018
214
Because you still forget that the BJT is a very nonlinear device. And that the voltage gain is not constant but will change with the input signal exponentially.

Av = Vout/Vin = - (Rc * Is)/Vt * exp^(Vin/Vt) (with CE capacitor across RE ressitor)

And this is why single stage BJT amplifier produces alot of distortion. We get about 1%THD per 1mV at the base (10% for Vin =10mV) .
https://www.researchgate.net/public...C_DISTORTION_IN_LOW_FREQUENCY_POWER_AMPLIFIER

http://www.kevinaylward.co.uk/ee/bipolardesign2/bipolardesign2.xht

You can improve the distortion by adding the emitter resistor (without CE capacitor). But to get "low THD" you need to pick RE >> re.
But it is impossible to get high voltage gain and low THD at low load resistance in such a simple circuit.
As you can see you're expecting too much from this simple circuit. And this is why no one uses this circuit as "low THD" amplifier.
I have run a simulation with the schematic changed to reflect standard component values (attached). I also
attached the resulting AC Analysis.

My hand-calculated analysis (based on Heathkit manual) is as follows.

Vin = 1mV @ 1 Khz
Iin = Is = Vin / Rin = 1mV / 4.22K = 237nA

ie = Vin / (RE + re') = 1mV / (470 ohms + (26mV / 3mA)) = 2.09uA
ib = ie / hfe = 2.09nA / 100 = 20.9nA
ic = ie - ib = 2.09uA - 20.9nA = 2.07uA

Pload = (VCEQ * 0.707) * (ICQ * 0.707) = (6V * 0.707) * (3mA * 0.707) = 9.00mW

Pin = Pload / Ap = 9.00mW / 100 = 90uW

Av = (RC * Is) / (Vt ^(Vin/Vt) = (2.5K * 237nA) / (0.7^(1mA/0.7V))
= 0.0006325 / 0.7^0.00142857 = 0.0006325 / 0.999 = 0.0006

I am confused with the difference between my results and the LTSpice ac analysis.

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#### ericgibbs

Joined Jan 29, 2010
12,495
hi 55,
Please post, a simulation with the schematic changed to reflect standard component values (attached).
E

#### elec_eng_55

Joined May 13, 2018
214
hi 55,
Please post, a simulation with the schematic changed to reflect standard component values (attached).
E
Sorry .... I forgot to do it.

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#### ericgibbs

Joined Jan 29, 2010
12,495
hi,
As you say thats well out.
Could you post a clip of the Heathkit manual that shows the equations.?
E

#### elec_eng_55

Joined May 13, 2018
214
hi,
As you say thats well out.
Could you post a clip of the Heathkit manual that shows the equations.?
E
They are strewn throughout the text I would have to scan a lot of pages and my
scan software is not working. Sorry E.

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#### Jony130

Joined Feb 17, 2009
5,230
What you are trying to calculated?

#### elec_eng_55

Joined May 13, 2018
214
What you are trying to calculated?

I have run a simulation with the schematic changed to reflect standard component values (attached). I also
attached the resulting AC Analysis.

My hand-calculated analysis (based on Heathkit manual) is as follows.

Vin = 1mV @ 1 Khz
Iin = Is = Vin / Rin = 1mV / 4.22K = 237nA

ie = Vin / (RE + re') = 1mV / (470 ohms + (26mV / 3mA)) = 2.09uA
ib = ie / hfe = 2.09nA / 100 = 20.9nA
ic = ie - ib = 2.09uA - 20.9nA = 2.07uA

Pload = (VCEQ * 0.707) * (ICQ * 0.707) = (6V * 0.707) * (3mA * 0.707) = 9.00mW

Pin = Pload / Ap = 9.00mW / 100 = 90uW

Av = (RC * Is) / (Vt ^(Vin/Vt) = (2.5K * 237nA) / (0.7^(1mA/0.7V))
= 0.0006325 / 0.7^0.00142857 = 0.0006325 / 0.999 = 0.0006

I am confused with the difference between my results and the LTSpice ac analysis.

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#### Jony130

Joined Feb 17, 2009
5,230
Ok I see

LTspice shows that β = 310; Ic = 3.17mA; hence re ≈ 8.2Ω

Rin = R1||R2||[ (β+1)*re+(RNF||RE)] ≈ 2.23kΩ

Vin = 1mV/Rin = 448nA

Ie = 1mV/ (re + (RE||RNF)) = 1mV/(8Ω + 8.2Ω) = 61.72uA
Ib = Ie/(β+1) = 200nA

Ic = Ie - Ib = 61.5μA

IL = Ic * Rc/(Rc + RL) = 61.72μA * (2.7/(2.7 + 5.6)) = 20μA ; So Vout = 20μA*5.6kΩ = 112mV

So, the Av is around 112V/V

Pload = (VCEQ * 0.707) * (ICQ * 0.707) = (6V * 0.707) * (3mA * 0.707) = 9.00mW

Pin = Pload / Ap = 9.00mW / 100 = 90uW

Av = (RC * Is) / (Vt ^(Vin/Vt) = (2.5K * 237nA) / (0.7^(1mA/0.7V))
= 0.0006325 / 0.7^0.00142857 = 0.0006325 / 0.999 = 0.0006
What you are trying to calculate here is beyond me.

I think that this is a time for you to throw away your "Heathkit manual" and start to learn form another source. Because what I see is that you are trying to fit the "equation from Heathkit manual" to every circuit you met. And this approach will not work in the long-term.
http://www.ittc.ku.edu/~jstiles/412/handouts/5.6 Small Signal Operation and Models/section 5_6 Small Signal Operation and Models lecture.pdf

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#### elec_eng_55

Joined May 13, 2018
214
Ok I see

LTspice shows that β = 310; Ic = 3.17mA; hence re ≈ 8.2Ω

Rin = R1||R2||[ (β+1)*re+(RNF||RE)] ≈ 2.23kΩ

Vin = 1mV/Rin = 448nA

Ie = 1mV/ (re + (RE||RNF)) = 1mV/(8Ω + 8.2Ω) = 61.72uA
Ib = Ie/(β+1) = 200nA

Ic = Ie - Ib = 61.5μA

IL = Ic * Rc/(Rc + RL) = 61.72μA * (2.7/(2.7 + 5.6)) = 20μA ; So Vout = 20μA*5.6kΩ = 112mV

So, the Av is around 112V/V

What you are trying to calculate here is beyond me.

I think that this is a time for you to throw away your "Heathkit manual" and start to learn form another source. Because what I see is that you are trying to fit the "equation from Heathkit manual" to every circuit you met. And this approach will not work in the long-term.
http://www.ittc.ku.edu/~jstiles/412/handouts/5.6 Small Signal Operation and Models/section 5_6 Small Signal Operation and Models lecture.pdf
Thanks again. I will study this info and you are certainly right about the Heathkit manual.

Jony:

Where does β = 310 come from? Can it be calculated without referring to the LTSpice simulation? Or is β just from a 2N3904 data sheet?

I am actually writing a program in VB to design circuits, the CE being the first. In fact, these are the last calculations to add to the program. I plan to use my program to design many different circuits, then run a sim, if the two are in agreement I would then run another sim using standard component values and then actually bread board the circuit for bench testing.

David

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#### elec_eng_55

Joined May 13, 2018
214
Hi Jony:

RNF = (RC//RL / Av) *re' = 9.5Ω ≈ 10Ω

CNF = 0.159 / (f1 * RNF) = 0.159 / (100Hz * 10Ω) = 150uF

THD = Vin * (re' / (RNF + re'))^2 = 1mV * (8.7Ω / (10Ω + 8.7Ω()^2 = 0.02% ⇒ LTSpice says THD is 0.65%

David

#### Jony130

Joined Feb 17, 2009
5,230
THD for 1mV at the input will be around 1 * (8.7Ω / (10Ω + 8.7Ω))^2 = 0.216%
But for 10mV it will become 10 * (8.7Ω / (10Ω + 8.7Ω))^2 = 2.16%
Try it in LTspice. And do not expect to much from this super simplified formula (2nd harmonic only). No one calculates the THD in more complicated amplifiers.

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#### elec_eng_55

Joined May 13, 2018
214
THD for 1mV at the input will be around 1 * (8.7Ω / (10Ω + 8.7Ω))^2 = 0.216%
But for 10mV it will become 10 * (8.7Ω / (10Ω + 8.7Ω))^2 = 2.16%
Try it in LTspice. And do not expect to much from this super simplified formula (2nd harmonic only). No one calculates the THD in more complicated amplifiers.
In LTSpice it says 0.65%.

The formula for calculating CNF doesn't even come close to a good value. I don't understand.

#### Jony130

Joined Feb 17, 2009
5,230
In LTSpice it says 0.65%.
Strange I see this results

The formula for calculating CNF doesn't even come close to a good value. I don't understand.
Which formula ??

This one RNF = (RC//RL / Av) *re ?

Try this one

RNF = (RC//RL / Av) - re

#### elec_eng_55

Joined May 13, 2018
214
Strange I see this results

View attachment 158857

Which formula ??

This one RNF = (RC//RL / Av) *re ?

Try this one

RNF = (RC//RL / Av) - re
RNF = (RC//RL / Av) - re is the one that I used and got 9.5Ω ≈ 10Ω

CNF = 0.159 / (f1 * RNF) = 150 uF

The THD is then 0.65% in LTSpice.

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#### ericgibbs

Joined Jan 29, 2010
12,495
hi,
My LTS result is close to Jony's
E

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