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Bypass / Coupling Capacitor Values

jdp

New Member
I have calculated capacitor values for the attached common-base amplifier.
Could someone advise if these values are correct. I have attached a schematic.
Frequency range is 20 Hz to 20,000 Hz.

C1 = 1 / (2 * π * f1 * RE * 0.1) = 80 uF @ 1 volt

C2 = 1 / (2 * π * f1 * Zin * 0.1) = 3,189 uF @ 1.7 volts

C3 = 1 / (2 * π * f1 * RL * 0.1) = 8 uF @ 6 volts

Thanks

 

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audioguru

Well-Known Member
Most Helpful Member
Without the "0.1" in the formula then the output level at the cutoff frequency is -3dB which is a small noticeable change but in numbers it is half the output power.
The circuit has two coupling capacitors (input and output) so using the formula without the "0.1" then the output at the cutoff frequency is -6dB which is fairly bad. Compromise with 0.3 or 0.5 in the formula.
 

ronsimpson

Well-Known Member
Most Helpful Member
I think the "RE" is much lower than you think. Maybe 40 ohms. So that is one of the problems.
We agree that 0.1 is wrong.
I have never used a common base amp like this. The input impedance is very low. The gain is very dependent on the transistor.
120534
 

jdp

New Member
Hi Ron:

For "RE" in the "C1 " formula I actually used the value of resistor RE = 990 ohms which I now realize is wrong.
r'e = 25.86 mV / IE = 25.86 mV / 1.01 mA = 25.60 ohms

C1 = 1 / (2 * π * f1 * r'e * 0.1) = 3,108 uF but I should use a different factor rather than 0.1 such as 0.5

C1 = 1 / (2 * π * f1 * r'e * 0.5) = 620 uF ?


Zin = r'e║RE = 25.60 ohms ║990 ohms = 24.96 ohms

The formula for the 0.1 factor came from the following site.
 

ronsimpson

Well-Known Member
Most Helpful Member
Back to post #3 by audioguru and the picture in #5. The 0.1 is too small depending on what you call bandwidth. Just one RC is drawn in #5. You have C1 and C3 so the graph will be different. The guru is suggesting is using 0.5 to get a -3db at the frequency you want, because of 2-caps. Or I use 1.0 to fine each RC and then because there are two caps I need to make the caps 2x bigger at the end. (same results) Note you can say -6db or -1db or any db you want, it is common to use -3.

Agree on C1.
C2 I think Rin=4700//990//(25 x 100) Where 25=Re, and 100= Bata or current gain of the transistor.
C3 R=10k

If you give me cap numbers I can test what the frequency response is.
 

jdp

New Member
Zin = 4700//990//(25*100) = 617 ohms

C1 = 1 / (2 * π * f1 * r'e * 0.5) = 620 uF ?
C2 = 1 / (2 * π * f1 * Zin * 0.5) = 26 uF ?
C3 = 1 / (2 * π * f1 * RL * 0.5) = 1 / (2 * π * 20 * 10K * 0.5) = 2 uF ?

I have a copy of LTSpice. I would like to know how to do a plot for each of these to gain a better understanding.

Thanks
 

ronsimpson

Well-Known Member
Most Helpful Member
I think this will work for you.
R6, C4 is to test what the output RC will do. Remove when you are done.
"Run" then click on the output to see a graph.
You can click on "E" to see what just C1 is doing.
 

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jdp

New Member
I am not sure what this graph is telling me. I got the same result with and without C4 and R6.
 
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