Bypass / Coupling Capacitor Values

[jdp]

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

 

[ronsimpson]

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

How did you find "RE"?

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

How did you find Zin?

C3 = 1 / (2 * π * f1 * RL * 0.1)

Where did you find the formula? "0.1"??

 

[audioguru]

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]

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.

 

[ronsimpson]

Here is the plot for 8uF and 10k only. -3db at 2hz.

 

[jdp]

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. What is a Bypass Capacitor? This is an article that explains what a bypass capacitor is. A bypass capacitor is a capacitor that shunts AC signals that may be on a DC signal to ground. In doing so, bypass capacitors filter out AC noise from a DC signal to produce a cleaner DC signal. www.learningaboutelectronics.com What is a Coupling Capacitor? This article explains what a coupling capacitor is and how it is used in an electronic circuit to couple AC signals and block DC www.learningaboutelectronics.com

 

[ronsimpson]

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]

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]

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.

 

[jdp]

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

 

[ronsimpson]

I got the same result with and without C4 and R6.

Yes, It is hard to see the effect of C3, R5 so I added C4, R6 (some values) so you can see what C3, R5 does if the amplifier was not there.