Common Emitter Amplifier Help

[Callo1234]

Hi, could you help?

In the following circuit

http://www.ecircuitcenter.com/Circuits/trce/trce.htm

Which capacitor would correspond to my high frequency cut off and which would correspond to my low frequency cut off? And also in each case what resistor values would I use to calculate the cut offs in each case.

Thanks for any help!

 

[audioguru]

Look at low pass filter and high pass filter circuits in Google.

This circuit has RS, C1 and the input impedance of the amplifier as a highpass filter as is told in the article that cuts low frequencies. It also has the RC, CL and RL as another highpass filter.
The capacitance of the transistor, stray capacitance of the pcb layout and the capacitance of the load determine the lowpass response which is its high frequency cutoff.

 

[Callo1234]

Ahh thanks, its clearer now

SO does that mean when i try to work out the capacitor vlaues I need.. via 1/2piRF

For CI, I just input into the equation the desired lower cut off frequnecy (40hz) and the parallel combination of (R1 AND R2) + Rs - as my value for R.

For CL, I just input into the equation my desired lower cut off frequnecy (40hz) again and the sum of the resistor values RC and RL?

 

[audioguru]

For the input filter calculation, you forgot to include the input impedance of the transistor.

You are correct with the output filter calculation.

 

[Callo1234]

For the input filter calculation, you forgot to include the input impedance of the transistor.

You are correct with the output filter calculation.

Thanks man, im glad Ive understood it, its just a case of summing input impendance of my CE into total R (its not parallel to my emiiter resistance)?

I have no clue how to calculate upper freq cutoff via my transistor but will try today, one thing that confuses me though is when i read up in the library about a similar circuit (the same basically just different component values), it mentioned

Add a .01 μF capacitor in parallel to the load resistor. This will simlate the effects of stray wiring capacitances and other capacitances in the amplifier circuit.

Do you think that Is a general idea for all circuits , and something that i should implement? or is the capacitor value likely to be specific to the values of the components in that particular circuit. In fact thinking about it, it will be another factor to my low freq cut off, which I think would just see me input 1/Rs as the value for R I think in 1/2piCR. That might not be worth putting into my multisim circuit at all.

Thanks for your help so far, its made things a lot lot clearer!

 

[audioguru]

The DC input resistance and AC low frequency input impedance of a common-emitter transistor amplifier is Beta times the internal emitter resistance. The internal emitter resistance is about 0.026 divided by the emitter current in amps.

The datasheet for the 2N3904 shows a graph of its low frequency input impedance at various currents.

The datasheet for the 2N2222 shows a max output capacitance of 8pF. Stray wiring capacitance is probably the same. The article you found says 0.01uF which is 625 times too high!

The high frequency cutoff frequency is when the reactance of the transistor's output capacitance plus the stray capacitance is equal the the value of collector resistor that is in parallel with the load impedance. 16pF and 10k ohms is 1MHz. 16pF and 1k ohms is 10MHz. At higher frequencies then the Miller effect of capacitive feedback from the collector to the base changes the calculation.

 

[Callo1234]

Ive just realised what you call Beta is what I call hfe , but I learned that

re (emitter resistance) = 1/40Ie

which is 1/ 40*10mA = 25 ohms



If I go by emitter resistance = 0.026 / 10 mA

I get re = 2.6 ohms


***Oh just checked and saw that 1/40 was 0.025 so I think you just meant emitter resistance is 0.025 x emiiter current in amps

 

[Callo1234]

< I just remembered that my Ce (which I am using to split my resistors to give me my midband gain) will have a low frequency cut off as well of 40hz.

And that Ce = 1/2 pi*f*re (re being what I just calculated)

Ce = 1/pi*40*25 = 318uF.

Strangely that had no effect on the lower cut off frequency of my graph (which stays at about 8hz!) and no matter what size capacitor I put there my lower freq cut off isnt effected when it has to be I would have thought, am I wrong? In fact i found that putting any value for CL is having no effect on my cut off frequency either, only C1 does!

For the input filter calculation, you forgot to include the input impedance of the transistor.

Ok so data sheet says
hfe (beta) // IC = 10 mA; VCE = 10 V // hfe = 75

So 75*25 =1875 ohms for my input impedance of my transistor, which is quite big. As it is in series with my emiiter resistance I just add that to my value of TOTAL R

SO C1 = 1/2pi*40*TOTAL R
Where Total R = Rs + R1//R2 + Input Impedance of transistor.

And my cut off has shot up to 20hz which is much closer

Sorry for all the detail, i just want to run my calcs here as i have no confidence in getting them right by myself.

The fact that it only goes up to 20hz is probably down to my CL and Ce not having any effect at all. In fact i even deleted my CL and my frequency plot from my bode plotter was unchanged, i dont know why, could that be to how the bode plotter is connected up?

 

[audioguru]

Your internal emitter resistance calculation makes the resistance increase as the current increases. That is backwards. Look at the graph of input impedance on the datasheet for the 2N3904. The impedance decreases as the current increases.

The Spice file shows the parts values. The emitter current is about 2.2mA, not 10mA.

The RE1 and RE2 resistor values are the same. CE is across RE2, not across the internal emitter rersistance. RE2 is 600 ohms. CE is 50 uF. So at extremely low frequencies where CE has no effect the gain is reduced only -6dB. So it is difficult to calculate a cutoff frequency when the cut is so small.

CL is 10uF and RL is 100k. So their cutoff frequency is 0.16Hz and has no effect on the much higher cutoff frequency of about 32Hz caused by C1.

 

[Callo1234]

Sorry my actual values are different to all those.

Im trying to get Voltage gain:

at least 20 dB at mid-band into 600 Ω
Frequency response:
Lower cut-off frequency: 40 Hz
Upper cut-off frequency: 1 MHz

Input impedance: 600 Ω
Output impedance: 600 Ω

DC supply voltage: 12 V
DC supply current: not more than 15 mA

Output power: not less than 2.0 mW into 600 Ω at mid-band.

CL is 10uF and RL is 100k. So their cutoff frequency is 0.16Hz and has no effect on the much higher cutoff frequency of about 32Hz caused by C1.

So what i dont get is, in placing 40hz as the lower cut off for all my high pass filters, i.e CL CE AND C1, i have been doing it wrong? Should CL have no effect on my lower cut off freq, same as Ce? And therefore I should make them have a tiny freq cut off. Its confusing because as you said above, that has no effect, so whats the point in them being there? This is tough. Ive been inputting 40hz for all those capacitors as thats the lower cut off i want, but it seems crazy that in the above circuit they have given CL a minute cut off frequency so that it has no effect on C1