the value of coupling capacitors

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
112
hi
i wanna ask is there a rule or an equation to find the appropriate value of a capacitor in a circuit ?
for example this circuit

it is a common emitter amplifier circuit and i choose the values of the three capacitors by trying different values but is there an equation that i might use to find their values ?
 

MrChips

Joined Oct 2, 2009
19,382
C1 is the input coupling capacitor.
C2 is the output coupling capacitor.

The values of C1 and C2 are determined by the desired low frequency response of the circuit.
If you were to model the AC behaviour of the input side, you will have a series source resistance, coupling capacitor C1, and an input load resistance. Altogether, the three in series create a high-pass filter that has a characteristic low frequency cut-off. You will select C1 to determine your cut-off frequency (knee or roll-off frequency).

Yes, there are equations to calculate this.

You do the same for the output capacitor C2.

CE is a different story.
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
112
Yes, there are equations to calculate this.
and what are they ? please can u tell me
i only know that it must be chosen that the lowest desired frequency can pass through
and why is ce different ? it is a bypass capacitor and its job is to pass the ac signal and not let it pass through the resistor re so it also must be chosen to pass the lowest desired frequency
 

Thread Starter

Ghina Bayyat

Joined Mar 11, 2018
112
no problem now i said i was because i read this equation in a book but didn't know what is R there was nothing about it
 

MrAl

Joined Jun 17, 2014
6,610
hi
i wanna ask is there a rule or an equation to find the appropriate value of a capacitor in a circuit ?
for example this circuit

it is a common emitter amplifier circuit and i choose the values of the three capacitors by trying different values but is there an equation that i might use to find their values ?

Hello there,

Actually the 3db down point is accomplished when C is the value:
C=1/(w*(R1+R2))

where
R1 is the equivalent input resistance (which may be the source resistance for example),
R2 is the output 'load' resistance,
C is the coupling cap value,
w is the angular frequency 2*pi*f with f the frequency in Hertz.
Units of resistance Ohms, capacitance Farads.

The reason for this is because the three components form a voltage divider and the output only appears across R2 the output resistor. When the above capacitance is used, at the frequency w we get an amplitude that is exactly 3db down from the infinite frequency level and that is often taken as the lowest operating frequency.
 

Picbuster

Joined Dec 2, 2013
979

MrAl

Joined Jun 17, 2014
6,610
Please read R as impedance there are next to resistors also capacitors and coils(wires act like one) in a transistor input circuitry.

Picbuster
Hello,

If "R" is actually an impedance then we would have to take the frequency response of that impedance into consideration also if we really wanted to get it right. The impedance may already be part high pass, or even work against us and be a low pass.
 

Picbuster

Joined Dec 2, 2013
979
Hello,

If "R" is actually an impedance then we would have to take the frequency response of that impedance into consideration also if we really wanted to get it right. The impedance may already be part high pass, or even work against us and be a low pass.
Correct it goes even more complicated:
The input impedance is subject to gain and output impedance.
Calculate for a negative input impedance using frequency,gain and output impedance generate an oscillator.
That's why you can't put a load on that type of oscillator without a buffer circuit ( fet /opamp).

Picbuster
 

MrAl

Joined Jun 17, 2014
6,610
Correct it goes even more complicated:
The input impedance is subject to gain and output impedance.
Calculate for a negative input impedance using frequency,gain and output impedance generate an oscillator.
That's why you can't put a load on that type of oscillator without a buffer circuit ( fet /opamp).

Picbuster
Hi,

What oscillator is that?

Very often the sources are resistive anyway though so it's a good place to start.
 

Picbuster

Joined Dec 2, 2013
979
Hi,

What oscillator is that?

Very often the sources are resistive anyway though so it's a good place to start.

If you have a box with an input and output you can specify that gain = output/ input
When output is phase shifted and connected to input the box will oscillate when gain >1.
The phase shift, for a particular frequency, is created with an impedance.
Regardless of the box content.
Put an amplifier, transistor or fet in the box and it will work.
Feed back phenomena is the nightmare of the amplifier builder ( known as an oscillator amplifies and an amplifier oscillate).

I will try to find my dictates about this subject if you are interested in the complete theory.
Let me know if I have to dive into my dungeons.

Picbuster
 

Picbuster

Joined Dec 2, 2013
979
Hi,

What oscillator is that?

Very often the sources are resistive anyway though so it's a good place to start.

If you have a box with an input and output you can specify that gain = output/ input
When output is phase shifted and connected to input the box will oscillate when gain >1.
The phase shift, for a particular frequency, is created with an impedance.
Regardless of the box content.
Put an amplifier, transistor or fet in the box and it will work.
Feed back phenomena is the nightmare of the amplifier builder ( known as an oscillator amplifies and an amplifier oscillate).

I will try to find my dictates about this subject if you are interested in the complete theory.
Let me know if I have to dive into my dungeons.

Picbuster
 
Top