# Common Drain - Source Follower JFET Amplifier Capacitors

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#### Itsec

##### Newbie level 3
Hallo! New to this forum. Although my main activities involve networking and IT consulting, i am attending courses for a telecommunications postgraduate. Among the courses, there is one concerning some basic circuit designing using breadboard. The previous labs were ok, but now i am stuck in a project about common drain - source follower amplifier using a 2n5458 JFET. I have calculated all the necessary values for resistors, but i am not able to find any equations in order to calculate the value of the capacitors (Cin, Cout, Cbs, Cdc).

Can anyone help me with it?

I also post an attachment with the spice circuit.

Vdd=15V. Idsq=Idss/2. Idss is 7.4mA.

Also, Ac amplitude is 25mV and we use sin values for transient analysis.

#### Attachments

• commondrain.JPG
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#### LvW

Hello Itsec,

if you know the task of each capacitor it will be no problem for you to find appropriate values.
I say "find" instead of "calculate" because for most applications it is not common (and not necessary) to exactly calculate the values.
Instead lower limits for the capacitances are found based on a "rule of thumb" that reflects the range of operating frequencies for the whole circuit.

Itsec

### Itsec

Points: 2

#### dselec

cdc= 100uf
the input 4.7uf for audio range
the last depends on gain u need 0.1uf-100uf

Itsec

### Itsec

Points: 2

#### Itsec

##### Newbie level 3
Thanks a lot for the quick replies. Well, the design of the circuit assures that we have current gain of 30. I think, this is solved by the choice of resistors. Also, we dont care about voltage gain. Considering now the range of frequencies, its up to me,so as a typical value I determine the fL( f-3db), lets say 1KHz. In previous labs with bjts, and in other assignments of this lab (common source amplifier), i found the values of the capacitors succesfully, using rules such as C=1/2*π*fL*R, but in this circuit, this calculations dont give me the frequency I am expecting.

#### FvM

##### Super Moderator
Staff member
I don't understand the purpose of RS1 = 82 Ohm and Cbs. Where did you get it?

#### Itsec

##### Newbie level 3
Its the circuit in our notes. Nothing changed. Only the values of resistors calculated from the equations given in our notes. RS1 is 8.2 ohm. In general, its the resistance Rs, which we calculate with this equation: Rs= (Vdd-Vdsq)/Idq, and then it gives us a type where we calculate RS1 in relation to RL, Current Gain=30, Rinput=100kΩ and gm of the jfet. Finally, RS2=RS - RS1. If you want i can attach the detailed procedure. As for the capacitor CBS, really i dont know. I am going to test some capacitor values and report back with my findings. Good morning!

#### FvM

##### Super Moderator
Staff member
If it's in your notes, there should be an explanation for RS1 and CBS. Perhaps you should show the original diagram and comments.

Normally, CBS stands for the bulk-source capacitance of an integrated MOS transistor. Obviously it doesn't apply for a discrete JFET.

In special cases, you may want an additional resistive load for high frequencies to prevent parasitic oscillations. But it's not regularly required for a source follower, and 8.2 ohm seems too low ohmic for this purpose.

P.S.: The "current gain" point needs explanation as well. You won't rarely define a current gain of a source follower. What's your definition of current gain in this case?

#### LvW

...........
Normally, CBS stands for the bulk-source capacitance of an integrated MOS transistor. Obviously it doesn't apply for a discrete JFET.
..........

The capacitor CBS acts as a shunt capacitor across RS2 and, thus, prevents large signal feedback.
Therefore, RS=RS1+RS2 provides dc feedback (stabilization of the dc operating point) and a small signal feedback is provided by RS1.

#### Itsec

##### Newbie level 3
The original diagram is the identical i attach above. In the notes, only the diagram exists and the equations in order to find the values of resistors. Nothing more. Moreover, during the courses, in the class, we have only dealt with BJTs. As for the current gain, we divide the current we take in RL(source current) with the initial current. In order to be more detailed, i attach the methodology given.

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