# Common source amplifier

1. Jun 21, 2005

### Soilwork

Just out of interest guys have I got this right?

Function of certain components of a common source amplifier:

Capacitors at input and output: They are coupling capacitors and simply block the DC voltage, but allow the AC voltage through.

Gate Resistance: Is the purpose of this to limit the amount of current reaching the gate of the JFET??

Source resistance: This reduces the gain of the circuit, which thereby reduces distortion.

Drain resistance: Limits the maximum voltage gain?

I think that the source and capacitor explanations are right, but I'm not that sure about the Gate and Drain resistors.

2. Jun 21, 2005

### SGT

The Gate resistor provides a ground reference to the Gate. The source resistor provides a positive voltage to the source. The combination of those factors makes VGS < 0. This is a requirement for the functioning of the FET.
The source resistor really reduces the gain, but if this is undesirable you can bypass it with a large capacitor.
The drain resistor really is responsible for the gain. If your source resistor is not bypassed the gain is approximately $$A_V = \frac {R_D}{R_S}$$. If it is bypassed the gain is approximately $$A_V = g_m R_D$$

3. Jun 21, 2005

### Soilwork

Cheers for that :)

Another question that I have been struggling finding an answer for is what effect a finite input impedance has on the response of an amplifier?

I mean I've been reading up on how to use op-amps and the differences between ideal and real, and can't find the reason for this one thing.
I did find somewhere that a finite input impedance puts an upper bound on the resistances in the feedback circuit...but how this effects the response of the amplifier I'm not sure.

Sorry for asking stupid questions, but I'm new at this :(

4. Jun 21, 2005

### Staff: Mentor

5. Jun 22, 2005

### SGT

A finite input impedance will be part of the feedback circuit.

6. Jun 22, 2005

### Soilwork

ahh k so it will change the amount of feedback in the circuit.
Cheers :)

Thanks for the book suggestion. I will probably order it next week.

7. Jun 26, 2005

### Rogue Physicist

Also, the input impedance should be much higher than the source impedance (internal impedance of the source). This not only guarantees that the voltage signal to be amplified is efficiently transferred in a useable form to the amplifier inputs, but also is critically important for lowering distortion of the waveform. It works by limiting the current draw from the source which can distort the voltage waveform and affect frequency response as well. If the current varies significantly in the input loop including the source of the signal, you begin to affect the signal you are trying to measure.

When discussing 'impedance matching' in amplifier stages (especially audio), what is meant is keeping the input impedance of the next stage high enough not to affect the circuit behaviour and signal of the preceding stage. For high frequency applications, stray capacitance from physical layout and stage coupling becomes a dominant issue.

Last edited: Jun 26, 2005
8. Jun 26, 2005

### SGT

I think you didn't understand clearly the post. Source here refers not to power source or signal source, but to the terminal of the FET called source.

9. Sep 18, 2006

### peace_lp

Finding bypass capacitors in common source amplifier

Hi,
I am new to this website. I am having troubles in designing a common source amplifier with a voltage gain, Av=26 dB (no bypass capacitor) and a gain of 40 DB (with a bypass capacitor) and a frequency of 100-1000HZ. Can anyone please guide me to calculate the value of the bypass capacitor? I searched everywhere for the methods of finding the capacitors values but still in vain.
thank you very much for all your help.

thanks,
peace

10. Sep 20, 2006

### dahlungril

Hey I have that book! allthough i have not read it yet...

11. Sep 20, 2006

### SGT

Basically, the impedance of the bypass capacitor at the lowest operation frequency (100 Hz) should be much smaller than the resistance of the source resistor (generally 10 times smaller).
If you have the source resistor, you can calculate the capacitance.