Need help understanding current source loading common source amplifier

Thread Starter

iinself

Joined Jan 18, 2013
98
For the circuit in the diagram, observe that Vgs is around 3V and so is Vds on the breadboard. I do not understand if it is necessary to keep the drain at 1/2 of V1 as it is done when using a drain resistor ( even in the case of using a current source). I am observing that the output is clipped for even around 6V p-p, there is another 10V to use. Basically not clear on how the current source works as a better drain resistor. Appreciate any help. Thanks
 

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sparky 1

Joined Nov 3, 2018
701
One way to establish an appropriate DC operating point is to use a constant current source.
The CC maintains the same bias, otherwise a finite change in bias can result in unwanted change in gain.

An appropriate DC operating point means a stable and predictable DC drain current.
Also having an appropriate DC voltage at the drain (Vds) will allow the MOSFET to operate in the
saturation region for changes in the AC signal levels. This will allow sufficient output signal swing.
 
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MisterBill2

Joined Jan 23, 2018
11,556
The regulated current source is indeed a way around a number of complications, just as Sarky says.
The one caution that I offer is that depending on how the current source is implemented it might be a noise source as well. That could be an unintended consequence, which sometimes happens.
 

Ian0

Joined Aug 7, 2020
4,821
The gain of the amplifier is gfs multiplied by the load impedance.
The impedance of a current source is very high. Therefore the gain can be very high, but you have to watch out for other resistances connected to the output which appear in parallel and reduce the impedance and thus reduce the gain.
 

Thread Starter

iinself

Joined Jan 18, 2013
98
The gain of the amplifier is gfs multiplied by the load impedance.
The impedance of a current source is very high. Therefore the gain can be very high, but you have to watch out for other resistances connected to the output which appear in parallel and reduce the impedance and thus reduce the gain.
Thanks

Yes I observed that in LTSpice I get extremely high gain but on the board I observe something much less (around 30). I don't have any other resistors other than for biasing (2M each)

Since there is constant current flowing through the MOSFET, the only way for change in voltage to occur at the drain is if the mosfet changes its resistance dynamically, it that a proper way to understand? Or since MOSFET is transconductance device we don't have to worry about voltage that much. I increased the supply voltage to 20V but did not see much increase in output power, but when I increased the current from the current source it increase the power substantially?
 
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