Welcome back to Electronics, this is Dr. Robinson.
In this lesson, we're going to look at the common source amplifier, and
in particular we're going to look at the DC Analysis of this amplifier.
In your previous lesson, we examined MOSFET characteristic curves and biasing
and our objectives for this lesson are to introduce the common source amplifier and
to analyze the common source amplifier at dc.
Now, to actually implement a gain block that has this functionality,
we have to build a circuit.
And you know one way of doing this.
So, for example, we could use a non-inverting op-amp amplifier or
an inverting op-amp amplifier to implement this functionality.
We designed the circuit so that they have a gain A.
But, in this lesson, what we're going to do is look at another way to do this.
We're going to look at how you can use a discreet MOSFET transistor to
implement this gain block.
Now, there are numerous single MOSFET circuits that can implement
this gain functionality, but during this lesson we're going to look at
one particular type called the common source amplifier or CS amplifier.
Now as we go through this lesson we should keep in mind the relationship between
a MOSFET's biasing, or its quiescent point, and its behavior.
Remember in the previous lesson, we looked at how by changing the external
voltages and currents of the MOSFET, we can change its behavior.
And in particular, we can change its region of operation.
Now you may also remember that for MOSFET to operate as an amplifier,
we must bias the transistor in its saturation region.
So at very low frequencies at DC voltages and DC currents
the voltages and currents that set the operating point of the MOSFET.
The impedance of the capacitor is very large.
So as omega goes to zero, in other words the quantities go to DC quantities,
this impedance does an open circuit.
So, in determining the Q point for this transistor,
which determines this circuit's gain characteristics, we can
treat all of the capacitors in the circuit as open circuits or large impedance.
So here I've redrawn the previous circuit with all of the capacitors made open
So we can analyze the DC performance of the common source amplifier or
determine it's cue point or quiescent point.
Now because of the structure of our MOSFET,
we know that the gate current into the MOSFET is zero.
IG, the gate current, is equal to zero and we also know that ID,
the drain current, is equal to IS, the source current.
Or in other words, the current going into the drain all comes out of the source, so
that both the drain current and the source current are equal to each other.
Or we could use Ohm's Law to determine the current through these two resistors and
then solve for the voltage here by adding the drop across R2 to the voltage VSS.
So in other words we can solve for
this current here, I,
by Vdd minus Vss divided by R1+R2.
We take that current I, and
multiply by the resistor R2 to get the voltage drop across R2.
And then we add that drop to VSS to find VG, the gate voltage.
VGS is equal to
the square root of
ID over K plus VTO.
Then we can substitute this expression for
VGS that we obtained from knowing that the MOSFET is
operating in its saturation region into this loop equation
to give us a single equation in known voltages and ID.
So this results, after substitution
of this equation into this equation,
a quadratic equation,
in terms of the square root of ID.
And we can then solve that equation for ID.
Now because it's a quadratic equation there'll be two solutions.
But it will turn out that one of the two solutions places the MOSFET in a region
of operation that is not allowed for it acting as an amplifier.
In other words, one value of ID would place the transistor in
it's cut-off region, while the other value for ID would place the transistor in it's
saturation region, the region it must in be in for it to act as an amplifier.
I've also solved the quadratic equation for ID to get this expression.
And I kept only the solution,
the results in the transistor operating in its saturation region.
And to simplify this expressions somewhat I defined a voltage V1.
That's given by VG minus VSS minus VTO.
So this value is calculated, then placed into this expression,
along with the circuit element values and
MOSFET parameters to calculate the drain current through the MOSFET.
Now I've written two additional equations, that allow us to solve for
the node voltage at the drain, or the drain voltage.
The voltage at the drain of the MOSFET, VD,
is equal to the voltage here VDD, minus the voltage drop across RD.
And that voltage drop, by Ohm's law, would be equal to IDRD.
So the VDD, a voltage we know, minus a voltage drop across the resister,
gets us to the node voltage VD, as shown by this equation.
And then similarly the voltage at the source of the MOSFET, VS,
we can find it by starting at VSS, this known DC voltage, and
then going up by one voltage drop across RS.
So VSS plus this voltage drop gives us an expression for
the source voltage of the MOSFET.
Now, these equations taken together allow us to solve for
the operating point, the DC operating point of a common source amplifier.
So in summary, during this lesson we introduced the common source or
CS amplifier and
we performed a dc analysis to derive the dc design equations for this amplifier.
In the next lesson we will continue our look at the common source amplifier.
We'll perform an AC analysis.
So thank you and until next time.
Провода от принтера лежали. «Должно быть, я оставила беретту на диване», - подумала. Кровь, вытекающая из головы, в голубоватом свечении казалась черной. На полу возле тела Хейла лежал листок бумаги.