This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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From the course by Georgia Institute of Technology

Introduction to Electronics

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

From the lesson

MOSFETs

Learning Objectives: 1. Develop an understanding of the MOSFET and its applications. 2. Develop an ability to analyze MOSFET circuits.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to Electronics, this is Dr. Robinson.

Â In this lesson, we're going to look at the common source amplifier, and

Â in particular we're going to look at the DC Analysis of this amplifier.

Â In your previous lesson, we examined MOSFET characteristic curves and biasing

Â and our objectives for this lesson are to introduce the common source amplifier and

Â to analyze the common source amplifier at dc.

Â Now you can think of this lesson as a practical application for

Â a MOSFET transistor.

Â What we want to do is look at a circuit that can be used to implement a gain stage

Â or a gain block in a system.

Â So at the system level, we can consider a gain block to be, let me draw the block.

Â And label with its gain A.

Â So to the circuit we apply an input voltage, Vin.

Â And we have an output voltage from this block, Vout.

Â Which is equal to the gain A times the input voltage.

Â Now, to actually implement a gain block that has this functionality,

Â we have to build a circuit.

Â And you know one way of doing this.

Â So, for example, we could use a non-inverting op-amp amplifier or

Â an inverting op-amp amplifier to implement this functionality.

Â We designed the circuit so that they have a gain A.

Â But, in this lesson, what we're going to do is look at another way to do this.

Â We're going to look at how you can use a discreet MOSFET transistor to

Â implement this gain block.

Â Now, there are numerous single MOSFET circuits that can implement

Â this gain functionality, but during this lesson we're going to look at

Â one particular type called the common source amplifier or CS amplifier.

Â Now as we go through this lesson we should keep in mind the relationship between

Â a MOSFET's biasing, or its quiescent point, and its behavior.

Â Remember in the previous lesson, we looked at how by changing the external

Â voltages and currents of the MOSFET, we can change its behavior.

Â And in particular, we can change its region of operation.

Â Now you may also remember that for MOSFET to operate as an amplifier,

Â we must bias the transistor in its saturation region.

Â Keep in mind as we go through the lesson that we can change the characteristics of

Â this amplifier by changing externally the DC bias.

Â In the following we're first going to introduce the common source amplifier

Â circuit, and then we're going to spend some time looking in particular at

Â the design equations necessary to buy us the transistor in its saturation region.

Â So, let's take a look at the circuit.

Â Now, in the circuit, VDD is a DC voltage,

Â a positive DC voltage, and VSS is a negative DC voltage.

Â These voltages, in combination with R1, R2, RD and

Â RS set the cue point of the MOSFET.

Â Now these components along with R3 and the capacitor values, control

Â the AC performance of this amplifier or effect the gain of this circuit.

Â Now we can treat this common source amplifier as a game block of the form,

Â Some block that has a gain A, we apply an input that has a AC voltage to

Â the block and then the output voltage, Is found

Â from the input voltage by multiplying the input voltage by the gain A.

Â A times VN.

Â But before we can treat this circuit as this block,

Â we must know the operating point of the transistor.

Â Because its operating point affects its gain.

Â Now remember that a capacitor's impedance Z sub c can be

Â written as 1 over J Omega C.

Â So at very low frequencies at DC voltages and DC currents

Â the voltages and currents that set the operating point of the MOSFET.

Â The impedance of the capacitor is very large.

Â So as omega goes to zero, in other words the quantities go to DC quantities,

Â this impedance does an open circuit.

Â So, in determining the Q point for this transistor,

Â which determines this circuit's gain characteristics, we can

Â treat all of the capacitors in the circuit as open circuits or large impedance.

Â So here I've redrawn the previous circuit with all of the capacitors made open

Â circuits.

Â So we can analyze the DC performance of the common source amplifier or

Â determine it's cue point or quiescent point.

Â Now because of the structure of our MOSFET,

Â we know that the gate current into the MOSFET is zero.

Â IG, the gate current, is equal to zero and we also know that ID,

Â the drain current, is equal to IS, the source current.

Â Or in other words, the current going into the drain all comes out of the source, so

Â that both the drain current and the source current are equal to each other.

Â Now what we want to determine is the Q point of this transistor.

Â In other words, what is it's drain current and

Â what are the drain voltage, the gate voltage,

Â and the source voltage for this transistor.

Â Now we can write an equation for the gate voltage in terms of R1, R2, VDD and VSS.

Â We know there's no current into the gate here, so we could,

Â one way of doing this would be to use superposition between VDD and

Â VSS to solve for the voltage here at the gate.

Â Or we could use Ohm's Law to determine the current through these two resistors and

Â then solve for the voltage here by adding the drop across R2 to the voltage VSS.

Â So in other words we can solve for

Â this current here, I,

Â by Vdd minus Vss divided by R1+R2.

Â We take that current I, and

Â multiply by the resistor R2 to get the voltage drop across R2.

Â And then we add that drop to VSS to find VG, the gate voltage.

Â So we can say that VG is equal to

Â VDD R2+VSS R1

Â divided by R1 + R2.

Â So here we have an equation that relates or

Â lets us calculate the gate voltage in terms of the passive component values and

Â the DC power supply values.

Â Now we can also write a loop equation from the gate voltage to the voltage VSS.

Â So what I'm going to do is start at this known voltage VG,

Â work my way around this loop to this known voltage VSS,

Â which is going to give us an equation in the gate to source voltage, VGS.

Â So let me write, VG, the voltage here, minus the gate to source voltage,

Â the voltage from the source to the gate, minus VGS,

Â would get us to here, then minus the drop across RS.

Â Must be equal to the voltage at this node VSS.

Â Now we know that ID = IS.

Â And at this transistor is operating as an amplifier,

Â we know it's operating in its saturation region,

Â where ID = K(VGS- VTO) squared.

Â Or, we can solve this equation for VGS.

Â VGS is equal to

Â the square root of

Â ID over K plus VTO.

Â Then we can substitute this expression for

Â VGS that we obtained from knowing that the MOSFET is

Â operating in its saturation region into this loop equation

Â to give us a single equation in known voltages and ID.

Â So this results, after substitution

Â of this equation into this equation,

Â a quadratic equation,

Â in terms of the square root of ID.

Â And we can then solve that equation for ID.

Â Now because it's a quadratic equation there'll be two solutions.

Â But it will turn out that one of the two solutions places the MOSFET in a region

Â of operation that is not allowed for it acting as an amplifier.

Â In other words, one value of ID would place the transistor in

Â it's cut-off region, while the other value for ID would place the transistor in it's

Â saturation region, the region it must in be in for it to act as an amplifier.

Â Here I've rewritten two of the equations we obtained on the previous slide.

Â The equation for the gate voltage, the DC gate voltage in terms of the DC

Â voltage sources and the circuit element values.

Â And this equation for

Â VGS, that results from the transistor operating in its saturation region.

Â I've also solved the quadratic equation for ID to get this expression.

Â And I kept only the solution,

Â the results in the transistor operating in its saturation region.

Â And to simplify this expressions somewhat I defined a voltage V1.

Â That's given by VG minus VSS minus VTO.

Â So this value is calculated, then placed into this expression,

Â along with the circuit element values and

Â MOSFET parameters to calculate the drain current through the MOSFET.

Â Now I've written two additional equations, that allow us to solve for

Â the node voltage at the drain, or the drain voltage.

Â From the equation, you can see how it's obtained.

Â The voltage at the drain of the MOSFET, VD,

Â is equal to the voltage here VDD, minus the voltage drop across RD.

Â And that voltage drop, by Ohm's law, would be equal to IDRD.

Â So the VDD, a voltage we know, minus a voltage drop across the resister,

Â gets us to the node voltage VD, as shown by this equation.

Â And then similarly the voltage at the source of the MOSFET, VS,

Â we can find it by starting at VSS, this known DC voltage, and

Â then going up by one voltage drop across RS.

Â So VSS plus this voltage drop gives us an expression for

Â the source voltage of the MOSFET.

Â Now, these equations taken together allow us to solve for

Â the operating point, the DC operating point of a common source amplifier.

Â So in summary, during this lesson we introduced the common source or

Â CS amplifier and

Â we performed a dc analysis to derive the dc design equations for this amplifier.

Â In the next lesson we will continue our look at the common source amplifier.

Â We'll perform an AC analysis.

Â So thank you and until next time.

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