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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Georgia Institute of Technology

317 ratings

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,

Â we're going to look at the AC behavior of the common source amplifier.

Â In the previous lesson, we introduced the common source amplifier, and

Â we introduced DC biasing of this amplifier.

Â Our objectives for this lesson are to introduce AC behavior of the common source

Â amplifier, and to analyze a common source amplifier circuit.

Â Now, in the previous lesson, we examined the DC performance of the amplifier.

Â We solved four equations that allow us to determine the amplifier's

Â quiescent point, or q point.

Â Now, to determine the AC behavior of a common source amplifier,

Â we need to perform what's called a small signal analysis.

Â Now, this analysis is somewhat more complicated than would be appropriate for

Â intro to electronics course, so

Â I'm going to just sketch out the method, and then give you the results.

Â Now, remember, the transfer characteristic for a mosfet or a plot of the mosfet drain

Â current versus its gate to source voltage VGS is a nonlinear relationship.

Â In fact, it's a square law relationship

Â where ID is related to the square of the gain resource voltage.

Â Now, let's say that we have set the operating point of the mosfet such

Â that its cue point lies on the curve at this point.

Â We can draw a tangent line at that point.

Â That has a slope of delta ID, delta VGS at this point,

Â or in terms of partial derivatives a change in ID,

Â the change in VGS at that point.

Â And set if equal to what we call GM or the transconductance.

Â Now, if we assume that the relationship between ID and VGS,

Â instead of this square law relationship is actually this linear relationship,

Â we can derive a model for the transistor.

Â And we assume that in the original circuit the capacitors are all short circuits, and

Â the only inputs to the circuit are AC voltages.

Â So, with these assumptions and with this linearized model,

Â we can derive an expression for the Midband Gain, which I've shown here.

Â Now, two things to notice here.

Â One is that this gain is an inverting gain we have a minus sign here, so

Â we have an inversion between the input and the output and another thing to notice is,

Â the Midband Gain is dependent on RS which is dependent on GM

Â which is dependent on the Q point of the market, so this common source amplifier's

Â AC performance is dependent on it's Q-point or it's DC behavior.

Â So here, I've redrawn the circuit schematic for the common source amplifier.

Â And we're going to use this schematic to determine the reason for

Â the negative sign in the gain expression of the previous slide.

Â Now remember, for DC currents and voltages, the currents and

Â voltages that's at the operating point of the transistor, the capacitors have

Â impedances that are large enough that we can consider them to be open circuits.

Â So, any DC currents flowing through RD, must flow in to the drain of the mosfet,

Â it can't flow through C2 to the output because of this large impedance.

Â Now, if we apply an AC signal here, a time-bearing sinusoidal voltage, when

Â this circuit is operating in its mid-band region, at that frequency, the impedance

Â of these capacitors is small enough that we can consider them to be short circuits.

Â So, an AC voltage here is connected by a short circuit to the output.

Â But DC currents here, this C2 presents an open circuit.

Â Now, we know, because of the relationship between the gate to source voltage and

Â the drain current, that as this AC voltage increases,

Â the drain current will also increase.

Â As the drain current increases,

Â the drop across RD increases, making the drain voltage smaller.

Â Because remember,

Â VD is equal to VDD minus IDRD.

Â Wherein this case, this current is due to both the DC sources and the AC source.

Â As the gate voltage goes up, the drain current goes up.

Â As the drain current goes up, this drop gets bigger, and

Â the drain voltage actually gets smaller.

Â So, increase in VG results in a decrease in VD, hence,

Â the negative sign in the gain expression.

Â Now, another thing to notice on a circuit is this resistor R3.

Â R3 is connected to the rest of this circuit through

Â this DC blocking capacitor C3.

Â So, R3 has no affect on the DC bias or

Â quiescent point because no DC current passes through R3.

Â But if you remember the expression for

Â the gain on the previous slide, it has an R3 dependence.

Â So, R3 allows us to adjust the AC behavior or

Â the gain of this circuit without effecting the quiescent point.

Â Now, I want to work an example problem where we analyze a given common source

Â amplifier to determine it's Q point or DC bias and

Â it's gain, given these values.

Â So, the intrinsic parameters for the mosfet, K and VTO are given here.

Â We have these passive circuit component values and

Â we have these DC power supply values, +15 and -15.

Â The first step in performing our analysis is to determine the key point or

Â operating point of our transistor.

Â Now, we can use our expression that we derived previously for VG or

Â we can simply solve for VG by superposition of VSS and VDD.

Â So, we can write that VG is equal to VDD times R2 plus VSS times

Â R1 divided by R1 plus R2.

Â And for these circuit component values, plus 15, minus 15,

Â 200 k and one mega ohm, we get the VG is equal to minus ten volts.

Â So, the DC voltage at the gate of our transistor is minus ten volts.

Â Now, let's determine the drain current

Â of the transistor using the expression that we derived previously.

Â Remember, in our drain current expression,

Â we had a parameter called V1 = VG-VSS-VTO.

Â So, for this case, we have a gate voltage of -10-VSS,

Â which is -15 minus a threshold voltage of 1.5 volts to get 3.5 volts.

Â Then, we have the ID, remember we solved the quadratic equation for

Â ID, to get the expression, ID is equal to.

Â The square root of 1 plus

Â 4K V1, our parameter,

Â RS -1 divided by 2 square

Â root of K RS all squared.

Â And if we substitute in the values, we get a DC drain current of

Â 0.86 milliamps.

Â Now, let's solve for.

Â We currently know the gate voltage, and we know the drain current of the mosfet.

Â Let's solve for the voltage at the drain of the mosfet VD.

Â And the voltage at the source, VS.

Â And remember, we can find VDD, or VD by

Â starting with VDD and subtracting the voltage drop across RD.

Â So, we can write that VD is equal to VDD.

Â The DC power supply voltage minus the voltage

Â drop across RD is equal to 2.13 volts.

Â And we can find VS by starting with the negative power supply voltage and

Â adding to it the drop across RS.

Â So VS, the voltage at the source,

Â is equal to VSS plus IDRS is equal

Â to negative 12.43 volts.

Â So now, we know the voltage, DC voltage, at every terminal of the mosfet and

Â the current into the mosfet.

Â So, we have determined its key point or operating point.

Â And let's verify that this mosfet is operating in a saturation region,

Â so that it can be used as an amplifier.

Â So, to be in saturation, we know that the drain source

Â voltage which is VDS must be greater than VGS minus VTO.

Â So here, we have VDS =

Â 2.13 minus a minus

Â 12.43 is equal

Â to 14.56 volts.

Â Then, we can solve for VGS minus VTO.

Â We know VG, we know VS and we know VTO.

Â So, VGS is equal to VG minus

Â VS which implies that

Â the VGS minus VTO is equal

Â to 0.926 volts.

Â So, we can see that VDS is greater than VGS- VTO, so

Â we know that we're operating in the saturation region.

Â Now, that we've determined the DC bias or the Q point of the transistor,

Â we can solve for the gain of this common source amplifier.

Â Because the small signal parameters are dependent on the key point.

Â So, our small signal parameter GM,

Â the transconducter, is equal to 2 square root KID.

Â And we know that RS,

Â the intrinsic source resistance of the mosfet is equal to 1 over GM.

Â And the gain of the common source amplifier,

Â V out over V in, is equal to negative RD in parallel

Â with RL divided by RS plus RS in parallel with R3,

Â so RS is dependent on GM and

Â GM is dependent on the DC bias of the transistor,

Â if we substitute in the values here,

Â we get that the gain of this common source amplifier is equal

Â to negative 14.53, a unitless quantity or

Â you can write it with units of volts per volt to

Â indicate an input voltage and an output voltage.

Â So, in summary, during this lesson you were introduced to the AC analysis of

Â a common source amplifier.

Â We also looked at an example,

Â where we analyzed a common source amplifier circuit.

Â Now, you should go back, and work through that example and

Â see if you can arrive at the same results that I did.

Â In our next lesson, we are going to look at a new type of transistor

Â known as a Bipolar Junction Transistor.

Â So, thank you and until next time.

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