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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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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Dec 09, 2016

Very nice course to startup with Analog Circuits and Electronics. Loved the way it was presented. The problems provided were also perfect for checking of understanding.

Mar 25, 2019

This course is very useful to learn basics of electronics. Assignments and quiz examinations are some tricky , so it increases our analytical skills.

De la lección

MOSFETs

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

#### Dr. Bonnie H. Ferri

Professor#### Dr. Robert Allen Robinson, Jr.

Academic Professional

Welcome back to Electronics, this is Dr. Robinson.

In this lesson,

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

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

we introduced DC biasing of this amplifier.

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

amplifier, and to analyze a common source amplifier circuit.

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

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

quiescent point, or q point.

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

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

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

intro to electronics course, so

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

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

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

In fact, it's a square law relationship

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

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

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

We can draw a tangent line at that point.

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

or in terms of partial derivatives a change in ID,

the change in VGS at that point.

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

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

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

we can derive a model for the transistor.

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

the only inputs to the circuit are AC voltages.

So, with these assumptions and with this linearized model,

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

Now, two things to notice here.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

the drain current, that as this AC voltage increases,

the drain current will also increase.

As the drain current increases,

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

Because remember,

VD is equal to VDD minus IDRD.

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

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

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

the drain voltage actually gets smaller.

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

the negative sign in the gain expression.

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

R3 is connected to the rest of this circuit through

this DC blocking capacitor C3.

So, R3 has no affect on the DC bias or

quiescent point because no DC current passes through R3.

But if you remember the expression for

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

So, R3 allows us to adjust the AC behavior or

the gain of this circuit without effecting the quiescent point.

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

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

it's gain, given these values.

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

We have these passive circuit component values and

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

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

operating point of our transistor.

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

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

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

R1 divided by R1 plus R2.

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

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

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

Now, let's determine the drain current

of the transistor using the expression that we derived previously.

Remember, in our drain current expression,

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

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

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

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

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

The square root of 1 plus

4K V1, our parameter,

RS -1 divided by 2 square

root of K RS all squared.

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

0.86 milliamps.

Now, let's solve for.

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

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

And the voltage at the source, VS.

And remember, we can find VDD, or VD by

starting with VDD and subtracting the voltage drop across RD.

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

The DC power supply voltage minus the voltage

drop across RD is equal to 2.13 volts.

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

adding to it the drop across RS.

So VS, the voltage at the source,

is equal to VSS plus IDRS is equal

to negative 12.43 volts.

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

the current into the mosfet.

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

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

so that it can be used as an amplifier.

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

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

So here, we have VDS =

2.13 minus a minus

12.43 is equal

to 14.56 volts.

Then, we can solve for VGS minus VTO.

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

So, VGS is equal to VG minus

VS which implies that

the VGS minus VTO is equal

to 0.926 volts.

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

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

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

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

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

So, our small signal parameter GM,

the transconducter, is equal to 2 square root KID.

And we know that RS,

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

And the gain of the common source amplifier,

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

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

so RS is dependent on GM and

GM is dependent on the DC bias of the transistor,

if we substitute in the values here,

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

to negative 14.53, a unitless quantity or

you can write it with units of volts per volt to

indicate an input voltage and an output voltage.

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

a common source amplifier.

We also looked at an example,

where we analyzed a common source amplifier circuit.

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

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

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

known as a Bipolar Junction Transistor.

So, thank you and until next time.

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