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.

Na lição

Bipolar Junction Transistors

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

In this lesson, we're going to continue our look at the BJT Common Emitter Amplifier,

but in this case, we're gonna look at its AC behavior rather than its DC behavior.

In our previous lesson, we introduced

the common emitter amplifier and we examined the DC biasing of the amplifier.

Our objectives for this lesson are to

examine the AC behavior of the Common Emitter Amplifier.

So here is a circuit schematic of the circuit we've been examining.

And remember last time,

we looked at how changing the external voltages and resistances in

this circuit can change the operating point or the behavior of this BJT.

So, as we change these resistors and change these voltages,

we can move along to transfer a characteristic curve,

picking different operating points.

So for example, if we chose our resistors and voltages so they were biased,

the transistor is biased at this point,

then we're operating in the cutoff region.

Or we can choose them so they were operating in this region,

either the active or saturation region.

Now, to see how this circuit behaves as an amplifier,

I want to first look at a simple model for

an amplifier and then we'll compare this model to an actual Common Emitter Amplifier.

Here I have what's called a transconductance amplifier.

The box represents the amplifier.

We apply an input voltage here to

the input side and then from this controlled current source,

a current is generated that's dependent on the input voltage.

You can see that the current is equal to gm times vin,

where gm is a parameter of the amplifier – it's called its transconductance.

Then that current, if we add an external load resistor to the amplifier – I'm calling it

RL – as that current flows due to the application of this input voltage,

we can generate an output voltage that,

by Ohm's law, is equal to the current times the resistance.

So, we can write that the output voltage Vout is equal to -i,

the current, times RL – the minus sign because

the current is flowing from minus to plus across RL rather than from plus to minus.

And then we know, from the controlled source,

that this current must be dependent on

the input voltage-- i is equal to gm times the input voltage (i = gm Vin).

Then we can substitute to get the overall gain equation

for this idealized model amplifier.

We can write the Vout is equal to minus gm Vin times RL (Vout = -gm Vin RL),

or the gain, the ratio of the output voltage to the input voltage,

is equal to minus

gm RL (Vout/Vin = -gm RL) So,

we have a gain that's dependent on this transistor parameter, this transconductance.

As we increase the transconductance,

we increase the gain of this idealized amplifier.

Now let's look at the I-V characteristics for that ideal amplifier.

And remember again, the current, the output current,

for that amplifier is related to the input voltage by the transconductance parameter gm.

And you can see that this equation has the form of y equals mx plus b,

that of a a straight line,

where gm is in the position of the slope.

I can solve this equation for gm.

You would see that gm is the ratio of current voltage.

On the I-V characteristic curve,

this is the current axis and this is the voltage axis.

So the slope here would be the rise over

the run or the ratio of current to voltage, which is equal to gm.

So, I have here the I-V characteristics for two separate ideal amplifiers,

one with a low gm – this would be a low gm – and one with a higher gm.

As I increase gm, I increase the slope here.

Now let's look at what happens when we apply an input voltage to the amplifier.

So, say I was applying

a sinusoidally varying voltage that varied between plus one and minus one volts.

So, the input voltage would be varying between plus one and minus one,

and the time it takes to move from plus

one to minus one would depend on the frequency of the sine wave.

So, faster frequency means I'm moving back and forth here more quickly.

So say we are applying that input voltage between

plus one and minus one to the low-gm amplifier.

Let me draw the limits here, or attempt to.

As we move back and forth in this voltage range,

we can determine what output current is generated by this particular amplifier by drawing

horizontal lines like this.

So as the input voltage varies between plus one and minus one,

we get an output current that varies between these two levels for the low-gm amplifier.

But if we use the higher-gm amplifier,

for the same input voltage,

we get a current range that varies from approximately here to approximately here,

a much larger range in output current.

And if we get a larger swing in current for the same input voltage,

that amplifier has a higher gain – it would produce

a higher output voltage for the same input voltage.

Now another thing to notice on these characteristic curves for this ideal amplifier is,

I can apply an input voltage anywhere and still have this output swing in current.

If I apply an input voltage between these two points,

I still get an output swing in current.

Same for here, same for any range.

Now let's look at how you would implement a real one of these amplifiers,

a real transconductance amplifier.

And the way you do it is build a Common Emitter Amplifier.

So here is the ideal amplifier we've been discussing,

and here is a real,

practical implementation of this amplifier,

where I've replaced the internal structure of the amplifier

by a BJT configured in this way.

Now, you can see the input voltage is

applied between the base and the emitter of the BJT,

and the output current of the amplifier is the collector current that

is produced due to this BE voltage.

Now you can compare that to the idealized amplifier.

Here, we apply a voltage to the input and we have

a controlled source that generates that current, the output current.

Now we know that in this ideal amplifier,

the relationship between input voltage and output current is linear.

We know that in this configuration, when we change vBE,

we can change the collector current,

but we know that it is very much not linear.

So let's look at the relationships between the I-V characteristics

of the ideal amp and those of the Common Emitter Amplifier.

Now here are the I-V characteristics for

both the ideal transconductance amplifier

here and for the amplifier using a BJT or the Common Emitter Amplifier.

Now remember, here, I can apply an input voltage anywhere and

get a swing in output current which could result to an amplification.

But for the common emitter,

the practical implementation of the transconductance amp,

that is obviously not true,

and this is where the biasing point or

the operating point of the transistor is so important.

For example, if we had this transistor biased at

this operating point and we were applying

an input voltage that swung around that operating point here,

we can see that no matter what the input voltage is,

the output current is equal to zero,

and it cannot possibly act as an amplifier.

But if instead, we use the external circuitry to bias the BJT in this region,

in its active region.

Then, as I swing an input voltage back and forth in this region,

we can obviously get changes in current and then we can use this BJT as an amplifier.

Now, one thing that's apparent,

even if we are operating in this region,

the equation for this curve is not a linear equation – I and

V are not related in a linear way as they are for the transconductance amp.

But, if we pretend that or we act as though

that – we're operating in a small range here of the input voltages.

If we're operating only in this small range,

then we can approximate this curve as though it were actually

linear and then model this transistor amplifier that is not linear as though it were.

So, say our operating point were right here,

then we could draw a tangent line at this curve,

of this curve at this point and act

as though the transistor is actually following

this linear characteristic that's tangent at the operating point.

And then the slope of this line,

this I-V characteristic, would be

the transconductance of the amplifier operating at that point.

And we could find the slope of this line by taking the derivative

of the I-V characteristic at that point.

So gm, the slope of the I-V characteristic,

would be equal to the partial derivative of Ic with respect to vBE.

Now you can show that this derivative is equal to this equation here: gm is equal to Ic,

the DC collector current,

divided by VT, the thermal voltage.

And remember, VT is typically assumed to be 0.0259 volts.

If this were the operating point,

then IC we could find by reading across here,

IC, the DC collector current.

So what we've done here is we have linearized the transistor,

we have modeled it as though it behaves linearly,

and its I-V characteristic is a line that is tangent to the operating point.

Then when we do that, we can treat the real implementation,

the Common Emitter Amplifier,

in the same way that we do the idealized implementation, this transconductance amplifier.

Now, let's review the circuit.

Here is, again, the circuit we've been looking at.

And you can see buried in this circuit is the simpler version

of the common emitter amplifier that we've been discussing in this lesson, right here.

The input is applied through this coupling capacitor to the base of the transistor.

Here's the load resistor connected to the collector of the transistor.

And then we have the external circuitry necessary to bias

the transistor in its on region or it's active region.

And the external circuitry also controls things about the amplification,

like the lower cutoff frequency.

When you include the effects of the external circuitry,

you get a gain equation that looks like this,

and we can compare that to the gain equation we got for

the simpler transconductance amplifier, here.

You can see that they're both dependent on gm,

though it's a little obscured here.

This parameter re is dependent on gm,

the transconductance parameter of the BJT.

beta is the base-to-collector current gain for the transistor,

and g_m again is the ratio of IC to VT.

So, by adding the external circuitry that is

necessary to bias the transistor in its active region,

we complicate the gain equation,

and you can show that it would be equal to this expression.

Now what I want to do is use the parameters for

this circuit that we've been analyzing and

calculate the gain and then we will simulate

the circuit and see if we get the same result.

So if you remember, we found previously that IC =

2.475 milliamps and this transistor had a β = 99.

So if you know this and you know the component values,

you can plug into these equations and solve for the gain.

You find that the gain Vout over

Vin is equal to -5.45 (Vout/vin = -5.45),

the negative sign indicating that there's an inversion.

When the input goes up, the output goes down.

So you should verify that you get the same answer here.

When I worked my way through this,

I had a gm = 95.6 milliamps.

So verify these results.

So on this slide, I'm showing you the result of a circuit simulation.

I am showing you a plot of the magnitude of

the gain versus frequency for the common emitter that we've been examining.

Now you can see that, overall,

it looks like a high-pass filter.

If we wanted to consider this circuit to be an amplifier,

we would operate or apply frequencies in it's midband region here.

So, I've placed a cursor here in the midband region,

and we can see that at about 652 hertz,

we have a gain of 5.33,

which corresponds well with the 5.45 that we estimated from our equation.

Now, it's a high-pass filter – that means

we have a lower minus three dB cutoff frequency.

We can find the magnitude at which we're down three dB by

multiplying the gain here in the midband,

which is 5.33, we divide that by the square root of two,

and we get a magnitude value of 3.77.

So we locate 3.77 here,

and that frequency would be our minus three dB frequency.

And I've done that here – at about 3.77,

we have a lower cutoff frequency of 38 hertz.

So, we're calling this an amplifier,

but, in reality, every circuit acts as a filter of some sort.

If you want it to behave as the amplifier that you've designed,

the one with a gain of 5.45 or so,

you operate it in its passband region,

but you need to be aware of the frequency response.

If, for example, you apply frequencies to

this amplifier below this lower cutoff frequency,

it won't act as an amplifier anymore

or it won't act as an amplifier with the gain you intended it to have.

And in fact, let's see,

we can locate one here – here's one,

the point where the gain is one.

If we're operating at frequencies below this frequency,

about 10 hertz, then the amplifier actually

acts as an attenuator – it has a gain less than one.

Above here, it has a gain greater than one,

but it's not until we get to the passband that we have the designed midband gain of 5.45.

So, in summary, during this lesson,

we examined the AC behavior of the Common Emitter Amplifier.

We saw that by linearized the common emitter amplifier,

we could treat it as an idealized transconductance amplifier.

So, thank you, and until next time.

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