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

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,

Â this is Dr. Robinson.

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

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

Â In our previous lesson, we introduced

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

Â Our objectives for this lesson are to

Â examine the AC behavior of the Common Emitter Amplifier.

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

Â And remember last time,

Â we looked at how changing the external voltages and resistances in

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

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

Â we can move along to transfer a characteristic curve,

Â picking different operating points.

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

Â the transistor is biased at this point,

Â then we're operating in the cutoff region.

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

Â either the active or saturation region.

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

Â I want to first look at a simple model for

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

Â Here I have what's called a transconductance amplifier.

Â The box represents the amplifier.

Â We apply an input voltage here to

Â the input side and then from this controlled current source,

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

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

Â where gm is a parameter of the amplifier â€“ it's called its transconductance.

Â Then that current, if we add an external load resistor to the amplifier â€“ I'm calling it

Â RL â€“ as that current flows due to the application of this input voltage,

Â we can generate an output voltage that,

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

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

Â the current, times RL â€“ the minus sign because

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

Â And then we know, from the controlled source,

Â that this current must be dependent on

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

Â Then we can substitute to get the overall gain equation

Â for this idealized model amplifier.

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

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

Â is equal to minus

Â gm RL (Vout/Vin = -gm RL) So,

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

Â As we increase the transconductance,

Â we increase the gain of this idealized amplifier.

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

Â And remember again, the current, the output current,

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

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

Â that of a a straight line,

Â where gm is in the position of the slope.

Â I can solve this equation for gm.

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

Â On the I-V characteristic curve,

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

Â So the slope here would be the rise over

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

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

Â one with a low gm â€“ this would be a low gm â€“ and one with a higher gm.

Â As I increase gm, I increase the slope here.

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

Â So, say I was applying

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

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

Â and the time it takes to move from plus

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

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

Â So say we are applying that input voltage between

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

Â Let me draw the limits here, or attempt to.

Â As we move back and forth in this voltage range,

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

Â horizontal lines like this.

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

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

Â But if we use the higher-gm amplifier,

Â for the same input voltage,

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

Â a much larger range in output current.

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

Â that amplifier has a higher gain â€“ it would produce

Â a higher output voltage for the same input voltage.

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

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

Â If I apply an input voltage between these two points,

Â I still get an output swing in current.

Â Same for here, same for any range.

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

Â a real transconductance amplifier.

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

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

Â and here is a real,

Â practical implementation of this amplifier,

Â where I've replaced the internal structure of the amplifier

Â by a BJT configured in this way.

Â Now, you can see the input voltage is

Â applied between the base and the emitter of the BJT,

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

Â is produced due to this BE voltage.

Â Now you can compare that to the idealized amplifier.

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

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

Â Now we know that in this ideal amplifier,

Â the relationship between input voltage and output current is linear.

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

Â we can change the collector current,

Â but we know that it is very much not linear.

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

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

Â Now here are the I-V characteristics for

Â both the ideal transconductance amplifier

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

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

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

Â But for the common emitter,

Â the practical implementation of the transconductance amp,

Â that is obviously not true,

Â and this is where the biasing point or

Â the operating point of the transistor is so important.

Â For example, if we had this transistor biased at

Â this operating point and we were applying

Â an input voltage that swung around that operating point here,

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

Â the output current is equal to zero,

Â and it cannot possibly act as an amplifier.

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

Â in its active region.

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

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

Â Now, one thing that's apparent,

Â even if we are operating in this region,

Â the equation for this curve is not a linear equation â€“ I and

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

Â But, if we pretend that or we act as though

Â that â€“ we're operating in a small range here of the input voltages.

Â If we're operating only in this small range,

Â then we can approximate this curve as though it were actually

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

Â So, say our operating point were right here,

Â then we could draw a tangent line at this curve,

Â of this curve at this point and act

Â as though the transistor is actually following

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

Â And then the slope of this line,

Â this I-V characteristic, would be

Â the transconductance of the amplifier operating at that point.

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

Â of the I-V characteristic at that point.

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

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

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

Â the DC collector current,

Â divided by VT, the thermal voltage.

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

Â If this were the operating point,

Â then IC we could find by reading across here,

Â IC, the DC collector current.

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

Â we have modeled it as though it behaves linearly,

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

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

Â the Common Emitter Amplifier,

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

Â Now, let's review the circuit.

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

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

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

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

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

Â And then we have the external circuitry necessary to bias

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

Â And the external circuitry also controls things about the amplification,

Â like the lower cutoff frequency.

Â When you include the effects of the external circuitry,

Â you get a gain equation that looks like this,

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

Â the simpler transconductance amplifier, here.

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

Â though it's a little obscured here.

Â This parameter re is dependent on gm,

Â the transconductance parameter of the BJT.

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

Â and g_m again is the ratio of IC to VT.

Â So, by adding the external circuitry that is

Â necessary to bias the transistor in its active region,

Â we complicate the gain equation,

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

Â Now what I want to do is use the parameters for

Â this circuit that we've been analyzing and

Â calculate the gain and then we will simulate

Â the circuit and see if we get the same result.

Â So if you remember, we found previously that IC =

Â 2.475 milliamps and this transistor had a Î² = 99.

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

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

Â You find that the gain Vout over

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

Â the negative sign indicating that there's an inversion.

Â When the input goes up, the output goes down.

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

Â When I worked my way through this,

Â I had a gm = 95.6 milliamps.

Â So verify these results.

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

Â I am showing you a plot of the magnitude of

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

Â Now you can see that, overall,

Â it looks like a high-pass filter.

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

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

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

Â and we can see that at about 652 hertz,

Â we have a gain of 5.33,

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

Â Now, it's a high-pass filter â€“ that means

Â we have a lower minus three dB cutoff frequency.

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

Â multiplying the gain here in the midband,

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

Â and we get a magnitude value of 3.77.

Â So we locate 3.77 here,

Â and that frequency would be our minus three dB frequency.

Â And I've done that here â€“ at about 3.77,

Â we have a lower cutoff frequency of 38 hertz.

Â So, we're calling this an amplifier,

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

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

Â the one with a gain of 5.45 or so,

Â you operate it in its passband region,

Â but you need to be aware of the frequency response.

Â If, for example, you apply frequencies to

Â this amplifier below this lower cutoff frequency,

Â it won't act as an amplifier anymore

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

Â And in fact, let's see,

Â we can locate one here â€“ here's one,

Â the point where the gain is one.

Â If we're operating at frequencies below this frequency,

Â about 10 hertz, then the amplifier actually

Â acts as an attenuator â€“ it has a gain less than one.

Â Above here, it has a gain greater than one,

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

Â So, in summary, during this lesson,

Â we examined the AC behavior of the Common Emitter Amplifier.

Â We saw that by linearized the common emitter amplifier,

Â we could treat it as an idealized transconductance amplifier.

Â So, thank you, and until next time.

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