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

325 ratings

Georgia Institute of Technology

325 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

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.

Â In this lesson we are going to look at a bipolar junction transistor used as

Â an electronic switch.

Â In the previous lesson, we examined the parameters of a BJT.

Â And our objectives for this lesson are to introduce the BJT switch.

Â Let's look at a schematic that I'm going to use to illustrate how we can use

Â the BJT as a switch.

Â In the circuit Vin acts as a control voltage to determine whether the BJT is

Â off or on.

Â When the BJT is on, current flows around this loop from collector to emitter.

Â When the BJT is off no current flows through this loop.

Â In this loop, the loop through Vin, through RB,

Â and across the base emitter junction, this Pin junction, we can analyze the behavior

Â here using techniques we learned about in our previous diode lectures.

Â Remember, if we forward bias the Pin junction we make Vin greater than

Â about 0.65 or 0.7 volts, then we know that current will flow

Â into the base of the transistor, around this loop.

Â And remember for a BJT in it's active region, IC is equal to beta IB.

Â So if we've made Vin large enough to forward bias this junction to cause

Â base current to flow, then we know that collector current will flow.

Â And we have turned the transistor on, so current is now flowing around this loop.

Â And we know this behavior to be true from our examination previously of

Â the EJT transfer characteristics.

Â If Vin is small, then VB is small, the transistor is off, and

Â no collector co, current flows.

Â IC is equal to zero.

Â But if we make Vin large enough, then IC is non-zero, and the transistor turns on.

Â Now let's look at how the voltage here at the collector,

Â VCE varies with changing, changes in Vin.

Â When Vin is small, we know that no base current flows, so

Â no collector current flows.

Â The transistor is in its cutoff region, and the voltage drop across RC,

Â given by ICRC is equal to 0.

Â So when Vin is equal to 0, the voltage here is equal to the voltage here,

Â is equal 12 volts.

Â But as we increase the Vin and we increase th Ic the voltage drop across this

Â resistor increases, causing the value of VCE to decrease.

Â So let's plot this behavior on the outlet characteristic curves of the BJT.

Â When Vin is equal to 0, we know the transistor is in it's cut off region,

Â along this red line, and the voltage VCE is equal to 12 volts, approximately here.

Â Then as I increase Vin which increases IB will move out of the cut off region

Â in this direction, because remember IB is increasing in this direction.

Â So each one of these curves represents a different, different IB and

Â it increases in that direction.

Â So, as Vin increases, IB increases.

Â But, as IB increases, IC inc, increases, which causes VCE to decrease.

Â So, VCE we're moving in this direction.

Â So, as Vin increases, we move from cutoff, and

Â we move along a line in this direction.

Â And you can imagine that as I continue to increase Vin,

Â we move from the cut-off region, this red line, through the active region,

Â eventually reaching the saturation region, where the transistor is maximally on.

Â The collector current is at its maximum.

Â So when operating the BJT as a switch we typically want to vary the region of

Â operation from cut off to saturation without ever entering the active region.

Â So the control voltage here typically has two possibilities.

Â A say, a small voltage of 0 volts to put the BJT in the cut off region,

Â where no current flows through this loop.

Â And say large voltage of, for example, 5 volts that puts

Â the transistor into the saturation region where the transistor is maximally on and

Â the current is at, at it's maximum.

Â So in this circuit we can control.

Â The behavior of this portion of the circuit by

Â varying the voltage in this portion of the circuit.

Â Now let's look at how IC varies specifically with Vin in the circuit.

Â When Vin is small we know that the junction here is not forward biased and

Â the transistor is off.

Â We're in this region here.

Â So for small Vin no collector current flows.

Â Then as the input voltage becomes larger, collector current begins to flow.

Â And you can see the exponential relationship that we

Â would expect here between Vin and IC.

Â Then if we continue to increase Vin the transistor enters saturation.

Â Where we can model the transistor as approximately a voltage source

Â of 0.2 volts.

Â And then once we're in the saturation region further increases in Vin do

Â not increase the collector current.

Â Now we can calculate the level of this collector current here knowing that

Â the voltage from collector to emitter of the transistor when it's in saturation is

Â about 0.2 volts.

Â So I can write here, that VCC, minus VCE,

Â divided by the resistor, RC, is equal to IC.

Â When the transistors in saturation we know that this is about .3 volts.

Â So I can write that 12 minus 0.2 volts divided by 1K is equal to IC,

Â is equal to about, or is it exactly 11.8 milliamps.

Â And you can see that on the scale,

Â that level of IC is approximately 11.8 milliamps.

Â So here I'm showing a plot of how VCE, the collector emitter voltage for

Â the BJT, varies with changes in Vin.

Â When Vin is small, we know the junction here is not forward biased.

Â So the transistor is in its cutoff region.

Â There's no drop across the resistor RC, so

Â the voltage here, is equal to the voltage here, is equal to 12 volts.

Â But as Vin increases, the transistor turns on.

Â Until we put the transistor in its saturation region of operation.

Â And in this region, we know that the voltage across the transistor is equal to

Â VCE SAT, is equal to approximately 0.2 volts.

Â So here, we can label this as VCE SAT.

Â Or saturation.

Â Is approximately equal to 0.2 volts.

Â Now you can see that it only takes a small change in Vin to move

Â the transistor from it's cut off region here to it's saturation region here.

Â And another thing to notice about the voltage between collector and

Â emitter, is that it has an inverting characteristic.

Â When Vin is small, the voltage output, or the voltage at the collector, is large.

Â But when Vin is large, the collector voltage is small.

Â So, if we thought of this as a circuit where the output is

Â taken at the collector and the input is here then we can consider this circuit to

Â be an inverting circuit or a BJT inverter.

Â So I wanted to show you an example circuit where we use an LED load on the switch and

Â we're going to determine the proper resistor values to have this func,

Â this circuit function correctly.

Â So, our input voltage, you could consider this to be an output from say,

Â a microcontroller, where the input to the microcontroller is a temperature sensor.

Â So say, the temperature is greater than some threshold.

Â Then we'd have the microcontroller put out a high voltage of 5 volts,

Â which would cause this BJT to turn on.

Â Which would cause current to flow in this loop, and have the LED illuminate.

Â The light emitting diode.

Â But if the temperature is less than the threshold determined by our sensor and

Â micro-controller combination, Vin would be equal to 0 volts,

Â the BJT would be off and the LED would not be lit.

Â Now, from the LED data sheet we can find that for

Â a diode current of about 20 milliamps.

Â We have good elimination, or good light output.

Â And we can also see from the data sheet that when the diode is on,

Â that the forward voltage drop across the diode VD is equal to about 1.8 volts.

Â So what I want to do is pick RC such that when the BJT is on or when the switch is

Â in its fully on position, that the current through the diode is 20 milliamps.

Â We know when the BJT is on, that the voltage here is equal to 0.2 volts.

Â So we can write that VCC, the voltage here,

Â minus a diadrop minus the drop across RC,

Â which would be ICRC must be equal to VCE.

Â So to solve this for RC we can write

Â VCC minus VD minus VCE over IC is equal to RC.

Â When the BJT is on we know that VCE is equal to VCE SAT of point 2 volts.

Â So I can write 12 minus 1.8 minus 0.2 divided

Â by the collector current that we want, 20 miliamps.

Â Will give us the value of collector resistor that we need.

Â It turns out to be 500 ohms.

Â So, this resistor value RC of 500 ohms limits the current to

Â 20 milliamps when the BGT is fully on.

Â Now let's look at how we would calculate a value for RB.

Â RB is here to limit the base current into the transistor.

Â And we know that when the transistor is on, this junction is for biased,

Â so the voltage here when the transistor is on, is about 0.7 volts So,

Â I can write here on the base circuit

Â that Vin minus 0.7 volts

Â divided by RB is equal to IB.

Â Now remember, in the active region, IC is equal to beta IB.

Â Now, IB is equal to IC over beta, and

Â if we assume the minimum value of beta for this transistor,

Â is 50, then we know at the verge between the active region and

Â the cut-off region that IB is equal to 20 milliamps.

Â Divided by 50, is equal to 0.4 milliamps.

Â So we can use this value of IB to determine a value of RB here.

Â But typically, when you operate a BJT as a switch, you want to

Â overdrive this transistor to take into account any changes in the load here.

Â So in other words, we put far more base current into the transistor than is

Â required to put the transistor in the saturation region.

Â And a typical overdrive factor is ten.

Â So we let, IB equal 10 times 0.4 milliamps.

Â Is equal to four milliamps.

Â And we use this base current to calculate the value of RB.

Â So we have RB is equal to Vin, which when the transistor

Â is on is 5 volts minus 0.7 divided by 4 miliamps,

Â which gives us a value of 1.1 KM's.

Â So a value of RC equal to 500 Ohms limits the diode current to 20 miliamps and

Â value for RB results in a base

Â current sufficient to put the transistor into the saturation region.

Â So in summary, during this lesson,

Â you were introduced to the BJT switch, and we examined BJT switch characteristics.

Â In our next lesson, we'll examine the BJT used as a common emitter amplifier.

Â So thank you, and until next time.

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