Demo 3.4: Transistors, from Switch to Amplifier
A diode has two terminals, and that is the whole limit of what it can do. You apply a voltage across it and it either conducts or it does not. There is nothing you can say to a diode that would change its mind, because there is nowhere to say it from.
Add a third terminal and everything changes. Now there is a place to apply a small signal that decides what a much larger current does, and a component that can do that can amplify, switch, oscillate, and hold a bit of memory. Almost every electronic device built since 1947 is an arrangement of these, and the phone in your pocket has several billion of them.
This demonstration takes the transistor the same way the previous one took the diode. First what is physically inside it, then the numbers that describe it, then the graph that is the honest description, and then the method for solving a circuit that contains one. The method turns out to be the load line again, which is the point.
What is actually in there
Section titled “What is actually in there”Two doped regions with a junction between them is a diode, and the previous demonstration was about what that junction does. Three doped regions in a row give you two junctions, and a bipolar junction transistor is exactly that: n-p-n or p-n-p, with a lead on each layer called the emitter, the base and the collector.
Wired as two separate diodes back to back, that arrangement would be useless. What makes it a transistor is two deliberate asymmetries in how it is built:
- The base is very thin, and lightly doped.
- The emitter is heavily doped, far more so than the base.
Now forward bias the base-emitter junction. Carriers pour out of the emitter into the base exactly as they would in a diode, and because the emitter is so much more heavily doped, almost all of the current across that junction is carriers going that way rather than the other. Once in the base they are minority carriers, surrounded by the opposite kind, and they will recombine. The only question is whether they recombine before they get anywhere.
They do not, because the base is thin. Most of them cross it and arrive at the second junction, and here is the part that seems wrong the first time you meet it: that second junction is reverse biased. A reverse-biased junction is supposed to block current. It does block the majority carriers on either side of it, and these arrivals are not majority carriers. They are exactly the kind of carrier a reverse field sweeps across rather than turns back.
So the collector collects almost everything the emitter sends, and the reverse bias helps rather than hinders. The few carriers that do recombine in the base have to be replaced through the base lead, and that replacement current is the base current. It is small, it is the only part of the arrangement you have any control over, and it is what the other current follows.
How to use it
Section titled “How to use it”Inside the transistor has two sliders, one for each junction. Watch the depletion layers and the arrows: forward bias the base-emitter junction and carriers start crossing, reverse bias the base-collector one and they get swept up. The four presets walk you through the three operating regions.
The three currents draws the emitter current splitting into collector and base as two pipes whose widths are the currents themselves, on a square-root scale, because on a linear one the base branch is invisible. Change the part and watch the split change while the base current stays where you put it.
The characteristic curves has both graphs. The input characteristic is a diode curve, because that is what it is. The output characteristic is a family of curves with the three regions shaded on it, and the family is the thing to understand: the base current chooses which curve you are on.
Load line and the Q-point is the payoff, and it is the diode demonstration’s method applied to a device with a knob. The axes are deliberately fixed so that changing something moves the line rather than relabelling the graph.
As a switch, and as logic sizes the base resistor for an LED, works out the overdrive factor, and then builds the NOT, AND and OR gates from the lecture out of the same stage.
Field effect transistors is the comparison the lecture ends on, with the square law, the pinch-off boundary, and a warning about the word “saturation”.
Transistors, from Switch to Amplifier
Three layers and two junctions, the three currents, the characteristic curves and the Q-point, and the one difference a field effect transistor makes.
Walkthrough
Section titled “Walkthrough”Step 1: Watch the two junctions separately
Section titled “Step 1: Watch the two junctions separately”Open Inside the transistor and press Nothing applied.
Two depletion layers, one at each junction, both built by diffusion alone exactly as in the diode. Notice how narrow the base is compared with the other two regions, and notice that the two layers are already eating into it from both sides.
Press Cutoff. The base-collector junction is now reverse biased at −5 V and its layer has grown considerably, but nothing is flowing. This is worth pausing on: the collector is doing everything it can to sweep carriers across, and it has nothing to sweep, because the emitter has not been switched on. The collector cannot start a current, it can only collect one.
Press Active. Now the base-emitter junction has 0.7 V across it, its layer has collapsed, and the arrows appear. Carriers are crossing the base and being swept up by the collector, and a small amber arrow shows the fraction leaking out of the base lead.
Press Saturation. The base-collector junction is now forward biased too, and the collector current has collapsed. Both junctions forward biased means the collector has stopped sweeping carriers away and started injecting its own, and the device has become a closed switch rather than an amplifier.
Then sweep the base-collector slider from −10 V up towards zero while in the active state. The collector current barely changes. That is the flat active region of the output characteristic, seen from the inside, and it is the reason a transistor behaves as a current source.
Step 2: Look at the size of the three currents
Section titled “Step 2: Look at the size of the three currents”Open The three currents with the 2N3904 selected and the base current at 20 µA.
The emitter pipe arrives carrying everything. It splits, and the collector branch takes 99.34 per cent of it while the base branch takes the remaining 0.66 per cent. Those two numbers are α and 1/(β+1), and the drawing is the only place in this demonstration where you can see at a glance how lopsided the split is.
Read the three lines of arithmetic underneath:
- I_C = β·I_B = 150 × 20 µA = 3 mA
- I_E = I_C + I_B = 3 mA + 20 µA = 3.02 mA
- α = I_C/I_E = β/(β+1) = 0.99338
The second line is Kirchhoff and it is always true. The first line is the transistor’s own contribution and it is only true in the active region. Keeping those apart matters, because the moment you saturate the device the first line stops holding and the second one carries on.
Now switch to the BC547. The base current has not moved, and the collector current has nearly doubled, because β is 290 instead of 150. Then switch to the BD139 and it drops to under half, because β is 60.
Step 3: Take β seriously as a problem
Section titled “Step 3: Take β seriously as a problem”Read the caution about β while you are here, because it is the single most practically useful thing on this tab.
β is not a design constant. The 2N3904 data sheet allows anything from 100 to 300 for the same part number out of the same bag. It also changes with collector current, and it rises by roughly half a per cent for every degree the device warms up.
So any circuit whose output depends on β will behave differently from one build to the next and drift as it runs. There are two standard answers, and you will meet both:
- For a switch, give the base several times the current the calculation says it needs, so that β cannot matter. That is Step 6.
- For an amplifier, arrange feedback so the operating point is set by resistors instead. That is a later module, and it is why the biasing arrangements you will see there look more complicated than they need to.
Step 4: Read both characteristics
Section titled “Step 4: Read both characteristics”Open The characteristic curves on Input.
This is a diode curve. It is a diode curve because the base-emitter junction is an ordinary forward-biased p-n junction and nothing more, so everything from the previous demonstration applies: nothing below half a volt, a knee near 0.6 V, and a decade more current for every extra 60 mV.
The practical consequence is the same too. You never set V_BE and hope for a current, because a few tens of millivolts of error is a factor of two in the answer. You put a resistor in series and set the current, then read off whatever V_BE happens to result. This is why every circuit on the following tabs has a base resistor in it, and why the familiar 0.7 V is a result rather than an input.
Now switch to Output.
One graph is not enough here, because a two-terminal device needs one curve and a three-terminal device needs a family. Each curve is one value of base current, and the base current chooses which curve you are on.
Three regions are marked:
- Cutoff, along the bottom axis. No base current, so no collector current at any V_CE.
- Saturation, the narrow band at the left. Both junctions forward biased, V_CE collapsed to a couple of hundred millivolts, and the current set by whatever is outside the device rather than by the base.
- Active, the wide flat region across the rest. I_C = β·I_B, and V_CE barely matters.
Drag the collector-emitter voltage slider along one curve from 1 V to 10 V and watch the current. It moves by about nine per cent across nine volts. That near-independence is what makes the collector a current source, and a current source with a resistor after it is an amplifier.
Then drag it down below 0.4 V and watch the current fall off a cliff. That cliff is the saturation knee, and notice how narrow it is: the shaded band is about a third of a volt out of ten. It is not drawn exaggerated. If you measure one, you should find it as thin as this.
Step 5: Solve the circuit with a load line
Section titled “Step 5: Solve the circuit with a load line”Open Load line and the Q-point. It starts on Mid-way bias: 10 V, 1 kΩ, 30 µA of base current.
The difficulty is the same one the diode presented. Kirchhoff gives you V_CC = I_C·R_C + V_CE, a straight line. The device gives you its characteristic. Both are true at once, and you cannot solve them together in any tidy way.
So do it graphically. Find the line’s two intercepts:
- Point A: set the current to zero. No collector current means no drop across the resistor, so the whole supply appears across the transistor. That is 10 V on the voltage axis, and it is also exactly what cutoff looks like.
- Point B: set the voltage to zero. Then the resistor has the whole supply across it, so I_C = V_CC/R_C = 10 mA on the current axis, and that is very nearly what saturation looks like.
Join them, then pick the curve for 30 µA of base current. Where they cross is the Q-point, and the demonstration reads it off at about 5.26 V and 4.74 mA.
Check it. 4.74 mA through 1 kΩ is 4.74 V across the resistor, and 4.74 + 5.26 is 10 V, which is the supply. Both laws hold, which is what made it the operating point.
Step 6: Move the Q-point three different ways
Section titled “Step 6: Move the Q-point three different ways”This is where the third terminal earns its keep, and it is why the axes are nailed down.
Press Larger collector resistor, which takes R_C from 1 kΩ to 1.5 kΩ. The line pivots about Point A, because Point A does not depend on R_C at all, and the Q-point moves to a new crossing.
Now look carefully at which number changed. V_CE dropped from 5.26 V to 3.04 V, and the collector current went from 4.74 mA to 4.64 mA. Half again as much resistance, and the current moved by two per cent.
Press Lower the supply and the same thing happens: the line slides sideways without changing its slope, V_CE falls to 2.39 V, and the current stays at 4.61 mA.
This is worth dwelling on, because it is the exact opposite of the diode. There, changing the resistor changed the current a great deal and left the voltage nearly alone, because a diode is close to a fixed voltage drop. Here, changing the resistor changes the voltage and leaves the current nearly alone, because a transistor in its active region is close to a fixed current source. Drag the resistance slider from 400 Ω to 2 kΩ, a factor of five, and the current moves by about seven per cent while V_CE swings from 8 V down to 1 V.
So what does set the current? Put the supply and resistor back and press More base current. The line does not move at all. The Q-point slides along it, and the current goes from 4.74 mA to 6.23 mA while V_CE falls from 5.26 V to 3.77 V.
That is the division of labour that makes the stage designable: the base current sets the current, and the supply and resistor set the voltage.
That third motion is the one a diode could not do, and it is the whole of analogue electronics in one picture. A small change in base current slides the Q-point along the load line, and a large change in collector voltage comes out. The voltage gain is set by the resistor, not by the transistor, which is a far more reliable thing to depend on.
Press Driven into saturation and watch the Q-point run into Point B and stop. Press Turned off and watch it sit at Point A. Those are the two ends, and they are where a switch lives.
Step 7: Design a switch properly
Section titled “Step 7: Design a switch properly”Open As a switch, and as logic and press A sensible switch.
Work down the design check on the right, because that column is the calculation in the order you would do it by hand:
- The LED and its resistor want about 14.5 mA.
- Divided by β, that needs 97 µA of base current at the absolute minimum.
- The 4.7 kΩ base resistor is actually supplying 895 µA.
- So the base is asking for 134 mA, which the collector resistor cannot possibly supply.
- The overdrive factor is therefore 9.2.
- V_CE has collapsed to about 115 mV.
- And the device is dissipating under 2 mW.
That overdrive of nine is not waste, it is the entire design. β could halve and this circuit would still work exactly the same, because the collector current is being set by the resistor and the LED rather than by the transistor. The rule of thumb is five to ten times the base current the calculation says you need.
Now press Only just saturated, which raises the base resistor to 22 kΩ. The overdrive drops to about 2.0. The circuit still works, and it is a bad design: a warmer day, or a different part out of the same bag, would take it out of saturation.
Press Not enough base drive at 47 kΩ and it does exactly that. The overdrive falls just below one, the device goes active, V_CE climbs to over half a volt, and the LED is dimmer than it should be because the transistor is now holding back voltage that ought to be across the resistor. Look at the power figure: it has gone up by a factor of four while doing less useful work.
Press Not driven at all and the device goes to cutoff, which is a perfectly good state. Then drag the input voltage slowly up from 0.4 V and watch it pass through the active region on its way to saturation. That transit is unavoidable, and it is why switching circuits care about doing it quickly.
Step 8: Build logic out of it
Section titled “Step 8: Build logic out of it”Press Building logic gates and step through the three.
NOT is one stage with a pull-up resistor. A high input turns the transistor on, which connects the collector to the emitter and so to ground, and the output goes low. A low input leaves the transistor off and the resistor pulls the output up. The inversion is not a design decision. It falls out of the fact that the transistor can only ever pull the output down.
That pull-up is the same component doing the same job as in the demonstration on pull-up and pull-down resistors, which is worth noticing: you sized one of those before you knew what was on the other end of it, and this is what was on the other end.
AND stacks two transistors in series between the supply and the output. Current can only reach the output if it gets through both, so both inputs must be high. OR puts the same two in parallel, so either will do.
Series means AND, parallel means OR. That association survives into every logic family since, including the CMOS in the chips on your board, and it is the single most portable idea on this tab.
Then read why nobody builds logic this way any more. These gates work and they waste current continuously, their output levels are not clean, and the stacked AND gate needs its upper transistor driven above its own emitter. CMOS fixed all of it, using the device on the last tab.
Step 9: Meet the other kind
Section titled “Step 9: Meet the other kind”Open Field effect transistors.
A FET has a gate, a source and a drain, and the current runs from source to drain through a channel. The gate does not touch that channel: in a MOSFET an oxide layer separates them, and an oxide layer is an insulator. So the gate draws no steady current at all. It works by the field it produces, which is where the name comes from.
The lecture’s water analogy is a good one. The source and drain are the pipe, V_DS is the pressure across it, and the gate is a valve. A valve does not consume any of the water it controls.
Drag V_GS up from 1.5 V. Nothing at all happens until you pass the threshold at 2.0 V, because below that there is no channel. Above it, the current follows the square of the overdrive V_GS − V_TH, which is far gentler than the diode’s exponential: doubling the overdrive quadruples the current, where a junction would multiply it by millions.
Now look at the dashed boundary. Unlike a BJT, whose knee sits at a couple of hundred millivolts whatever the base current, a FET’s knee moves right as the gate is driven harder, at V_DS = V_GS − V_TH. Left of it the channel reaches the drain and the device is a resistor the gate controls. Right of it the channel is pinched off near the drain and the current stops depending on V_DS.
Finish on the comparison table. Everything in it comes back to one line: a BJT is controlled by a current and a FET by a voltage. That is the sentence the lecture ends on, and it is the reason almost every transistor manufactured today is a MOSFET.
Check your understanding
Section titled “Check your understanding”A transistor's base-emitter junction is forward biased and its base-collector junction is reverse biased. Which region is it in?
You are designing a transistor switch and calculate that 97 µA of base current will just barely give you the collector current you need. What base current should you actually design for?
On the output characteristic, why are the curves in the active region almost horizontal?
Match each item to what it is
Which of these are true of a MOSFET but not of a bipolar transistor? Select all that apply.
A common-emitter stage has V_CC = 10 V and R_C = 1 kΩ. Where does Point B of the load line sit, and what does it correspond to?
Wrap-up
Section titled “Wrap-up”Six things to take away.
- A transistor is three doped layers with a thin base between two junctions. Forward bias one, reverse bias the other, and the reverse-biased one collects almost everything the forward-biased one injects. Every number that follows is bookkeeping about that.
- I_E = I_C + I_B is Kirchhoff and always true. I_C = β·I_B is the device, and only true in the active region. Confusing the two is the commonest error in saturation problems.
- β is not a design parameter. It varies threefold between parts with the same number and drifts with temperature. Design so that it does not matter.
- The output characteristic is a family of curves, one per base current, with three regions on it. A switch works at the two ends, an amplifier in the middle, and the middle is where the heat is.
- The load line method is the same one that solved the diode circuit, with one addition: the base current chooses which curve you cross, so the Q-point can be moved without touching the supply or the resistor. That is amplification.
- A FET does the same three jobs with a voltage instead of a current, because its gate is insulated from the channel. Watch the word “saturation”, which means the closed-switch state for a BJT and the flat amplifying region for a FET.
Two of the ideas here point forward. Series-means-AND and parallel-means-OR is how every logic gate you meet from now on is actually built, so the gates in the digital half of this module are no longer black boxes. And the load line with a movable operating point is the construction used to bias every amplifier stage you will design, where the Q-point is chosen deliberately rather than merely observed.
© 2026 Derek Molloy, Dublin City University. All rights reserved.