In applications where we use a transistor or a BJT to amplify an AC signal, it becomes crucial to set its base bias voltage correctly. This allows the transistor to sit comfortably in its “active” zone, which means we are working within the linear range of the output characteristic curves.
The exciting part is that both NPN and PNP bipolar transistors can switch gears and function as “ON/OFF” solid-state switches. By adjusting the base terminal bias, we can easily turn them into switches!
Solid-state switches are one of the primary functions of transistors. They allow us to toggle a DC output “ON” or “OFF.” For instance, some output devices like LEDs do not require much current, just a few milliamps at logic-level DC voltages, so we can drive them directly from a logic gate without any issues.
However for high-power applications like motors, solenoids, or lamps, significantly more power is needed than a standard logic gate can provide. That is where our reliable transistor switches come in.
When we use a bipolar transistor as a switch, regardless of whether it is NPN or PNP, we configure the biasing to allow operation across both sides of the “I-V” characteristic curves mentioned earlier.
When we discuss transistors to be used as switches, we mainly focus on two key areas, which are the Saturation Region and the Cut-off Region. This means that, we do not have to deal with the complex Q-point biasing and voltage divider circuits required for amplification.
Instead we can simply toggle the transistor between its “fully-OFF” state (cut-off) and its “fully-ON” state (saturation).
It is like flipping a light switch on and off, making the use of transistors as switches incredibly straightforward!
Operating Regions of a Transistor (BJT)
Understanding the Transistor Regions: Cut-off and Saturation
Referring to the graph above with its curves, we notice a pink area at the bottom, referred to as the “Cut-off region”.
To the left, there is a blue area known as the “Saturation” region of the transistor. Both regions play a crucial role in understanding transistor operation. Let us learn more what each of these regions entails!
Cut-off Region
Let us start with the cut-off region. In this zone, the transistor is essentially inactive. Here, there is no input base current (IB), no output collector current (IC), and the collector voltage (VCE) is at its peak.
Because of this, a large depletion layer forms, indicating that no current flows through the transistor. This means the transistor is completely “Fully-OFF.” Here’s a more detailed breakdown:
- The input and base remain at ground level (0 volts).
- The base-emitter voltage (VBE) is less than 0.7 volts.
- The base-emitter junction is reverse-biased.
- The base-collector junction is also reverse-biased.
- Consequently, we can assert that the transistor is “fully-OFF” in this cut-off region.
- No collector current is present (IC = 0).
- VOUT equals VCE, which matches VCC, and we can denote that as “1.”
- Essentially, the transistor behaves like an “open switch.”
So, in short, the cut-off region or the “OFF mode” for a bipolar transistor acting as a switch, features both junctions reverse-biased, with VB less than 0.7 volts and IC at zero.
For a PNP transistor remember that the emitter should be negative compared to the base.
Saturation Region
Now let us shift our focus to the saturation region! In this section of the graph, the performance of the transistor improves significantly as the base current reaches its maximum level.
This results in a significant flow of collector current, minimizing the collector-emitter voltage drop.
This keeps the depletion layer small while allowing maximum current to pass through the transistor. So, what does it indicate when our transistor is in this zone? It means it is switched “Fully-ON.” Here are some key highlights:
- The input and base are connected directly to VCC.
- The base-emitter voltage (VBE) exceeds 0.7 volts.
- The base-emitter junction is now forward-biased.
- The base-collector junction is also forward-biased.
- Thus, our transistor is “fully-ON” in this saturation region.
- Maximum collector current flows (IC = VCC/RL).
- Ideally, VCE drops to 0 during saturation.
- VOUT equals VCE, resulting in “0.”
- In this state, the transistor functions like a “closed switch.”
So, when we are talking about the “saturation region” or what we like to call the “ON mode” for a bipolar transistor acting as a switch, we can think of it like this, both of the junctions are forward biased… which means we have got our base voltage (VB) sitting above 0.7 volts, and the collector current (IC) is at its maximum level.
Just a little side note for those dealing with PNP transistors: in that case, the emitter has to have a positive potential compared to the base.
Now when our transistor is in this saturation region, it is functioning like a “single-pole single-throw” (SPST) solid-state switch.
Here is how it works: when there is no signal hitting the base of the transistor, it effectively turns “OFF,” which is like having an open switch, no current flows through the collector at all.
But here is where it gets interesting! When we apply a positive signal to the base of the transistor, it toggles the switch to “ON” acting like a closed switch. This allows maximum current to flow through the circuit.
It is like we are giving the transistor the green light and it just lets all that current zoom through… And that is how we can think about the saturation region and how our bipolar transistor behaves as a switch.
When we need to switch moderate to high levels of power, one of the best methods we can use is a transistor with an open-collector output.
In this setup, we connect the emitter terminal of the transistor directly to ground. What this does is allow the transistor’s open-collector output to “sink” an external voltage down to ground, which gives us the ability to control any load that we have connected.
Let us take a look at an example: imagine we are using an NPN transistor as a switch to operate a relay. This is a pretty common application!
Now, when we are dealing with inductive loads like relays or solenoids, there is something important we need to keep in mind.
We should place a flywheel diode across the load. Why? Well, when the transistor switches “OFF,” it generates back EMF from the inductive load, and that can potentially damage our transistor.
The flywheel diode helps dissipate that back EMF, protecting our transistor from any nasty surprises.
Now if we are working with loads that have really high current or voltage requirements—think motors, heaters, and things like that—we can control that load current using a suitable relay.
This way we can manage those higher power demands safely and effectively. And there you have it! That is how we can use transistors in a straightforward way to control power and protect our components while doing it!
Analyzing a Basic NPN BJT Switch Circuit
So, if we take a look at this circuit, it looks kind of similar to that Common Emitter circuit we checked out in those earlier tutorials.
But this time around we are using the transistor as a switch. That means we need to make sure the transistor is either completely “OFF” (which we call cut-off) or totally “ON” (that is what we refer to as saturated).
Now if we think about what an ideal transistor switch would be like, it would have this magical ability to create infinite resistance between the Collector and Emitter when it is in that “fully-OFF” state.
This would mean that no current could leak through at all—zero, nada! And then when it is in that “fully-ON” state, it would have absolutely no resistance between the Collector and Emitter, allowing the maximum amount of current to flow through like a breeze.
However in the real world, things do not always work out perfectly. When we flip the transistor to the “OFF” position, there are these tiny little leakage currents that still manage to trickle through.
And when we switch it to “ON,” the transistor does not quite become a superhero, instead, it has a low resistance that leads to a small saturation voltage (we call this VCE) across it.
So yes, while our transistor is not a flawless switch by any means, it does keep power loss to a minimum when it is in both those cut-off and saturation states.
Now, for us to get that Base current flowing, we need to make the Base input terminal a bit more positive than the Emitter.
To do this, we have got to boost it up above 0.7 volts, which is what a silicon device needs to get going. So as we tweak around with this Base-Emitter voltage, which we call VBE, we are also changing the Base current.
This little adjustment directly affects how much Collector current is zipping through the transistor, just like we talked about before.
Now when we reach the point where the maximum Collector current is flowing, that is when we say the transistor is in a state called Saturation. It is like hitting that sweet spot!
The cool part is that the value of the Base resistor plays a huge role in determining how much input voltage we need and what kind of Base current we require to flip the transistor into that fully “ON” mode. So it is all connected and super important for getting everything to work just right!
Solving a BJT Switch Circuit Problem #1
Let us consider a transistor or a BJT having it current gain β = 150, collector current Ic = 5 mA and base current Ib = 22 uA. We want to calculate the value of the Base resistor (Rb) which may be right to switch the collector load fully “ON” as soon as the input terminal voltage goes beyond 3 V.
RB = (Vin – VBE)/IB
= (3 – 0.7)/22 * 10-6)
= 104545 Ω = 104 kΩ, the nearest standard value being 100 kΩ, which ensures that the transistor (BJT) switch is always saturated.
Solving a BJT Switch Circuit Problem #2
Let us consider the same values as explained in the above problem and then calculate the minimum Base current which will sufficient to switch the BJT “fully-ON” (saturated) using a load rated with 250 mA current when we increase input voltage to 6 V. We will also calculate the resultant value of the base resistor Rb.
To calculate the base current, we use the following formula:
IB = IC/β = 250/150 = 1.66 mA
Next, we calculate the value of the necessary base resistor:
RB = (Vin – VBE)/IB
= (6 – 0.7)/(1.66 * 10-3)
= 3193 = 3.193 kΩ
So, we can use BJTs specifically to connect or interface big, powerful loads that need a lot of current or high voltage, for example like motors, relays, or lamps with low voltage digital integrated circuits (ICs) or logic gates, like AND gates and OR gates.
When we get an output from a digital logic gate, it usually only able to generate out about +5 volts. But the device we want to control might need a higher voltage supply, like 12 volts or even 24 volts!
And if we are dealing with something like a DC motor, we might want to control its speed using a technique called Pulse Width Modulation (PWM), which involves sending a series of pulses.
That is where transistor switches really become so useful, they let us do all this much faster and with way less hassle than if we were relying on those old-school mechanical switches.
Interfacing a Digital Logic Input with an NPN BJT Switch
As we can see in the above diagram, a small current and voltage output from the NOT gate which is a digital IC is easily interfaced with an NPN BJT to toggle large high power loads like motor or lamps.
Interfacing a Digital Logic Input with an PNP BJT Switch
As can be seen in the above image, a PNP BJT can be interfaced with any digital logic input to toggle high power loads, just like an NPN version, however here the polarity of the PNP BJT becomes just the opposite to the NPN polarity.
How to Use a Darlington BJT as a Switch
Now, sometimes we find that the DC current gain in a bipolar transistor is just not enough to directly handle the load current or voltage we need.
This is where things get a bit interesting: instead of relying on just one transistor, we often use multiple switching transistors to get the job done. Basically we have a smaller input transistor that helps us switch a much bigger output transistor on and off, which can handle a lot more current.
To really boost our signal gain, we connect these two transistors in what’s called a “Complementary Gain Compounding Configuration.” But do not worry if that sounds complicated, it is more commonly known as a “Darlington Configuration.”
In this setup, the amplification factor we get is actually the product of the gains from both transistors working together.
Let us break it down a bit more. Darlington Transistors are made up of two separate bipolar transistors, either NPN or PNP types.
The two BJTs are hooked up in such a way that the current gain from the first transistor gets multiplied by the current gain of the second one. This means that we end up with a device that behaves like a single transistor but has an impressively high current gain, all throughout needing just a miniscule amount of base current to turn ON.
When we talk about the total current gain, which we refer to as Beta (β) or hfe for a Darlington device, it is simply the product of the individual gains from both transistors. So we can express it using the following formula, like this:
β(total) = β1 * β2
When we talk about Darlington Transistors, one of the best things is that they can achieve really high β values and handle much larger collector currents compared to just using a single transistor switch.
Let us try to understand this with an example.
Imagine we have our first input transistor that has a current gain of 100. Then we have got a second switching transistor with a current gain of 50.
When we put those two together, the total current gain skyrockets to 100 multiplied by 50 which gives us a whopping 5000!
Now let us say our load current is 250 mA, based on what we discussed earlier. With this amazing Darlington setup, the base current we need is only 250 mA divided by 5000, which turns out to be just 40 microamps (40uA).
That is a massive drop from what we would have needed with a single transistor, where we would be looking at around 1 mA which is a lot bigger compared to 40 microamps.
Isn’t that impressive how much more efficient we can be with these Darlington Transistors? They really help us do more with less!
Analyzing Two Basic Darlington Transistor Configurations
There are two basic types of Darlington transistor configurations which are given below. Let’s try to understand each one of these.
Darlington with NPN BJT Output
Darlington with NPN/PNP Complementary BJT Output (Sziklai)
In this Darlington configuration, we have got the collectors of both the transistors connected together, while the emitter of the first transistor is connected to the base terminal of the second one.
This means that the emitter current from the first transistor actually becomes the base current for the second transistor, which is what switches it “ON.”
So, here is how it works. Our first transistor, often called the “input” transistor, gets an input triggering signal at its base. It does its usual job of amplifying that signal and then uses this boosted signal to drive the second larger “output” transistor.
This second transistor takes things a step further by amplifying the signal again, giving rise to an hugely amplified total current gain.
One of the standout features of Darlington Transistors is just how much higher their current gains are compared to regular single bipolar transistors.
But that is not all! In addition to their ability to handle increased current and voltage switching, another big benefit of using a “Darlington Transistor Switch” is their fast switching speeds.
This makes them perfect for applications like inverter circuits, lighting systems, and controlling DC or stepper motors.
However there is a little something we need to keep in mind when we are using Darlington transistors instead of your typical single bipolar types.
When we are using Darlington transistors as switches, the input triggering voltage across the Base-Emitter (VBE) has to be a bit higher—around 1.4 volts, compared to 0.7 V for single BJTs, because of how those two PN junctions are connected in series.
Conclusions
So here is a quick summary of what we need to keep in mind when we are using a transistor as a switch. Here are the key points:
Firstly transistor switches are extremely useful for controlling all sorts of electrical devices, for example like lamps, relays, and even motors.
When we are using a bipolar transistor in a switch configuration, it is important that they operate in one of the two states. They have to be either “fully-OFF” or “fully-ON,” and never hang in between.
When a transistor is fully “ON,” we say that it is in its Saturation region, which means it is doing its job of allowing current to flow freely. On the flip side, when it is fully “OFF,” we call that the Cut-off region, where no current is flowing at all.
The cool part about using transistors (BJTs) as switches is that a tiny Base current can control a much larger Collector load current. This makes them really efficient for switching applications.
Now if you are dealing with inductive loads—like relays or solenoids—then you need to insert in a “Flywheel Diode.”
This little component helps manage any back EMF (electromotive force) that can occur when the load is being switched off, protecting our circuit and the driver BJT from potential damage.
Lastly, if we find ourselves needing to control larger currents or voltages, that is where Darlington Transistors come into play. These are perfect for those situations because they can handle more power while still being efficient.
References:
Which transistor should be used as a switch for a load of 5 V, 100 mA