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How to Use NPN Transistor? Function Analysis

Author: Apogeeweb
Date: 22 Jun 2021
 1040
npn transistors diagram

Introduction

The transistor is one of the basic semiconductor components, which has the function of current amplification in electronic circuit. It is made of two PN junctions very close to each other on a semiconductor substrate. Two PN junctions divide the entire semiconductor into three parts: The middle part is the base area, and the two sides are the emitter and the collector. 

What is NPN Transistor? For Beginner

Catalog

Introduction

Ⅰ NPN Transistor Arrangement and Symbol

Ⅱ How Do NPN Transistors Work?

Ⅲ NPN Transistor Uses: A Controllable Valve


Ⅰ NPN Transistor Arrangement and Symbol

Before explaining the principle, let's first understand the basic structure and symbols of the NPN transistor. To identify the NPN transistor pins, it will be Collector (c), Base (b) and Emitter (e).

NPN Transistor Structure and Symbol

Figure 1. NPN Transistor Structure and Symbol

NPN transistor is composed of two N-type semiconductors and one P-type semiconductor. Generally, an NPN transistor has a piece of P-type silicon (the base) sandwiched between two pieces of N-type (the collector and emitter). The arrangement is shown in the Figure 1.

 

Ⅱ How Do NPN Transistors Work?

Here is the main description to illustrate the basic principle and function of NPN transistors.
1) Current Amplification
The following analysis is only for NPN silicon transistors. As shown in the figure above, we call the current flowing from the base B to the emitter E the base current Ib; the current flowing from the collector C to the emitter E is called the collector current Ic. The directions of these two currents are both flowing out of the emitter, so an arrow is used on the emitter E to indicate the current direction.
The amplification function of the transistor is: the collector current is controlled by the base current (assuming that the power supply can provide a large enough current to the collector), and a small change in the base current will cause a large change in the collector current: the change in the collector current is β times the change in the base current, that is, the current change is amplified by β times, so we call β the magnification of the transistor (β is generally much larger than 1). If we add a changing small signal between the base and the emitter, it will cause a change in the base current Ib. After the change in Ib is amplified, it leads to a big change in Ic. If the collector current Ic flows through a resistor R, it can be calculated according to the Ohm's Law formula U=R*I, and the voltage on this resistor will change greatly. According to the voltage on this resistor, so we can get the amplified voltage signal. In short, the change satisfies a certain proportional relationship.
2) Bias Circuit
When the transistor is used in the actual amplifier circuit, it is also necessary to add a suitable bias circuit. There are several reasons for this. First of all, due to the non-linearity of the transistor's BE junction (equivalent to a diode), the base current must be generated after the input voltage reaches a certain level (for silicon tubes, 0.7V is often used). When the voltage between the base and the emitter is less than 0.7V, the base current can be considered as zero. However, in practice, the signal to be amplified is often much smaller than 0.7V. If no bias is applied, such a small signal is not enough to cause a change in the base current (because when it is less than 0.7V, the base current is all 0).
Add a suitable current to the base of the transistor (called the bias current, and the resistor in the figure used to provide this current, is called the base bias resistor). When a small signal follows this bias current are superimposed together, a small signal will cause a change in the base current, and the change in the base current will be amplified and output on the collector. Another reason is meeting the requirement of the output signal range. If there is no bias, then only those increased signals will be amplified, but the decreased signals will be invalid (because the collector current is 0 when there is no bias, and it cannot be reduced). With bias, let the collector have a certain current in advance. When the input base current becomes smaller, the collector current can be reduced; when the input base current increases, the collector current increases. Both the reduced signal and the increased signal can be amplified.
3) NPN Transistor Switch
Let's talk about the saturation mode of the transistor. As shown in the figure above, because of the limitation of resistance Rc (Rc is a fixed value, then the maximum current is U/Rc, where U is the power supply voltage), the collector current cannot increase indefinitely. When the base current increases and the collector current cannot continue to increase, the transistor enters a saturated state. The general criterion for judging whether the transistor is saturated is: Ib*β>Ic.
In a saturation state, the voltage between the collector and the emitter of the transistor will be very small, which can be understood as a switch. In this way, when the base current is 0, the collector current is 0 (this is called the triode cut-off), which is equivalent to the switch off; when the base current is large, it is equivalent to the switch on. In cut-off and saturation state, a transistor is equal to a switch.
4) Operational State
If we replace the resistor Rc with a bulb in the above figure, then when the base current is 0, the collector current is 0, so the bulb is off. If the base current is relatively large (greater than the current flowing through the bulb divided by the magnification β), the transistor will saturate, and the bulb will light up. Since the control current only needs to be a little larger than β of the bulb current, a small current can be used to control the on and off of a large current. If the base current increases slowly, the brightness of the bulb will also increase (which is a saturation process).
The figure below is a basic transistor switch circuit. The base should connect a base resistor (R2), and the collector connects with a load resistor (R1).

NPN

Operational Mode

NPN

Cut-off

Une<Uon

Uc>Ub

Active

Ube>Uon

Uc>Ub

Saturation

Ube>Uon

Uc<Ub


NPN transistor uses the B-E current (IB) to control the C-E current (IC). The E pole has the lowest potential, and usually the C pole has the highest potential during normal amplification, that is, VC>VB>VE.
NPN base extremely high voltage, the collector and emitter are short-circuit and low-voltage, and the collector and emitter are open-circuit.
NPN is suitable for two situations:
If the input is a high level and the output needs a low level, NPN is better.
If the input is a low level and the output needs a high level, NPN is better.

2N2222 NPN Transistor

2N2222 NPN Transistor Pinout

Ⅲ NPN Transistor Uses: A Controllable Valve

NPN is a component that uses b (base) current Ib to drive the current Ic flowing through CE, and its working principle is much like a controllable valve.

a controllable valve

Figure 2. A Controllable Valve

The blue water flow in the thin pipe on the left impacts the lever to open the valve of the large water pipe, allowing the larger red water flow to pass through the valve. The larger the blue water flow, the greater the red water flow in the big pipe. If the magnification is 100, then when the blue water flow is 1 kg/hour, then 100 kg/hour of water is allowed to flow through the large pipe. The principle of the transistor is the same. When Ib (base current) is 1mA, a current of 100mA is allowed to pass through Ice.

NPN Transistor Diagram

Figure 3. NPN Transistor Diagram

Let's analyze this circuit. If its magnification is 100, and ignore the base voltage. The base current is 10V÷10K=1mA, so the collector current should be 100mA. According to Ohm's law, the voltage on Rc is 0.1A×50Ω=5V. Then the remaining 5V is on the C and E poles of the transistor. Now if we let Rb be 1K, then the base current is 10V÷1K=10mA, according to the magnification of 100, is Ic 1000mA? If it is really 1A, then the voltage on Rc is 1A×50Ω=50V. The power supply voltage has been exceeded, and the transistors have become generators? This is not the case. See below:

NPN Transistor Compared to A Valve

Figure 4. NPN Transistor Compared to A Valve

Continue the metaphor. When the control current is 10mA, the valve on the main water pipe is opened to allow 1A current to flow, but can 1A be realized? No, because there is a resistor on it, it is equivalent to a fixed valve. It is stringed on top of the main water pipe. When the opening of the lower controllable valve is greater than the opening of the upper fixed resistor, the water flow will not increase any more, but will be equal to the water flow passing through the fixed valve opening above. Therefore, it is useless to open the lower transistor to a large opening. Therefore, we can calculate the maximum current of the fixed resistor 10V÷50Ω=0.2A, which is 200mA. That is to say, in the circuit, the base current increases and the collector current also increases. When the base current Ib increases to 2mA, the collector current increases to 200mA. When the base current increases again, the collector current will no longer increase, and it will not move at 200mA. At this time, the upper resistor also acts as a current limiter. 

 

Let us understand the status of the IO in the microcontroller.

AT89S51/52

Figure 5. AT89S51/52 

The circuits with 24 IO ports of P1-P3 in the single-chip microcomputer are as shown in the figure above. Usually the purpose of using electronic circuits is to allow devices to obtain a certain current to make them work. For example, to make light-emitting diodes bright, a current of more than 1mA is generally required. However, the single-chip microcomputer is a smart chip. It can make logical analysis and judgments by detecting the voltage value of each IO port, and outputs high or low voltage as the result signal. Therefore, it can be seen that the IO ports of the single-chip microcomputer focus on voltage, not the current flowing through R and the transistor. Here what is the relationship between the voltage and current of the IO port in the single-chip microcomputer? 

 

Continue the water pipe example.

water valve

Suppose we let the valve of R open larger and let the control valve below be fully closed. At this time, as shown in Figure 6, it can be seen that the pressure at point P is the same as the water tank. When we fully open the following control valve, as shown in Figure 7, the water will flow through the pipeline with a large flow, and the pressure at point P is 0 at this time. This principle is very similar to electronic circuits. The logic quantity measured at the output point P is 1 (power supply voltage) or 0 (0 potential) by transistor turning off or on. However, there is a problem with this process, that is, when the output of point P is required to be 0, the transistor will be turned on very large, and the current flowing through it will be very large. There are 32 IO ports on the single-chip microcomputer, which consumes a lot of power. 

wvf

Look at Figure 8. If we close the upper valve R very small and close the lower control valve fully, then the pressure at point P will still the same as the water tank, which is the same as in Figure 6 above. When we open the control valve greatly, as shown in Figure 9, although the pressure at point P is also 0, the flow of water passing through at this time is greatly reduced. In this way, we can either output 1 or 0. So very little water is consumed. The circuit in the single-chip microcomputer does exactly this. The resistance R on it is about 50K, and the maximum current is 5V÷50K=0.1mA. In other words, when P outputs 1, no current is consumed, and when P outputs 0, the current consumed is 0.1mA. Because of its large pull-up resistance R, for beginners, it is necessary to have certain methods to directly drive LEDs or other loads. Here to share the various situations when the IO port is connected to the load.

AT89S51/52 & 74HC373

Figure 10. AT89S51/52 & 74HC373

Let's take a look at the situation of connecting TTL devices first. When P1.0 is connected to an input pin of 74HC373, and the input impedance of TTL is very high, about a few hundred K to M ohm level. We assume 500K resistor to P1.0 to ground. In this way, when the transistor is turned on, the P1.0 point is at a low level, and a current of 0.1mA flows through Rc and then through the transistor to the ground, and no current flows through Ri. When the transistor is cut off, the current flows through Rc and then flows to the ground through Ri. Due to the resistor voltage divider effect, there are partial voltages on Rc and Ri, and the voltage at point P1.0 is the divided voltage of Rc and Ri. Total current is 5V÷(50K+500K)=0.009mA, then the voltage at point P1.0 is 0.009mA×500K=4.5V. TTL stipulates that output 2.4~5V is high level. So this connection is correct. Now let's take a look at the situation of using S51 to drive the LED.

AT89S51 correct connection

AT89S51 Correct Connection

Let’s take a look at the situation in Figure 11. Obviously, only P1.0 is a high potential to light the luminous tube, so the transistor must be cut off. In this case, the current flows through Rc to the luminous tube and then to the ground. To make the luminous tube turn on, there must be a threshold voltage exceeding 2.1V at both ends of the luminous tube. Therefore, the current flowing through the luminous tube is (5V-2.1V)÷50K=0.058mA, which is too weak to conduct.
Look at Figure 12. It can be seen from the figure that P1.0 must be at a low potential if the luminous tube turned on. The transistor of the P1.0 port must be turned on. At this time, the current flows all the way through Rc to the transistor and then to the ground. The other way consumes 2.1V on the luminous tube. Then current flows through with almost no resistance, but the maximum current of the triode of the IO port cannot exceed 15mA. If it exceeds, the triode will be burned out, so this connection method is incorrect. So how can these two connections be able to drive the light-emitting tube? See below:

 

AT89S51 incorrect connection

AT89S51 Incorrect Connection

Looking at Figure 13, a resistor Ri is connected between P1.0 and Vcc. When the transistor is turned on, two currents will flow through its c, e pole, one is the 0.1mA current on the internal R, and the other is the current on Ri. In order to prevent the transistor from over-current and burn out, we must make sure the resistance value, Ri=5V÷15mA=0.333K, which is about 330 ohms. At this time, the current flowing through the transistor is about 15mA, and the light-emitting tube is not bright at this time. When the transistor is turned off, both currents will flow through the luminous tube. The current flowing through the internal resistance of S51 is (5V-2.1V)÷50K=0.06mA, which is so small that we can ignore it. The current flowing through Ri is (5V-2.1V)÷330Ω=0.0087A, which is 8.7mA. However, the current consumed when the luminous tube is off is greater than the current consumed when the luminous tube is on. If many IO ports are used to light up many LEDs, such a circuit is not economical.
Look at Figure 14, after connecting a resistor in series with the luminous tube between Vcc and P1.0. When the transistor is turned on, the two currents will flow through the c, e after confluence. The current on the internal resistance is still 0.1mA. The current on the ce should be less 15mA. If exceeds 15mA, the resistance is determined as (5V-2.1V) ÷ 15mA = 0.193K, which is about 200 ohms. In this way, the current flowing through the luminous tube is about 15mA, and the luminous tube is on. When the transistor is cut off, it blocks the paths of these two currents, so no current is consumed. Low level P1.0 directly drives the light-emitting tube. It can be seen that this circuit consumes 15mA of current when the light-emitting tube is on, and does not consume current when it is off, so this circuit is effective. S51 direct drive digital tube generally also uses this principle.

 

Frequently Asked Questions about NPN Transistor

1. What is meant by NPN transistor?
An NPN transistor is the most commonly used bipolar junction transistor, and is constructed by sandwiching a P-type semiconductor between two N-type semiconductors. An NPN transistor has three terminals– a collector, emitter and base. The NPN transistor behaves like two PN junctions diodes connected back to back.

 

2. How do NPN transistors work?
The NPN transistor is designed to pass electrons from the emitter to the collector (so conventional current flows from collector to emitter). The emitter "emits" electrons into the base, which controls the number of electrons the emitter emits. ... The transistor is kind of like an electron valve.

 

3. What is a NPN transistor used for?
NPN transistors are mainly used in switching applications. Used in amplifying circuit applications. Used in the Darlington pair circuits to amplify weak signals. NPN transistors are used in the applications where there is a need to sink a current.

 

4. Which is better PNP or NPN transistor?
A NPN transistor has electrons as majority charge carriers whereas the PNP transistor has holes as majority charge carrier. ... mobility of electrons is more than hole,so as a result npn transistor are faster than pnp that's why they are preferred.

 

5. What does NPN mean?
NPN stands for Negative, Positive, Negative. Also known as sinking.

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