In this part of my electronics tutorial I’ll focus on transistors and also review what I’ve covered in previous parts of this series. If you haven’t watched the previous videos watch them first or you’ll be confused.
A transistor is a semiconductor component that is used to amplify or switch electronic charges. I’ll show how they are used as switches and amplifiers. We’ll see how changing voltages at the base effects the transistor. We’ll cover the Darlington configuration. Then we’ll create a fun voltmeter. All of the circuit diagrams follow the video below.
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Transcript From the Video
A. (Slide 1) A transistor is a semiconductor component that is used to amplify or switch electronic charges. We can switch charge flow on and off and we can amplify small signals to large ones required to use other components. It uses a small current to control a larger current.
B. An advantage of transistors is that many of them can fit on a small piece of silicon. There discovery lead to the discovery of Integrated Circuits.
C. The 2 most common transistors are Bipolar Junction Transistors and Field Effect Transistors.
D. (Slide 2) Bipolar Junction Transistors
a. BJTs contain P type, which contains positive charge carriers, and N type semiconductors, which contain negative charge carriers.
b. NPN transistors contain a P type between to N types.
c. PNP transistors contain a N type between to P types.
d. By changing the voltage on the base and emitter you control the flow of the current through the transistor. The examples will show how this works in practice.
E. (Slide 3) Field Effect Transistors
a. Contains either a N or P type semiconductor from top to bottom which conducts a charge when the opposite N or P type on the sides allows for conductivity.
b. When the Gate has a voltage applied to it charge flows from the source to the drain.
F. (Slide 4)To use a water analogy a transistor is like a faucet that controls the flow of water with a shutoff valve. The flow can be on, off, or partially on. In this way it acts almost like a variable resistor.
G. (Slide 5)In this model of an NPN Transistor
a. Current flows through the Collector and Base and flows out the Emitter
b. V BE represents a diode
c. V CE is the voltage that will change depending on the amount of charge flowing into the Base
d. The current that flows out of the Emitter equals the sum of the current from the Collector and Base
e. If the voltage flowing into the Base is below a certain level (normally .7 V) no current will flow through the transistor.
f. As the Base voltage increases the resistance is reduced incrementally and the current through the Collector increases. When the resistance is at 0 the current through the Collector will be equal to the current through the Emitter.
H. (Circuit) Here I'll demonstrate how the transistor will break the circuit if a current isn't applied to the Base.
a. SPDT Switch
b. 10 kΩ Resistor
c. 1 kΩ Resistor
d. Red LED
e. NPN Bipolar Junction Transistor
f. 9 V Battery
I. (iCircuit PNP Transistor) Demonstrate the same with a PNP Transistor and NPN transistor
a. We are using the transistor as a switch here by cutting off the charge to the base. As long as the voltage to the base is below .7 V in the design it is still essentially a switch.
b. Transistors make great switches because they are small, require little power and can turn on and off billions of times per second. Those are just a few reasons why they are used in integrated circuits.
J. Amplifying Transistor
a. When a transistor is used as an amplifier you provide just enough current to flow to the base to make it functional. Or, to put it another way the small signal you want to amplify flows into the base and the transistor turns it into a large signal.
a. (Circuit) Here we use 2 LEDs to show how a transistor amplifies. When the transistor receives a low voltage that just turns it on fluctuations in the signal received by the bas get amplified. You want to use resistors to make sure the transistor isn't saturated so that the signal stays stable.
1. NPN Bipolar Junction Transistor
2. 100 kΩ Resistor
3. 1 kΩ Resistor
4. SPDT Switch
5. 2 Red LEDs
6. 9 V Battery
b. (iCircuit Amplifying Signals)
1. Notice how low the current is going into the left LED
K. Transistor Ratings
a. When choosing the transistor for your circuit here are some things you should know.
1. Maximum Collector Current : The maximum DC current your transistor can handle
2. DC Current Gain (hFE) : The ratio of collector current to base current. This provides you with information on how much you can amplify a signal.
L. (Circuit) Here we slow the charge and discharge of the LED with a capacitor.
a. Here the 680 kΩ Resistor controls how long it takes to charge the resistor and the 470 Ω Resistor controls the discharge. When the switch is flipped the charge goes towards charging the capacitor. When the capacitor voltage reaches .7 V the transistor turns on the LED. The current rises through the LED because of the transistors current gain. This allows for a slow charge time, but still a bright LED.
b. When the switch is turned off the capacitor discharges and the LED dims until the voltage to the transistors base goes below .7 V and then the LED turns off.
c. 680 kΩ Resistor
d. 470 Ω Resistor
e. 3.3 kΩ Resistor
f. SPDT Switch
g. 100 μF (microfarad) Capacitor
h. NPN Bipolar Junction Transistor
i. Red LED
j. 9 V Battery
M. (iCircuit Capacitor Plus Transistor)
N. (Circuit) This circuit is the same as the previous except this time I'll use 2 transistors in the Darlington configuration. Here the current flowing through the emitter of the top transistor flows to the base of the bottom transistor. So the current is being amplified twice. Since we are using 2 transistors we require a voltage of 1.4 V to power the LED.
a. 100 kΩ Resistor
b. 470 Ω Resistor
c. 3.3 kΩ Resistor
d. SPDT Switch
e. 2 NPN Bipolar Junction Transistors
f. 100 μF (microfarad) Capacitor
g. Red LEDs
h. 9 V Battery
O. (iCircuit Darlington)
P. (Circuit) Voltmeter
a. A Voltmeter measures voltage difference between 2 points in a circuit. The circuit representing the voltmeter lies to the left of the number 25 row. We will measure the voltage on the right of the 25 row. We can adjust the resistance of the 50 kΩ potentiometer which is connected to the base of the first transistor. As we do this the LEDs change in brightness because the resistances at the bases of both transistors isn't equal.
b. If we turn the dial on the potentiometer until the LEDs have an equal brightness we can determine the voltage by subtracting the percentage turned on the potentiometer from 100 and then multiply that number times .09.
c. With 2 resistors of 10 kΩ and 33 kΩ we get a voltage of
1. (Slide 6)
V = (R b / (R t + R b)) * V bat
= 33 kΩ / (10 kΩ + 33 kΩ) * 9 V
= 6.9 V
2. I measured 6.7 V with the multimeter
3. (Slide 7)
I guessed the dial was at around 30 %
V = (100 - Dial %) * .09
6.3 V = 70 * .09
4. (iCircuit Voltmeter) The LEDs seem equal at 250 with a max of 950. So 250 / 950 = .26
100 - 26 = 74
74 * .09 = 6.7 V