Electronics Shatin Tsung Tsin Secondary School By Mr. C.K. Yu.

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Presentation transcript:

Electronics Shatin Tsung Tsin Secondary School By Mr. C.K. Yu

Learning Approach ( 學習方式 ) A black box approach ( 黑盒方式 ) We don’t care what happens inside the box and how it works, we do know the behaviour of the box and know when and how to use it. (e.g. we don’t know how the refrigerator works, but we know the behaviour of the refrigerator and we do know when and how to use it.)

Learning Approach Things to do when we have a new electronic device: 1) Find out how it works, but not what happens inside. 2)Think how we may use this device.

Electronics( 電子學 ) Electronic device : Light Emitting diode (LED) Current can flow in correct connection. At high voltage (e.g. 5V), the LED comes on. We use ‘ ’ to represent this high voltage. 1 0 At low voltage (e.g. 0V), the LED comes off. We use ‘ ’ to represent this low voltage. Remember: voltage represents p.d.

Electronics( 電子學 ) Electronic device : Light Emitting diode (LED) LED can be used to test whether the voltage is high (5V) or low (0V) if one end is connected to 0V

Digital Electronics Basic digital devices, logic gates, have two operation states, a)high state (or logic level ‘1’) and b)low state (or logic level ‘0’). We shall use ‘1’ and ‘0’ to represent the two logic states.

NOT gate a. Logically, NOT means opposite. e.g, not True is False. b. The symbol of a NOT gate is. c. The output, O, of a NOT gate is 1 (High) when its input, A, is 0 (Low), and vice versa. d. The rule of the logic operation can be found as: O =

NOT gate Truth table A (input)O (output) 01 10

OR gate a.The output, O, of an OR gate is 1 (High) if any of its input (A/B) is/are 1 (high). b.The rule of the logic operation of OR can be found as: O = A B (Note : Result bigger than 1 is still 1, high is still high)

OR gate c. Truth table A(input)B (input)O (output) There are four combinations for two inputs + O = A + B

AND gate a.The output,O, of an AND gate is 1 (High) only if both inputs (A/B) are 1 (high). b.The rule of the logic operation of AND can be found as: O = A x xx x B (Note : Result bigger than 0 is 1)

AND gate c. Truth table A (input)B (input)O (output) x O = A x B

NOR gate a.The output,‘O’, of an NOR gate is 0 (High) if any of its input (‘A’/’B’) is/are 1 (high). b.The rule of the logic operation of OR can be found as: O = A B (Note : Result bigger than 0 is 1)

NOR gate c. Truth table A(input)B (input)O (output) O = A + B

NAND gate a.The output,‘O’, of an NAND gate is 0 (High) if its two inputs (‘A’/’B’) are 1 (high). b.The rule of the logic operation of AND can be found as: O = A x xx x B (Note : Result bigger than 0 is 1)

NAND gate c. Truth table A (input)B (input)O (output) x O = A x B

Use of Logic Gates To be omitted In the circuit of logic gate, connections to the supply rails (dashed lines) are usually omitted.

Use of Logic Gates What will the output of the diagram be if the input is connected to the supply rail of high voltage, 5V? Ans : The output is _____voltage, ___V, or logic state of ___. LOW0 ‘0’

Example Connect the positive and negative supply rails to 5 V and 0 V respectively. Connect the output O of the gate to an LED and plug a ‘flying’ lead into the input.

Example Result: 1.Observation : If the flying lead is not connected to anything, is the LED on? __________

Example 2. What is the input of the NOT gate when the flying lead is not connected to anything? High/Low ===

Example 3. While connecting the ‘ flying ’ lead to the 5 V supply rail to make the input high, the LED is not lit. The output is HIGH/LOW or the output state is ____. === 1

Example 4. While connecting the ‘ flying ’ lead to the 0 V supply rail to make the input high, the LED is lit. The output is HIGH/LOW or the output state is ____. === 0

Experiment session P. 59 – P. 60 and P.61 – P.62

Application of Logic Gates Simple burglar alarm and Simple water heater

Simple burglar alarm What gate is being used? NAND gate LDR

Simple burglar alarm The buzzer sounds when the output of the logic gate is ___ V or ___ (state). 5‘1’

Simple burglar alarm Under which situations will the output of the logic gate is 5 V or ‘1’?

Simple burglar alarm If the switch is not pressed, input A is a flying lead. Its state is high (‘1’). If no light shines on the LDR, input B is also a flying lead. its state is also high (‘1’). Then the output is __.‘0’ The buzzer will not/will sound. ===

Simple burglar alarm Under which situations will the output of the logic gate is 5 V or ‘1’? Either input is LOW

Simple water heater low resistance Thermistor If covered by water, input B becomes high Resistance becomes high at low temperature, and input A becomes high

Simple water heater The heater is on when water covers the contacts and is at low temperature.

Simple water heater When water is at low temperature. V DE + V EF = 5V I (R DE + R EF )= 5V If R EF >>R DE then V EF ≈ 5V e.g. if R EF =1000 Ω, R DE =5 Ω Then I = 5V/1005Ω= A V EF = I x R EF = 4.98 V The potential at A is 4.98 V and A is at high (‘1’) state D E F

Simple water heater When water covers the contacts, the resistance is 0 Ω. The input B is connected to the 5V supply directly. The potential at B is 5 V and B is at high (‘1’) state D E F

Future thinking Should we add a switch to make sure that the water heater only works when the switch is also on? How should we design the logic circuit for this purpose? This question will not be tested in the final examination.

The End Thank you very much