THE 555
![Picture](/uploads/6/7/3/2/6732348/2446123.gif)
The 555 is everywhere. It is possibly the most-frequency used chip and is easy to use.
But if you want to use it in a "one-shot" or similar circuit, you need to know how the chip will "sit."
For this you need to know about the UPPER THRESHOLD (pin 6) and LOWER THRESHOLD (pin 2):
The 555 is fully covered in a 3 page article on Talking Electronics website (see left index: 555 P1 P2 P3)
But if you want to use it in a "one-shot" or similar circuit, you need to know how the chip will "sit."
For this you need to know about the UPPER THRESHOLD (pin 6) and LOWER THRESHOLD (pin 2):
The 555 is fully covered in a 3 page article on Talking Electronics website (see left index: 555 P1 P2 P3)
When drawing a circuit diagram, always draw the 555 as a building block with the pins in the following locations. This will help you instantly recognise the function of each pin:
![Picture](/uploads/6/7/3/2/6732348/7973306.gif)
Note: Pin 7 is "in phase" with output Pin 3 (both are low at the same time).
Pin 7 "shorts" to 0v via the transistor. It is pulled HIGH via R1.
Maximum supply voltage 16v - 18v
Current consumption approx 10mA
Output Current sink @5v = 5 - 50mA @15v = 50mA
Output Current source @5v = 100mA @15v = 200mA
Maximum operating frequency 300kHz - 500kHz
Faults with Chip:
Consumes about 10mA when sitting in circuit
Output voltage up to 2.5v less than rail voltage
Output is 0.5v to 1.5v above ground
Sources up to 200mA but sinks only 50mA
HOW TO USE THE 555
There are many ways to use the 55.
(a) Astable Multivibrator - constantly oscillates
(b) Monostable - changes state only once per trigger pulse - also called a ONE SHOT
(c) Voltage Controlled Oscillator
Pin 7 "shorts" to 0v via the transistor. It is pulled HIGH via R1.
Maximum supply voltage 16v - 18v
Current consumption approx 10mA
Output Current sink @5v = 5 - 50mA @15v = 50mA
Output Current source @5v = 100mA @15v = 200mA
Maximum operating frequency 300kHz - 500kHz
Faults with Chip:
Consumes about 10mA when sitting in circuit
Output voltage up to 2.5v less than rail voltage
Output is 0.5v to 1.5v above ground
Sources up to 200mA but sinks only 50mA
HOW TO USE THE 555
There are many ways to use the 55.
(a) Astable Multivibrator - constantly oscillates
(b) Monostable - changes state only once per trigger pulse - also called a ONE SHOT
(c) Voltage Controlled Oscillator
Inside The 555 IC
![Picture](/uploads/6/7/3/2/6732348/9782926.gif)
It is having 2 comparator and 1 flip flop 1 discharging transistor and last but not least output transistor pair
ASTABLE MULTIVIBRATOR
![Picture](/uploads/6/7/3/2/6732348/5342618.gif)
The graph applies to the following Astable circuit:
The capacitor C charges via R1 and R2 and when the voltage on the capacitor reaches 2/3 of the supply, pin 6 detects this and pin 7 connects to 0v. The capacitor discharges through R2 until its voltage is 1/3 of the supply and pin 2 detects this and turns off pin7 to repeat the cycle.
The top resistor is included to prevent pin 7 being damaged as it shorts to 0v when pin 6 detects 2/3 rail voltage.
Its resistance is small compared to R2 and does not come into the timing of the oscillator.Using the graph:Suppose R1 = 1k, R2 = 10k and C = 0.1 (100n).
Using the formula on the graph, the total resistance = 1 + 10 + 10 = 21k
The scales on the graph are logarithmic so that 21k is approximately near the "1" on the 10k. Draw a line parallel to the lines on the graph and where it crosses the 0.1u line, is the answer. The result is approx 900Hz.
Suppose R1 = 10k, R2 = 100k and C = 1u
Using the formula on the graph, the total resistance = 10 + 100 + 100 = 210k
The scales on the graph are logarithmic so that 210k is approximately near the first "0" on the 100k. Draw a line parallel to the lines on the graph and where it crosses the 1u line, is the answer. The result is approx 9Hz.
The frequency of an astable circuit can also be worked out from the following formula:
The capacitor C charges via R1 and R2 and when the voltage on the capacitor reaches 2/3 of the supply, pin 6 detects this and pin 7 connects to 0v. The capacitor discharges through R2 until its voltage is 1/3 of the supply and pin 2 detects this and turns off pin7 to repeat the cycle.
The top resistor is included to prevent pin 7 being damaged as it shorts to 0v when pin 6 detects 2/3 rail voltage.
Its resistance is small compared to R2 and does not come into the timing of the oscillator.Using the graph:Suppose R1 = 1k, R2 = 10k and C = 0.1 (100n).
Using the formula on the graph, the total resistance = 1 + 10 + 10 = 21k
The scales on the graph are logarithmic so that 21k is approximately near the "1" on the 10k. Draw a line parallel to the lines on the graph and where it crosses the 0.1u line, is the answer. The result is approx 900Hz.
Suppose R1 = 10k, R2 = 100k and C = 1u
Using the formula on the graph, the total resistance = 10 + 100 + 100 = 210k
The scales on the graph are logarithmic so that 210k is approximately near the first "0" on the 100k. Draw a line parallel to the lines on the graph and where it crosses the 1u line, is the answer. The result is approx 9Hz.
The frequency of an astable circuit can also be worked out from the following formula:
555 Astable Frequencies
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Turining OFF a 555 ic
![Picture](/uploads/6/7/3/2/6732348/136966950.gif)
The only problem with this circuit is the gradual lowering in volume as the electrolytic discharges. The 1,000u to 4700u determines the length of time the circuit is activated AFTER the Bell-Push is pressed. The circuit drops to zero current (the only current is the leakage of the 1,000u electrolytic).