THE 555 IC

Page 3 

The next circuit shows one way to turn off a 555 after a period of activation:

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).

In the following circuit the first 555 gates the second 555.  

The second 555 is not turned off. The circuit inside both 555's are always drawing current.
It is not practical to "turn off" a 555 as shown in the next diagram:

The output of a 555 is 1.7v less than rail voltage. This means the second 555 is receiving 10.3v from the output line of the first 555 if the rail voltage is 12v. The maximum output of the second 555 will be 10.3 - 1.7 = 8.6v  This may be too low for many output devices and the result may be disappointing.   
The Schmitt Trigger can be gated too. 
The first point to note is the hex Schmitt trigger IC contains 6 identical gates and the chip is normally permanently connected to the supply rail. If any of the unused inputs are tied HIGH, the particular gate draws very little current (less than 1uA), making the total for the chip about 6uA. 
There are two ways to GATE a Schmitt Trigger and prevent it from oscillating. 
The diagrams below show a Schmitt Trigger being gated so that the output is:
1. LOW,
2. HIGH.

Mouseover the animations below and see how the GATING LINE inhibits the oscillator:

Mouse-over to INHIBIT the Left Oscillator

For the left circuit, if the gating diode is taken HIGH, the capacitor charges quickly. This inhibits the operation of the oscillator and the output goes LOW. If a load is connected to the output of this gate, it will not be driven and the gate will consume the least current. 
For the right circuit, when the gating diode is taken HIGH it does not have any effect on the operation of the circuit and the oscillator continues to operate.

Mouse-over to INHIBIT the Right Oscillator

For the left circuit, if the gating diode is taken LOW, it does not have any effect on the operation of the circuit and the oscillator continues to operate.
For the right circuit, if the gating diode is taken LOW, the capacitor is discharged and the oscillator is INHIBITED. The output goes HIGH and the load will be driven. The circuit will draw maximum current. 

If NO LOAD is connected to the output, an inhibited gate will draw more current than when it is oscillating. Both arrangements will draw a similar current when inhibited. The current taken will be about 1uA for the gate plus the current through R.  

The 555 can be used as a timer up to 10 minutes. This circuit is also called a DELAY. 
To start timing, the START button is pressed briefly and the output of the chip goes LOW. At the expiration of 10 minutes, the output goes HIGH and the red LED illuminates. 
A simple application may be for a cooking operation in a shop. 
If a product needs to be cooked or heated etc, the button can be pressed and the LED illuminates when the time has expired.   

When calculating the time-duration for the circuit above, the capacitor charges from 0v to 2/3 rail voltage. 

The output drive-current for a 555 is 200mA maximum. The output voltage is 1.7v less than rail voltage. 
A driver transistor can be connected to the output pin to improve the output current to 1amp (or more) and deliver an output voltage that is near rail voltage. Globes are a typical example of a high-current load. They require up to 6 times the normal current when starting. This is due to the cold filament having a very low resistance. The same applies to motors. They have a high start-up current requirement. 
Any driver transistor can be fitted as shown in the diagram below:

Use a driver transistor for loads greater than 200mA

The CMOS 7555 has an output current capability of 50mA and will need a driver transistor for currents above 80mA.

Here are some mistakes in a 555 circuit. They are "technical." Can you spot them?

1. Pin 2 must be taken LOW for it to activate the 555. The circuit shows a positive voltage being applied to pin 2. This will do nothing. 
2. Connecting pin 2 to pin 4 and leaving them "open" as shown in the diagram above is very dangerous. These are fairly high-impedance pins and the consequences of leaving them open will be unpredictable. 
3. Pin 2 cannot be left "open."  The 10u will initially charge to 2/3 rail voltage and the voltage will be detected by pin 6. Pin 7 will then discharge the 10u and wait for a low to be detected by pin 2. Pin 2 will actually have no voltage on it but the pin requires a very small current (about 500 nano-amp) to activate the chip.  If a static charge delivers this current, the chip will cycle. The outcome is unpredictable. 
4. Pin 4 cannot be left "open." For pin 4 to reset the chip, it must be taken below 0.7v and supplied a current of 100uA. If it is left open you cannot guarantee the chip will operate. 

The only way to trigger a 555 with a positive pulse is to have the output sitting HIGH.  A positive pulse on pin 6 will change the output to LOW. The voltage on pin 6 must be greater than 2/3 supply voltage. When the 555 detects the trigger pulse, the output goes low and a transistor inside the 555 takes pin 7 to the 0v rail. This starts to discharge the 10u via the 1M resistor and when pin2 detects a LOW, the output goes HIGH again. See Positive and negative triggering on P1 of this discussion.



In the Knight Rider circuit, the 555 is wired as an oscillator. It can be adjusted to give the desired speed for the display. The output of the 555 is directly connected to the input of a Johnson Counter (CD 4017). The input of the counter is called the CLOCK line. 
The 10 outputs Q0 to Q9 become active, one at a time, on the rising edge of the waveform from the 555. Each output can deliver about 20mA but a LED should not be connected to the output without a current-limiting resistor (330R in the circuit above). 
The first 6 outputs of the chip are connected directly to the 6 LEDs and these "move" across the display. The next 4 outputs move the effect in the opposite direction and the cycle repeats. The animation above shows how the effect appears on the display. 
Using six 3mm LEDs, the display can be placed in the front of a model car to give a very realistic effect. The same outputs can be taken to driver transistors to produce a larger version of the display. 


The circuit runs 9 LEDs and the 10th output goes to the 555 via a transistor. The 10th output takes the 100u HIGH and this turns on the BC547 to inhibit the 555 for approx 3 seconds. 
The 100u gradually charges via the 10k and the voltage on the base of the BC547 drops to a point were the transistor turns off and the 555 oscillates 10 cycles and halts again.


The Light Detector circuit detects light falling on the Photo-cell (Light Dependent Resistor) to turn on the 555. Pin 4 must be held below 0.7v to turn the 555 off. Any voltage above 0.7v will activate the circuit. The adjustable sensitivity control is need to set the level at which the circuit is activated.  When the sensitivity pot is turned so that it has the lowest resistance (as shown in red), a large amount of light must be detected by the LDR so that its resistance is low. This produces a voltage-divider made up of the LDR and 4k7 resistor. As the resistance of the LDR decreases, the voltage across the 4k7 increases and the circuit is activated. 
When the sensitivity control is taken to the 0v rail, its resistance increases and this effectively adds resistance to the 4k7. The lower-part of the voltage-divider now has a larger resistance and this is in series with the LDR. Less light is needed on the LDR for it to raise the voltage on pin 4 to turn the 555 on. 


For the Dark Detector circuit above, when the level of light on the photo-cell decreases, the 555 is activated. Photo-cells (Photo-resistors) have a wide range of specifications. Some cells go down to 100R in full sunlight while others only go down to 1k. Some have a HIGH resistance of between 1M and others are 10M in total darkness. For the circuit above, the LOW resistance (the resistance in sunlight) is the critical value.  
More accurately, the value for a particular level of illumination, is the critical. The sensitivity pot adjusts the level at which the circuit turns on and allows almost any type of photo-cell to be used.


The Police Siren circuit uses two 555's to produce and up-down wailing sound. The first 555 is wired as a low-frequency oscillator to control the VOLTAGE CONTROL pin 5 of the second 555. The voltage shift on pin 5 causes the frequency of the second oscillator to rise and fall. 


The 555 can be used as an amplifier. It operates very similar to pulse-width modulation. The component values cause the 555 to oscillate at approx 66kHz and the speaker does not respond to this high frequency.  Instead it responds to the average CD value of the modulated output. The output is low but demonstrates the concept of pulse-width modulation


The Infra-Red Transmitter circuit produces a low for about 40uS and has a duty-cycle of 90% HIGH and 10% LOW. It delivers a pulse of 150mA at a frequency of about 2kHz to the infra-red LED.


A 555 can be wired as a Schmitt Trigger to clean up noise signals.  


The Touch Switch circuit will detect stray voltages produced by mains voltages and electrostatic build-up in a room. Pin 2 must see a LOW for the circuit to activate. The circuit can be made 100 times more sensitive by adding a transistor to the front-end as shown in the diagram below:

A 555 contains a flip flop along with a couple of comparators. When the output at pin 3 is 'high', C1 slowly charges through R1 up to 12V; when it is 'low' C1 discharges through R1 down to 0V. Pressing switch S1 upsets the 6v balance between R2 and R3 on pins 2 & 6 for a split second, triggering the flip flop and changing the state of the output from 'high' to 'low' or vice-versa. The wide hysteresis of the 555 (between 1/3 supply voltage and 2/3 supply voltage) limits false switching from switch bounce.

A negative supply can be generated with a 555 operating in astable mode. The generated voltage is approximately 3v less than the rail voltage due to pin 3 rising to about 1.7v below rail voltage, plus the loss in the power diode. A small loss is produced by the electrolytic in the diode-pump design, creating an overall loss of approx 3v. 
The output current should be kept to below 50mA, otherwise the output voltage will drop further. 

This is a request from Daniel, one of our subscribers.  
He needs to flash "turn indicators" using a 555 and a single 20 amp relay. Here is our suggestion. The timing resistor needs to be selected for the appropriate flash-rate. 

Flashing the "TURN INDICATORS"

This circuit will alternately flash two 4.5v globes.

The 1.5v LED Flasher circuit will flash a LED at approx 1 flash per second. The ZSCT1555 IC from Zetex is designed to operate from a single 1.5v cell and will continue to operate to a terminal voltage of 0.9v. The circuit was originally presented by A J De-Guerin.
The first thing we note is the 1.5v supply voltage. From our previous discussions we know that a LED will not illuminate until the voltage across it reaches 1.7v. This means the circuit will not work unless the output is generating a voltage higher than 1.7v.
And that is exactly what happens. 
The circuit charges the 47u via the 330R resistor when the output of the chip is HIGH. This puts a voltage of 1.5v on the electrolytic. 
We will now replace the electrolytic by a 1.5v battery to make the discussion easier to understand. 
We have a 1.5v battery connected to the output of the chip with the positive of the battery connected to the output of the chip.
When the output goes LOW, the left-hand-side of the battery falls by 1.5v and this causes the right-hand-side to fall by the same amount. 
The negative lead of the "battery" is actually 1.5v BELOW the 0v rail and this puts a total of 3v (in theory) across the LED. 
We have already learnt that the voltage across a LED will never rise above 1.7v (for a red LED) and so the voltage on the electrolytic creates a voltage higher than 1.7v across the LED and the energy flows into the LED to make it flash. The circuit is called a VOLTAGE DOUBLER. 

I have just received a comment from a design engineer. He has built a similar circuit to the one above and it has failed to oscillate. 
He comments:
I've even tried different values for the components (lower values for the capacitor and higher resistor values) with the same results. When I turn it on, the output stays low and the voltage on the capacitor stays at around 1.1-1.2v with a supply voltage of 1.5v. I've replaced the ZSCT1555 with another one, with the same results.  
With this type of uncertainty about reliability, I would give this circuit a MISS!   The problem could be the low voltage. The circuit may work at a higher voltage but this would negate its purpose. A 1.5v LED flasher circuit can be created with a "3909" chip (LM 3909 - now obsolete, but you may find one in a junk store) or by using the discrete 3-transistor oscillator/driver circuit, shown below:

LM 3909 Flasher circuit           3-Transistor Flasher circuit on 1.5v supply      

Here is another 1.5v flasher circuit using 2 transistors:


Hundreds of projects have been designed around the 555 and its dual and quad versions. 
Some of the projects will be presented in this e-magazine from time to time but in general the consideration of the author is to design a circuit with a chip that consumes less current. The 7555 meets this requirement but has a very low drive capability. 
The 555 (7555) is suitable for a circuit needing a single building-block but if it requires more than one "block," it it suggested to go to the 74c14 (Hex Schmitt Trigger IC) where 6 separate gates are available. For an equivalent cost you get six times more value!
The versatility of the 74c14 Hex Schmitt Trigger will be covered in a future issue.