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#1
12-02-2011, 05:55 PM
 BershaM Junior Member Join Date: Aug 2011 Posts: 15
Astable multivibrator Using 555

Hi all , it's my first post here and i'am really hoping for some help

The circuit simulation using multisim works fine but on the breadboard Both LEDs are on and its not the required from the circuit and i don't know whats wrong :/

i'd really appreciate it if someone helped.
#2
12-02-2011, 06:00 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

It's working; it is just that the LEDs are flashing on and off so quickly that it looks like they are on all of the time. You can't tell if they are blinking if the blink rate is faster than about 16Hz.

Increase R4 to perhaps 300k to 500k Ohms.

If LED2 does not turn off all the way, add a 330 Ohm resistor from U1 pin 3 to +9v.
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#3
12-02-2011, 08:52 PM
 Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,044 Blog Entries: 5

Math is your friend, as Wookie suggested. The formula's for the 555 work extremely well. For a symmetrical square wave R1 should be as low as you can go, the lower value it is, the closer the duty cycle is to 50%.

F = 1.44 / ((7KΩ + 2 X 4KΩ) X 1E-6F) = 96 Hz

Try bumping the capacitor to 100µF.

F = 1.44 / ((7KΩ + 2 X 4KΩ) X 100E-6F) = 0.96Hz (which is visible)

Duty Cycle = (7KΩ + 4KΩ) / (7KΩ + 2 X 4KΩ) = 11KΩ / 15KΩ = 73%

************************************************** *

So if R1 = 1KΩ, R2 = 7KΩ, C = 100µF then

F = 0.96 Hz, Duty Cycle = 53.3%

********************************

If R1 = 1KΩ, R2 = 680KΩ, and C1 = 1µF, then

F = 1.06Hz, Duty Cycle = 50.04%
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"Good enough is enemy of the best." An old engineering saying, Author unknown.

General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.

Last edited by Bill_Marsden; 12-02-2011 at 09:04 PM.
#4
12-02-2011, 09:13 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Just to help avoid confusion, you don't have to increase BOTH R4 AND C2 - just increasing one or the other will slow the flashing down enough so that you can see them blink.

A caution: don't make R1 less than 1k Ohms, or you will start burning up lots of power in R1, and your battery will become discharged very quickly.

If you want to make the duty cycle close to 50%, then use a diode connected across R4, cathode towards C2, and make R4 and R1 about equal. With this modification, you will probably need to make R1 somewhat larger than R4 to get a true 50% duty cycle.
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#5
12-02-2011, 10:16 PM
 BershaM Junior Member Join Date: Aug 2011 Posts: 15

SgtWookie and Bill_Marsden i am really grateful .
#6
12-03-2011, 04:42 PM
 BershaM Junior Member Join Date: Aug 2011 Posts: 15

One more thing if you allow me , sgtWookie.

How does using a diode as you suggested make the duty cycle closer to 50%? ( iam a little curious )

And i've read somewhere that both R1 , R2 shouldnt go less than 2.2k ohms as it may damage the discharge transistor at pin 7 , is that true??
#7
12-03-2011, 07:54 PM
 Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,044 Blog Entries: 5

Quote:
 Originally Posted by BershaM And i've read somewhere that both R1 , R2 shouldnt go less than 2.2k ohms as it may damage the discharge transistor at pin 7 , is that true??
No, it is not true. We have seen a continuous spec of 15ma, which would be 600Ω for 9VDC. It can go higher, and does when a capacitor discharges through it, as with a monostable.

The main reason for larger resistor is as Wookie suggested, wasted power. When pin 7 is grounded you have a resistor that is doing absolutely nothing put pulling electricity. A pure waste. There are other reasons why you might want to make it smaller though, so it is a balancing act.

The diode was not to change the duty cycle, it was to turn off the LED completely. This is not normally a problem, unless you have an older LED that drops 1.5VDC. This is because the output of a standard 555 (not CMOS) has a pair of transistors that will drop 1.3V between Vcc and pin 3 when it is on. I don't know how knowledgeable you are, but they are called a Darlington pair.

I have a cousin of this circuit with my 555 Hysteretic Oscillator article.

The 555 Projects
__________________
..
"Good enough is enemy of the best." An old engineering saying, Author unknown.

General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.

Last edited by Bill_Marsden; 12-03-2011 at 09:43 PM.
#8
12-03-2011, 08:06 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Quote:
 Originally Posted by BershaM How does using a diode as you suggested make the duty cycle closer to 50%? ( iam a little curious )
With the traditional astable multivibrator configuration of R1/R2/C1, the timing capacitor C1 charges via R1 and R2 in series, but discharges to pin 7 via R2 only. The output pin 3 is high when C1 is charging. This means that the charge cycle will always be larger than the discharge cycle, as R1 must be greater than zero.

If a diode is added across R2, then C1 charges via R1 and the diode, and discharges via R2 to pin 7. The values for R1 and R2 are now independent. Before, you had a duty cycle range of about 52% to ~ 98% or so; now it can go from ~3% to ~98%.

Quote:
 And i've read somewhere that both R1 , R2 shouldnt go less than 2.2k ohms as it may damage the discharge transistor at pin 7 , is that true?
I have a "rule of thumb" that R1 should never be less than 100 Ohms per volt of Vcc. That limits R1's current to 10mA or less. It's conservative, but it's good to be conservative - circuits are more reliable when you are conservative.

Pin 7 has internal current limiting to somewhere between 35mA and 55mA if it is the original design. However, if you get too close to those numbers, your timing calculations will be off, and you will be dissipating a fair amount of power in the timer. If power dissipation becomes excessive, you can burn up the timer.

You will find 555 designs on the web that have simply a potentiometer between Vcc and pin 7. This is definitely an oversight, as if the pot is turned nearly all the way down, you will very likely burn up the potentiometer, or damage the 555.
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#9
12-03-2011, 08:09 PM
 Bill_Marsden Super Moderator Join Date: Mar 2008 Location: Dallas, TX (GMT-5 w/ DST) Posts: 19,044 Blog Entries: 5

The problem with this schematic, which is what Wookie is talking about, is it is not very accurate.

With the first schematic, the one you used, there is a strong decoupling between power supply and how the circuit acts. It should be the same at 5V as at 15V.

With this design it is very dependent on power supply, since the diode drops a relatively fixed voltage. It is one reason I don't recommend it more. Certain types of diodes do improve performance. A schottky diode, for example, can drop 0.2V or less with low current, with is much better than silicon 0.6V to 0.7V drop. The equations I show are also skewed due to the diode adding its flavor to the mix, it is not nearly so predictable.

However, R1 and R2 do not interact, each controls the time they are assigned on/off. This is a strength.

I keep these schematics and a few more here...

My Cookbook
__________________
..
"Good enough is enemy of the best." An old engineering saying, Author unknown.

General info:
If you have a question, please start a thread/topic. I do not provide gratis assistance via PM nor E-mail, as that would violate the intent of this Board, which is sharing knowledge ... and deprives you of other knowledgeable input. Thanks for the verbage Wookie.

Last edited by Bill_Marsden; 12-03-2011 at 08:14 PM.
#10
12-03-2011, 08:14 PM
 SgtWookie Expert Member Join Date: Jul 2007 Location: In the vast midwest of the USA; CST Posts: 22,038

Quote:
 Originally Posted by Bill_Marsden No, it is not true. We have seen a continuous spec of 15ma, which would be 600Ω for 9VDC. It can go higher, and does when a transistor discharges through it.
I'm sure you meant to write "when a capacitor discharges through it."

Quote:
 The diode was not to change the duty cycle, it was to turn off the LED completely.
Oops - it WAS to provide the ability to change the duty cycle from almost zero to almost 100, or at least obtain an actual 50% duty cycle when R1/R2 were nearly equal.

Quote:
 This is not normally a problem, unless you have an older LED that drops 1.5VDC. This is because the output of a standard 555 (not CMOS) has a pair of transistors that will drop 1.3V between Vcc and pin 3 when it is on. I don't know how knowledgeable you are, but they are called a Darlington pair.
If LED2 won't go off, you can connect a 1k resistor in parallel with it. That should allow it to go off without affecting the brightness too much.

When the 555 timer output is high, current is supplied via a Darlington voltage follower; so there are the equivalent of two diode voltage drops from Vcc to the output. These "diode drops" are the base-emitter junctions of the Darlington arrangement.
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