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.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??
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.How does using a diode as you suggested make the duty cycle closer to 50%? ( iam a little curious )
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.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'm sure you meant to write "when a capacitor discharges through it."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.
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.The diode was not to change the duty cycle, it was to turn off the LED completely.
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.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.
Thanks for the warning i was about to use a potentiometer instead of fixedYou 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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by Luke James