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# LED FLASHER

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This circuit from http://www.uoguelph.ca/~antoon/circ/flash2.htm is a led flasher, my question is what is its frequency, write the expression used to calculate it? at what maximum freq it can operate? thanks ##### Share on other sites This circuit from http://www.uoguelph.ca/~antoon/circ/flash2.htm is a led flasher, my question is what is its frequency, write the expression used to calculate it?

I'm sorry I don't know, the frequency won't be very predictable, it will vary with the power supply voltage, the Hfe or the transistors, the turn-on voltage of the transistors  (both are also temperature dependant)

If you want more stability then why don't you use a 555 timer or even a quatz crystal oscillator?

at what maximum freq it can operate? thanks

In theory it depends on the Ft of the transistors, but in practice relaxation oscillators are fairly low frequency devices and as far as I'm aware most designs limited to <3MHz. For example the normal 555  is limited to just 500kHz but the TS3V555 will work up to 2.7MHz when run from 5V - see datasheet.

If you want higher frequencies then use LC oscilators which are easy to build >300MHz - the highest I've managed is 2.5GHz but it wasn't stable, however you can buy RF oscillator modules in this frequency range.

TS3V555.pdf

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I calculate the flash rate to be 2.012 Hz.
The attached image and formula was used to calculate the flash rate.
Reducing the value of the 10uF capacitor and increasing the value of the 39k
resistor will increase the flash rate. ##### Share on other sites Well you learn something new everyday, I alway thaught the frequency was quite unpredictable as it has altered when I adujusted the power supply on circuits I've built, just as few questions:

How did you come up with this? Did you find it some where on the Internet?

What about the 2ln(2) term? Where does it come from? Is it a constant? Would it change if you used MOSFETs or darlingtons in this circuit?

Are the power supply, gain of the transistors and base threasholds factors or are these just myths?

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A supply voltage more than about 8.5V might mess-up the timing of the multivibrator if the LEDs are only 1.8V, because the capacitors drive the bases negative to a voltage exceeding the absolute max of 6V rating of the transistor, when the reverse-biased base-emitter will have avalanche breakdown similar to a zener diode. When I worked for Philips, I was told to never allow the base-emitter of a transistor to reach avalanche breakdown.

Without LEDs then the supply voltage restriction is at only about 6.7V.
A diode in series with each emitter to ground will prevent avalanche breakdown. ;D

BTW. Increasing the value of the 39k resistors will reduce the flash rate. If you want a faster flash rate then decrease the values of the capacitors, 39k resistors or both. ;D

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Audioguru is correct, I was having a brain cramp.  Increasing the value of the
base resistor(39k) will decrease the flash rate.

The 2ln term actually means 2 times the natural log of the result of  the  RC calculation.
The circuit and formula is from my notes.

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First let me apologize to Alun and others for any confusion my post of the
formula to calculate the freqency.  As it turns out the formula is incorrect.
When I did the calculation I got a negative number.  I knew that could not be
right so I ignored the - sign.
I did some checking and find that the calculation is a little more complex, but
the attachment shows a general formula that is extremly close when the Vcc is
9V.
Audioguru is correct about the reverse-biased transistor.  In the case of the 2N3904 my current data books did not list the maximum negative bias for the subject transistor, so I actually measured the avalanche and found it to be
very close to 7.5 volts.  Actually measuring the - bias of the circuit my scope
showed 5.2 volts. ##### Share on other sites Thank you for your contribution, where did you get 0.602 from? it looks familular but I can't remember where I've seen that value before, is it 1RC time constant?

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this might help a little if R1=R2=R and C1=C2=C the multivibrator is then symetrical and generates square waves of equal mark space ratio then the period of oscillation is given by the formula T=1.38RC  therefore the free running frequenct of such a multivibrator is given by the formula

F=1/T
=1/1.38RC

1.38 is 2 x 0.69 where 0.69 is the log    Vo/Vt

where Vo= initial voltage and Vt is the voltage after a time t

THIS EQUATES TO

t=RC  log Vo/Vt

I hope this sheds some light on the subject

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• 5 months later...

I am playing with this exact same circuit.

May I ask if somebody could be kind enough to give me a brief explanation of what is happening the moment the circuit is turned on?

I have a brief idea of whats going on but im not sure if its correct.

Is R1 used for current limiting and also as part of the RC time circuit with C1?
Is the path to ground for charging C1 through the base of Q2? I thought the transistor would not turn on until approx .7 volt is at the base. Will there still be a slight leakage of current to emitter when the base is below the .7v?
Also what happens to the voltage stored in the caps when they have stopped charging, does the voltage pass through the collector to emitter?

Thanks
Graham

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Is R1 used for current limiting and also as part of the RC time circuit with C1?

Yes and no. R1 limits current through the LED and determines the very fast charge time of C1. The slow RC time that C1 is discharged by R3 plus the RC on the other side determines the timing.

Is the path to ground for charging C1 through the base of Q2? I thought the transistor would not turn on until approx .7 volt is at the base.

When C1 has finished discharging by R3 then it begins charging in reverse by R3 when Q2 turns on.

Will there still be a slight leakage of current to emitter when the base is below the .7v?

Only if the base voltage approaches its breakdown voltage of about -6V.

Also what happens to the voltage stored in the caps when they have stopped charging, does the voltage pass through the collector to emitter?

When C1 is fully charged by R1 then Q2 is on and Q1 is off. R3 keeps Q2 on until C2 has finished slowly discharging by R2.

A guy on another site made this circuit and it stayed with both LEDs on. He used only a 3V battery.
The fix was resistors to the positive supply from the positive wire of each capacitor to allow them to charge to a higher voltage.
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so the time constant for one side would be r*c = 470*10uf = 4.7 ms ? The components affecting the current would be the led, capacitor and resistor? if this was flowing through c1 and q2 was turned on then would that be the case as if q1 was off then would you need to consider that in the path?

r1 is connected to the pos side of c1 and current would flow and charge c1, what affect would r3 have on the neg side c1? would this affect the time constant/current flowing through c1? and how do you determine the saturation current for a transistor eg bc108?

is it the case that across an led that is not on there is a voltage drop of approx .7v then when it is turned on approx 1.4v drop, for a standard led?

how would you solve for the impedance of this circuit or would that not apply to this? would this be classed as an ac circuit with the current changing direction?

so many questions I know but once I get my head around how this works properly then I dont think I will have many probs understanding similar circuits.

many thanks

graham

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so the time constant for one side would be r*c = 470*10uf = 4.7 ms?

No.
The capacitors each charge very quickly. Then when a transistor conducts, the positive terminal of the capacitor goes to the +0.1V saturation voltage of the transistor (look on the datasheet) and the negative wire of each charged capacitor is driven way below ground to a negative voltage, then is slowly discharged by the base resistor. It is the slow discharge times that are added that determine the oscillator's frequency, plus the negligible charge times. It is a little more complicated to determine a capacitor's discharge time constant because it stops when the transistor's base conducts that is connected to it.

how would you solve for the impedance of this circuit or would that not apply to this? would this be classed as an ac circuit with the current changing direction?

This circuit is two capacitors that are charged and discharged by DC voltages. The impedance of an ouput coupling capacitor is about the only impedance that must be considered.

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