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Coin Toss: Easy Question

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Electric Rain

New Member
OK, I’m working on a “Coin Toss” project. And I have two schematics:

The pictures are at the bottom.

# 1 Uses more parts so not only will it be more complicated, but it will cost more too. The only good thing, is that it runs off of 5V and # 2 runs off of 9V. However, that’s even worse in my case, because they don’t have 5V batteries, as we all know. So, I’ll have to use a voltage regulator, which is big, and needs too be heat sinked. (And of course, will be even MORE money.)

# 2 I would much rather use this one, because… well, I won’t go into a big explanation, so let’s just say it’s because the reasons above do not apply. But the ONLY problem with #2, is that I don’t know what transistor to use. I know it has it right there under the transistor it the schematics, but I can’t really find that one. I use Digi-Key, and their catalog doesn’t show part numbers. So can someone tell me a good transistor to use and give me the Digi-Key part number please? Thank you in advance because I know you guys will be able to figure it out. :wink:
 

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Phasor

Member
Here's some tips:

The transistor: Any UJT (unijunction transistor) will do. 2N2646 and 2N2647 are the most common.

The first schematic: You CAN run it off 9V, and you don't need the 7473. Here is a revised schematic.
 

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mechie

New Member
Are you sure?

Phasor
Are you sure your revised circuit will work as intended?
won't it always stop with the red LED lit?
 

Someone Electro

New Member
Relay strange it realy works!? :shock:

I simulated the circut in Crocodile Tehnology and it worked!???
 

kinjalgp

Active Member
Yes, it will work. I don't see any problem with Phasor's circuit except for that 1k resistor. It should be atleast of 5k. While designing, that resistor is always taken greater than 5k and accordingly all other component values are calculated.

In practice, capacitor is assumed first and R1 and R2 are calculated on basis of that. But if R1 comes out to be less than 5k, capacitor value is altered to match the requirements.
 

nettron1000

New Member
Yep , Phasor's circuit works fine, i just tried it out on my prototyping board.

But i made some slight changes ;Perhaps its just the type of LEDs im using, but I noticed that when one LED is on the other LED ,which was suppose to be off, glows dimly. So i replaced the 470 ohm resistors with 1K resistors.

I also replaced the .1 uF capacitor with a 4.7uF, so that you can see the LEDs flashing at a high rate when the pushbutton is held down. This makes it abit more interesting.
 

Roff

Well-Known Member
If you use the "no-latch" circuit, won't the duty cycle asymmetry cause more "greens" than "reds" after many flips? With a CMOS 555, this can be remedied by not using pin 7 or the 1k resistor, instead returning the top end of the 10k to pin 3. The bipolar 555 "high" level on pin 3 does not reach the supply voltage, so asymmetry would still exist.

With either CMOS or bipolar, I suspect that leakage will eventually cause the output to flip on its own. This won't happen with a latch (7473 or 4017).
 

Phasor

Member
If you use the "no-latch" circuit, won't the duty cycle asymmetry cause more "greens" than "reds" after many flips?
That's why I chose the 1k much smaller than the 10k, to make the duty cycer close to 50%. But statistically speaking, yes, there will be more greens.

I suspect that leakage will eventually cause the output to flip on its own
This is true also. Perhaps a low-leakage cap should be used.
 

nettron1000

New Member
Ive revised the circuit again to account for the duty cycle problem. This one will give a nearly perfect 50/50 duty cyle if you match the timimg resistors as close as possible.

I used 10 K resistors for Ra and Rb( metal film type ) and 4.7Uf for C1, D1 and D2 are 1n914 diodes, power supply was 9 volts.

 

professoryak

New Member
nettron1000 said:
Ive revised the circuit again to account for the duty cycle problem. This one will give a nearly perfect 50/50 duty cyle if you match the timimg resistors as close as possible.

I used 10 K resistors for Ra and Rb( metal film type ) and 4.7Uf for C1, D1 and D2 are 1n914 diodes, power supply was 9 volts.

I thought that the equation for duty cycle was (Ra+Rb)/(Ra+2Rb) therefore, by making the resistors equal the duty cycle becomes 66.7%, the original design would have given a duty cycle of 52.4%, surely much better
 

Hero999

Banned
This is an old thread but your post is usefull so it doesn't matter.

You really need 50% duty cycle which could be done with a CMOS CD4060 timer oscillator and counter IC.

If you want a cheap circuit then try this circuit, if you use close tollerance capacitors and resistors then the duty cycle should be near enough 50% not to matter.
 

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