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Author Topic: Monostable by 2 transistors  (Read 5722 times)

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Marcus

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Monostable by 2 transistors
« on: April 28, 2012, 17:19:43 PM »
Reference: Monostable Multivibrator circuit (http://pcbheaven.com/wikipages/Transistor_Circuits/)

Hi,

I am new in this forum and I'm post my question in this section as suggested by the webmaster.
The monostable timer, realized by two transistord, is only a little part of my project (that involve, 3 or 4 timers working in chain) and everything have to work in a 12V DC environment.
I have found ther two transistors monostable circuit very interesting, because:
1) I can trig by a positive singal
2) it is not retriggerable
3) I can trig start signal for a time longer than the output "high" period without side effects when I remove the trig signal.

The circuit schema, like the picture in the web page metioned above, show the values only for some of the resistors involved, but
1) there is no value(s) explained for R1, R3, R4,
2) no identification for the two transistors (are they BC508?)
3) how is the Vs range for this circuit? As I mentioned above, my project have to work in a 12V DC environment.

So, I need your precious help to determine the values of the resistors and the suitable transistor type.

Thanks in advance

kam

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Re: Monostable by 2 transistors
« Reply #1 on: April 28, 2012, 20:44:53 PM »
Hello Marcus,

let me give you first one advice. The circuit is made with transistor, which means that you need to calculate more than the simple T formula.
I use 2n2222 tranistors, R1 and R3 are 1k and R4 are 10K.

I strongly suggest that you use the same circuit implemented with 555 timer:
555 Circuits

It is more reliable and easier to calculate.

Marcus

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Re: Monostable by 2 transistors
« Reply #2 on: April 28, 2012, 21:33:53 PM »
Hi Kam,

thank you for your reply.

I just seen more than one monostable circuit 555 based, and I see the circut by the link you provide, but... there are two main issues:
1) every 555 based monostable need to be triggered by a transition from "high to low" logic level: no documents say me that "no connection to pin 2 means high level by default".
2) a simple 555 monostable circuit does not implement an important (for me) feature: I need that the trig period could be less, equal or greather than T(high-output) period. I read that, by 555, I can use only T(high-output) period greather than the trig period. The 555 circuit, that implement the behaviour I need, has a lot of external components (I could seen what I write in a schema found in the web): at this moment, I think that is not the best solution for me.

Of course, the R1, R3 and R4 need to be calculated, but I think that the 2N2222 transistors are suitable for 12V DC circuits, don't so?

Anyway, some formulas to calculate R1, R3 and R4 can be very useful for me.

Thanks a lot
« Last Edit: April 28, 2012, 21:36:07 PM by Marcus »

kam

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Re: Monostable by 2 transistors
« Reply #3 on: April 28, 2012, 22:13:41 PM »
Ok i see what you mean.
2n2222 can stand easily 12v. You can start with the values that i provided before. Just keep in mind that you might need to interface this circuit with another transistor at the output. You know how to do this?

Marcus

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Re: Monostable by 2 transistors
« Reply #4 on: April 28, 2012, 23:54:56 PM »
Quote
You know how to do this?

Yes of course! I'll use a BC437 or equivante with a base resistor of 10K, just to drive a mini 12V DC realis, without forget to put a diode in paralles with the realis coil!

Next monday I'll go to buy the components and test them mounted on a breadboard.
I hope will be no issues.

Thank for your support.

kam

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Re: Monostable by 2 transistors
« Reply #5 on: April 29, 2012, 11:22:08 AM »
the BC347 has minimum hfe 40. A 10k resistor will have base current = 12/10000 = 1.2mA, so Ic will be 1.2*40 = 48mA. I'm not sure if this is enough for your relay. Maybe you should provide some 300mA to be sure, which reacquires base current 300/40 = 7.5mA, which can be achieved with a base resistor = 12/7.5 = 1K6. So better use a 1500 Ohms resistor instead.

Marcus

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Re: Monostable by 2 transistors
« Reply #6 on: April 30, 2012, 18:43:15 PM »
Hi,

today I discover and use the circuit simulator LTSpice IV (Linear Technolgy Corporation) and I make some simulation using the value of R2 (1500K) and C1 (1uF) for the time of 1 second. Unfortunately the simulation say that 1500K for R2 is too much for Q2 activation. So I try to reduce R2 to 150K and C1 to 10uF, and it work fine as expected.
Now, is clear for me, that the value of R2 has to be mantained in certain range, even if I don't know which them are.

kam

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Re: Monostable by 2 transistors
« Reply #7 on: May 01, 2012, 11:04:23 AM »
excellent marcus ;) hope this will work for you

Marcus

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Re: Monostable by 2 transistors
« Reply #8 on: May 05, 2012, 19:56:56 PM »
Done, it works in excelent manner, but.... there is a little problem: when I supply or remove the 12V to the circuit, the output goes ON for a very short time (less than 1 second). I think that the cause lies in some overcurrent transient. The next week I'll buy the components for the "classic" 7812 fixed voltage regulator, in the hope that this can make remove the tansients occours when supply or remove the 12V (just for note, the 12V origin is a car battery).

Anyway, any kind of tip is appreciated.

kam

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Re: Monostable by 2 transistors
« Reply #9 on: May 05, 2012, 20:32:27 PM »
maybe it is a start-up problem that won't be solved with a regulator. Maybe you need a circuitry to disable the output for some 2 seconds upon startup. Something like a transistor with a large capacitor and resistor

Marcus

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Re: Monostable by 2 transistors
« Reply #10 on: May 06, 2012, 12:01:51 PM »
Mmh,
find attached the LTSpice simulation result (very near to the real situation).
I modify the 12V source to a PWL, to simulate the supply/remove of the main power to the circuit. The supply start at 0.5 sec and remove at 4.8 sec (green line). The blue line is the external impulse and the red line is the output.
Note that the output has a little peak (about 0.5 Volt) at 0.5 sec and an high peak (about 11 Volt) at 4.8 sec.
So, even if I disable the output for 2 seconds at startup I have no control for the peak at the end.
I'd try to put a little capacitor (0,1uF) between the output and ground: it remove the peak at startup but only reduce the peak to the end.
It's become a "not very little" problem...
Any suggestion?

kam

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Re: Monostable by 2 transistors
« Reply #11 on: May 06, 2012, 23:36:30 PM »
There is generally a problem with such circuits which host capacitors and transistors together. The solution is again the output transistor that you used to control the output to remain off for 2 seconds regardless of the monostable circuit output. This final controlling transistor must therefore have 2 functions: First is that it will shadow the output upon startup and for 2 seconds, and second, it will immediately shadow if no power is applied.


Edit:
You may consider using an AND port for these functions... If one input of the port comes from the transistor circuit and the other comes from the supply, then problem is solved (no supply=no output at all).
Then you use another AND port (from a chip with 4 AND ports) with an RC circuit, Calculate RC to become higher than the HIGH threshold of the input of the port in 2 seconds. This way, upon start up, the output will not appear until the RC becomes higher than the AND input threshold (in 2 seconds).
« Last Edit: May 06, 2012, 23:38:55 PM by kam »