A sleep Timer

[walid1]

Hi
The following circuit is function as a sleep timer:

from: http://www.redcircuits.com/Page4.htm
I edited it and put the parts values and some other information

The writer wrote the following about it:

{After turn-on by P1 pushbutton, the LED illuminates for around 25 minutes, but then it starts to blink for two minutes, stops blinking for two minutes and blinks for another two just before switching the lamp off, thus signaling that the on-time is ending. If the user want to prolong the reading, he/she can earn another half-hour of light by pushing on P1. Turning-off the lamp at user's ease is obtained by pushing on P2.}

I calculate the clock pule duration T = RC/0.455 = (1M*100n)/ 0.455 = 0.22 sec
Now the clk o/p (pin 9) is ON - OFF at rate of 0.22 sec
And according to the values at the above schematic (the colored one), I think that the LED is illuminate from 0 sec up to 15 minutes, then it blinking for anther 15 minutes before the lamp off.
WHO IS TRUE THE WRITER OR I'M?
thank you in advance

 

[walid1]

Hi

Three days and there was no answer or response, this means that what I said is true and the original writer of the article was wrong.
I have another question:
We know that Pin 3 of 4060 turns from Low to the High after half an hour.
This High signal is applied to the emitter of Q2 through C3
The subcircuit composed of Q1 and Q2 and related components is responsible for connecting or disconnecting the operating voltage from the rest of the electronic circuit.
Can you please explain the mechanism of how these two transistors do their job.
thank you very much.

 

[audioguru2]

Hi Walid,
Sorry, I didn't calculate the CD4060 timer.

The two transistors form a thyristor. The push button turns on Q2 which turns on Q1 They are connected together so both stay turned on. Applying a positive pulse to the emitter of Q2 turns it off then Q1 no longer has anything to keep it on so it also turns off.

 

[walid1]

Hi guru
First, thank you very much

The two transistors form a thyristor

A thyristor has 3 terminals; A1, A2 and G. Can you please mark these terminals on the schematic (Q1 & Q2).
I think they are as shown in the following fig.thyq1q2-001

In the above Fig, and if my idea is true, this thyristor is triggered when a 0 volt signal is applied to its gate.
when a high or a VCC signal is applied, it is turned OFF.
please guru correct me.

The push button turns on Q2

there are two push button, P1 and P2, I think you mean P1 which when we pushing it we triger or RESET the counters to begin counting.
pin3 of 4060 o/p is zero for 30 minute, and so the thyristor is conducting. at the end of the 30 minutes, pin3 go HIGH so turn off the thyristor.

In fact I try to understand the mechanism of turning this thyristor ON and OFF but can't. I used SWCAD III but still unclear. please guru explain it for me.

The push button turns on Q2 which turns on Q1 They are connected together so both stay turned on. Applying a positive pulse to the emitter of Q2 turns it off then Q1 no longer has anything to keep it on so it also turns off

Before pushing P1, the lower terminal of C3 was float.
When pushing P1, this applying a negative pulse to the emitter of Q2 turns it ON which turns on Q1.
After 30 minutes, the o/p of pin3 is changed to HIGH this applying a positive pulse to the emitter of Q2 turns it OFF which turns OFF Q1.
OK, I can keep this in my mind, but i'm sure that i didn't understand how this be done. I'm waiting hope GURU explain it.

the last point: Is C2 (=100n) is a part of the thyristor equivalent circuit.

thank you GURU very very much.

 

[audioguru2]

Hi Walid,
Pushbutton P1 is connected to the positive supply. When it is pushed, it applies a positive voltage to the base of Q2 through D3. Then Q2 turns on and it turns on Q1. Q1 keeps Q2 turned on and Q2 keeps Q1 turned on.

A positive pulse to the emitter of Q2 turns off Q2 and then Q1 also turns off.

P2 turns off Q1 then Q2 also tuns off. I don't know why C2 is there.

 

[walid1]

At the outset extend deep thanks to Mr. AUDIOGURU for his effort and interest and wonderful answers.
In truth, I thought the circuit was working in a different way, but now the picture clarified somewhat.
Still I have some questions:
(1)I am confident that your true 100%, but I can not imagine. Note the following figure: 

please comment.
(2)

Pushbutton P1 is connected to the positive supply. When it is pushed, it applies a positive voltage to the base of Q2 through D3. Then Q2 turns on

I know that: to turn on the npn transistor, its collector must be connected to +ve batt (through a resistor) and its emitter to ground, then when applying a voltage to its gate it passing current from C to E.
In our Q2 its collector didn't connected to any Vcc, so how to be ON.
(3)

I don't know why C2 is there.

Are concluded from this talk that it could be removed from the circuit without any difference?
thank you guru

 

[audioguru2]

I know that: to turn on the npn transistor, its collector must be connected to +ve batt (through a resistor) and its emitter to ground, then when applying a voltage to its gate it passing current from C to E.
In our Q2 its collector didn't connected to any Vcc, so how to be ON.

The emitter-base of the PNP transistor is a forward-biased diode from the positive supply to the collector of the NPN transistor.

Are concluded from this talk that C2 could be removed from the circuit without any difference?

I think so. It might keep Q1 turned off when the circuit is first powered.

The "gate" supplies power to the IC because Q2 is an emitter-follower and its base and emitter rise to almost the supply voltage.