555capacitortimer
As a beginner to electronics I'm having a tough time understanding circuits that have a combination of different components, including the following timer circuit. Its based around a LM555 timer IC. I get the potentiometer and the resistors, but a couple of other things go over my head.
1) why's there a connection from between R1 and R2 to pin 7? Is it working as a voltage divider?
2)Whats the purpose of the capacitors C1 and C2?
As Russell says, your 555 timer circuit looks a bit off. Search for 'astable 555 circuit' to get your square wave generated. There are online calculators to set the values of the resistors and capacitors.
To get 0.2Hz you will need about 50K for R1 and R2 and 47uF for C:
Now the reason that you are just seeing a flash when you changed to a MOSFET is that their gain is so high. So even if you are using your R1 and C1 to generate a ramp, the on/off voltage of the MOSFET is probably crossed immediately.
What you need instead is an NPN transistor that can handle the current that you need for your LEDS. Just find out how much current you are drawing for the LEDs and check the datasheet of the transistor to see if it will work.
Here's a circuit that I simulated that gives a pretty good ramp up and down at 20mA. Not sure how much current you need but that might be a pretty good start.
I used a 0.2Hz square wave instead of the 555 just for simplicity.
To slow down the ramp up/down time increase R1 or C1. Because R1 is connected to the base, this will also effect the brightness of the LED. So it might be better to choose a good value there first and then change C1 for the timing.
What if you put R10 to 10KOhm and inserts an 82KOhm instead of C4? The time between starting and the light switch is controlled by IC7, so the connection from IC7 pin 3 to IC6 pin 4 is only to turn on the light, there is probably used a 555 because it is easy, as IC6 really only works as a flip flop.
Update:
This is from a simulation. Hope it gives you an understanding of the signals.
0 = LED off
1 = LED on
S4 = enabled
S3 = enabled
Yellow = IC7 pin 3
Blue = IC6 pin 4
Red = IC6 pin 3
Green = IC7 pin 6/7
Best Answer
In astable mode, the 555 timer puts out a continuous stream of rectangular pulses having a specified frequency. Resistor R1 is connected between VCC and the discharge pin (pin 7) and another resistor (R2) is connected between the discharge pin (pin 7), and the trigger (pin 2) and threshold (pin 6) pins that share a common node. Hence the capacitor is charged through R1 and R2, and discharged only through R2, since pin 7 has low impedance to ground during output low intervals of the cycle, therefore discharging the capacitor. Also that the circuit configuration above does not permit a duty cycle of less than 50%, because the time-constant for charging C1 is always greater than for discharging. To achieve any arbitrary duty cycle, R2 can be moved to be in series with pin 7, the discharge pin. The duration of the high-output interval (during the charging of C1) is then 0.693(R1C1), and the low-output interval (while discharging C1) is 0.693(R2C1). The total time period, T, is 0.693(R1+R2)C1