IC controlled emergency lamp with charger

[walid1]

Hi all
In the attached file below u can find the circuit diagram of the IC controlled emergency lamp with charger
my question is what the purpose of using D5 and D8, what they do in the circuit??

thanks

 

[audioguru2]

Hi Walid,
The article says that D5 adds to the zener's voltage for a 6.9V reference which is higher than the battery.
D8 bypasses R4 when the 555 timing capacitor charges so it charges quickly.

 

[walid1]

Hi guru

The article says that D5 adds to the zener's voltage for a 6.9V reference which is higher than the battery.

I am very sorry, I did not notice it. How wonderful it for a zener non-existent

D8 bypasses R4 when the 555 timing capacitor charges so it charges quickly.

My understanding is that the cap charged through only R3, but now I do not know how dischargting, or to any point, is it to  pin is 6 or 7 of the 555 IC.
and why he do so, what shape of the o/p pulse he want? please explain it for me.

 

[audioguru2]

The Mosfets are a push-pull amplifier that drive the transformer. They work best if the waveform is a 50-50 square-wave.
A 555 without the diode produces a rectangular waveform because the capacitor is charged slowly by R3 and R4 in series, but is discharged quicker by only R4. The diode shorts R4 for most of the charge time so that the charge time is the same as the discharge time.

 

[walid1]

Hi my best guru

They work best if the waveform is a 50-50 square-wave.

I can't discribe my feeling, you really the MAESTRO of this great site
u make the information appeared in the photo below, you make it very clear.
thank u guru, I wish you all the best in the world
yours walid

 

[walid1]

Hi guru

The diode shorts R4 for most of the charge time so that the charge time is the same as the discharge time.

unless R3 = R4 the charge time will not = the discharge time
in the circuit R3=820 and R4=1000, why not equal?

thank you

 

[audioguru2]

Hi Walid,
Without the diode a 555 will not have equal charge and discharge times.
The capacitor charges through R3 plus R4 which is a higher resistance that the discharge resistance of only R4 by itself.
With the diode then the capacitor begins charging slowly by R3 plus R4 until the voltage across R4 (and across the diode) reaches about 0.7V. Then the diode conducts and only R3 charges the capacitor quicker.

Since part of the charging time is slower than the discharge time by only R4 then the value of R3 must be less than the value of R4 for the charge time to be equal to the discharge time.