soft start help

Hi All

I have this simple soft start scheme which I modify to include a LED. My question is the resistor 220R sufficient for the LED?

Thanks
 

Attachments

  • delay with LED.jpg
    delay with LED.jpg
    47.8 KB · Views: 284
Measure the voltage of the power supply before you install the LED.
Make sure that the relay coil is in the circuit or the voltage will be very high.
What is the d.c. resistance and operating voltage of the relay coil? That will determine the resistance of R1 and R2.
I'm guessing that 20k is too much resistance for the LED.
 
Hi Frank

Below is the spec:
R1, R2*.............................470K 0.25W 1% mf
R3*..................................220R 0.25W 1% mf
R4, R5, R6, R7...................10R 5W
C1...................................330n 250V (for direct mains connection)
C2, C3............................. 470uF 40V
B1.................................. B250C1500
Re1.................................24V (250V-8A contact)
F1*..................................2A slow blow
 

metal

Member
2004-01-29 5:25 pm
I see, then we have to calculate the bypass resistors values before we use the circuit because resistors might end up flamed if wrong values used. For a transformer, the power rating is enough to be used for these calculations?

Also what happens if I use 1000uF/50V capacitor instead of two 470uF/40V capacitors?
 
Pm,
be very careful with this circuit.
It is connected directly to the mains.
Don't touch any parts while it is plugged into the mains supply.

The 470k+470k do not supply much current.
If your mains is 230Vac and you use 24V for the relay drive, you have 215Vac to drive current through the 940k, I=V/R = ~0.23mA

The main current to drive the circuit comes through the capacitor, it must be X rated.
Find their AC impedance at your mains frequency and using the same formula find out how much AC current passes the capacitor.
Is that sufficient to drive the LED and to drive the relay?