is it okay to use an X1/X2 film cap in this circuit...

[jarthel]

http://mitglied.lycos.de/Promitheus/delay.jpg

I'm looking to use the film cap in place of the .33uF 250V cap.

thank you for the reply

 

[smithy666]

You should use an X1 or X2 rated cap for this circuit. BTW, this is from Elektor. I have used this successfully for years in an amp. If you are interested I have a PCB for this which also incorporates a remote power on circuit (you can use any switch to power on). Let me know if you want more info.

 

[jarthel]

You should use an X1 or X2 rated cap for this circuit. BTW, this is from Elektor. I have used this successfully for years in an amp. If you are interested I have a PCB for this which also incorporates a remote power on circuit (you can use any switch to power on). Let me know if you want more info.

quite interested. send me a mail.

Thank you

 

[smithy666]

drop me an e-mail @ atomicnirvana@gmail.com and I'll send you schematic + pcb layout

 

[jarthel]

is the cap value across the 2x 470Kohm important?

thanks again

 

[smithy666]

Yes, crucial. Its impedance is about 10k at 50Hz, which drops most of the mains voltage, effectively reducing it to around 24V (less the diode drop and small drop from the series resistor) to drive the relay coil.

 

[jarthel]

email sent.

thank you.

 

[jarthel]

Yes, crucial. Its impedance is about 10k at 50Hz, which drops most of the mains voltage, effectively reducing it to around 24V (less the diode drop and small drop from the series resistor) to drive the relay coil.

so in the elektor circuit, a 24V relay is used?

thanks again

 

[smithy666]

Yes, it is a 24V relay. I've sent through the files to you

 

[jarthel]

Yes, it is a 24V relay. I've sent through the files to you

can you tell me (or point me to guide) on how to calculate the resulting voltage by changing the value of the 330nF cap?

 

[smithy666]

The impedance of the cap is determined by 1/(2*pi*f*c), where pi=3.141, f=50Hz and c=capacitance in farads (you need to divide the 0.33uf by 1000000). For 0.33uF, this produces an impedance of about 9.6k. Calculate the current drawn by the relay coil (24v coil divided by whatever the coil resistance is), then use ohms law to calculate the voltage drop across the cap. By changing the value of the cap it changes the impedance and hence changes the voltage drop for a given current.

 

[jarthel]

The impedance of the cap is determined by 1/(2*pi*f*c), where pi=3.141, f=50Hz and c=capacitance in farads (you need to divide the 0.33uf by 1000000). For 0.33uF, this produces an impedance of about 9.6k. Calculate the current drawn by the relay coil (24v coil divided by whatever the coil resistance is), then use ohms law to calculate the voltage drop across the cap. By changing the value of the cap it changes the impedance and hence changes the voltage drop for a given current.

I assume the 9.6K value is then paralleled to the 2x 470K to get the final resistance equivalent?

why not just use resistance to drop the voltage?

thanks again

 

[mzzj]

I assume the 9.6K value is then paralleled to the 2x 470K to get the final resistance equivalent?

why not just use resistance to drop the voltage?

thanks again

Using only resistor would result pretty high power dissipation for small circuit like that (around 5W for 230v operation)