help with digital dimmer

Tsalagi

Mar 15, 2010
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Now I'm trying to make a digital dimmer I found in a known magazine. The thing is that the circuit in the magazine is for 230vac/50hz. My house mains are 110vac/60hz. What changes will I need to rework in the circuit? What calculations will be necessary and how can I make them. Will this affect the firmware? See tha attached schematic. Thanks in advance for your help, electronics friends.
 

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Resqueline

Jul 31, 2009
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Depending on the software in the PIC and the power supply reserves you might not have to change anything at all.
You may however have to rewrite the software to suit the 8.33ms timing instead of 10ms.
Hardware changes: you might possibly need to increase the size of the 220nF cap.
A capacitor that will provide the same current: 220/110*230/1.2=383nF (use 390nF or less).
 

(*steve*)

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That's from an article in Silicon Chip magazine.

You can email a letter to the editor and any which are published appear on their web site (the main articles are behind a paywall though)

If the code is available, it too is *IN FRONT* of the paywall - i.e. you don't have to pay.

If I can find more information for you I'll edit this post.

Edit:

OK, if this is the device, then this is the PIC code.

They include source. You will need to make changes for 60Hz, I would suggest changing the values for

Code:
PULSE_O		equ	H'23'	; operational pulse value
DIVID_O		equ	H'24'	; operational divide value

and would draw your attention to the following code:

Code:
; POSITION counter timer 250 counts = 10ms (count 20 is 14 degrees for max brightness and count 
; 220 is 158 degrees or min brightness start point. 

COUNTER	

	incf	POSITION,f	; position in half wave starting from zero crossing point

; adjust interrupt rate with counter

	movf	DIVIDE,w	; divide value
	movwf	DIVID_O		; operating divide value

Even if you make changes here, it may still be worthwhile contacting them as there may be other modifications they might suggest (particularly in the power supply area)
 
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Oh, it's VITALLY important that you note the different designations of the Earth and ground symbols and be aware that they're not connected!
 

Tsalagi

Mar 15, 2010
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Thanks, Resqueline, for your answer. But can you please help me understand the numbers in your formula to get the cap value? Thanks again
 

Tsalagi

Mar 15, 2010
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OK, got it. Thank you for your help, Steve. I'll take your advice and try to contact them.
 

Resqueline

Jul 31, 2009
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220nF*(230V/110V)-20%
The voltage has more than halved (110/230=0.4478) so the capacitance has to more than double (230/110=2.09).
But then the frequency has increased by 20% (60/50=1.2) so the capacitance has to decrease by 20%.
You may very well get away with a 330nF cap. if the original design has some built-in reserve.
Capacitor Current = Voltage / (1 / (2 * pi * frequency * capacitance)) = 16mA (of which only half is available due to half-wave rectifying).
 
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