I'm having issues with a CanaKit. I need to buy a transformer for it. I didn't realize that the kit didn't include it. The kit in question is the CK185 - 3-Channel IC Colour Organ with Built-In Mic and Line-In. It requires a transformer - 18V AC C.T. / 100 mA. I'm searching through DigiKey and cannot find a transformer with these specifications. Can anyone suggest a transformer I could use (from Digikey preferably)? Thanks in advance!
The Jameco wall wart does not have a C.T. (center-tap) so it cannot be used.
The required transformer has two outputs with a peak of 12.7V. Use a 12VDC wall-wart.
Buy a wall-wart with an output current of at least 300mA so the voltage isn't too high.
A 100mA wall-wart might have a voltage higher than 24VDC at the low current used by the project. But maybe Cana-kit wants the circuit to operate at 24VDC.
Thanks for the help everyone!! I did a little more searching online and found a place that sold the kit with a wall wart (9V 300mA).
With my limited knowledge in electronics I l decided to look at the circuit diagram. It looks like it converts the incoming AC signal into DC. It literally connects both "taps" together. I'm not sure what the theory is behind this but whatever the case I tested the circuit with a 9V 300mA wall wart and it works just fine. All I had to do is connect the positive lead of the wart to one of the "taps" which left the other "tap" unconnected.
I don't like guessing at these sort of things. Could someone explain to me how to calculate the DC voltage and current when converting from AC to DC? In this case the diagram calls for a 18V AC C.T. / 100 mA transformer. What does the Voltage and Current end up being when you convert that to DC?
An 18V C.T. /100mA transformer is tiny and cheap. It has very thin wire that has a high resistance. So it might have an output that is much higher than 18V (maybe 36V) at no load or a low current load. Its resistance drops the output to 18VAC when it has a 100mA load.
Each side of the center-tap produces 9VAC. The peak of 9VAC is 12.7V. A rectifier diode will drop the voltage to 12VDC pulses and a filter capacitor will smooth the pulses into 12VDC.
If you use a 9VDC adapter then the circuit gets 9V -0.7V (one rectifier)= 8.3VDC and still might work.
According to that guide the VDC will be approximately 0.45x18VAC = 8.1VDC (minus the rectifier or course). Does this make any sense or did I completely miss the boat?
According to that guide the VDC will be approximately 0.45x18VAC = 8.1VDC (minus the rectifier or course). Does this make any sense or did I completely miss the boat?
The average voltage occurs when the filter capacitor is removed. Then the output pulses from 0V to the peak of 12.7V over and over.
DC is supposed to be filtered and smooth. Then the filter capacitor charges to the peak minus a diode drop and makes 12VDC.