Power for the Spinneret
David Betz
Posts: 14,516
I'm finally getting around to powering up the Spinneret that I bought ages ago and I'm trying not to fry it. The documentation says that you can supply between 4 and 9 volts to J6 pin 2. I have a wall wart that says it outputs 9 volts but when I measure it not connected to any load I get 14.55 volts. Is there any chance this will damage the Spinneret or will the in-circuit voltage drop down to the 9 volts that the wall wart says it produces? Secondly, is there any danger in powering the Spinneret at the extreme 9 volt end of its range or is it better to choose something a bit lower?
Comments
However- this is a linear regulator so the higher the input voltage the more heat it will disperse. So the best rule of thumb is the lower the better. I would suggest 5V. I use cell phone USB changer wall warts that I find at garage sales for $1 or less. These are well regulated at 5V and usually around 500 - 1000 mA.
Hope that helps.
The NCP1117-3.3 regulator spec says up 20V input and the 10uF capacitor on the input is listed in the BOM as >=25V.
That said, if the voltage stays above 9V with the circuit connected I would change to a lower voltage wall wart.
Main issue is power dissipation of the regulator at high input voltages.
C.W.
I'm wondering why the 10V out with a 6V spec?
Mike- it all depends whether or not the power supply is regulated. It may only be a transformer, bridge rectifier and a capacitor in the 6V wall wart. The 10V is the OCV (open circuit voltage) or no-load voltage. It's saying that up to 1000mA the voltage will probably go down to 6V whereas the chargers for cell phone are a switching regulated power supply so they are regulated to 5V. Measuring 4.99 is ok
The label should indicate they type of supply it is.
-Phil