B
Ben Gustave
- Jan 1, 1970
- 0
This high voltage is due to the fact that the circuit is unloaded. When you
hook up the pump the voltage will drop, according to the current draw of the
pump(and the capacitance of the cap). Chances are you don't have a big (10+
watt, 25 ohm) resistor to load the supply while you test for voltage (if you
do have something like that, run the numbers through P=E^2/R to see if it's
too small). If you hook the resistor up and test the voltage, you'll see
what I'm talking about. If you do, be careful, because the resistor will get
hot if it's powered for a while.
OT: I absolutely hate how much $$ I can waste at Radioshack (The Source,
here, actually) during a project....absolutely despise that place.
-ben
hook up the pump the voltage will drop, according to the current draw of the
pump(and the capacitance of the cap). Chances are you don't have a big (10+
watt, 25 ohm) resistor to load the supply while you test for voltage (if you
do have something like that, run the numbers through P=E^2/R to see if it's
too small). If you hook the resistor up and test the voltage, you'll see
what I'm talking about. If you do, be careful, because the resistor will get
hot if it's powered for a while.
OT: I absolutely hate how much $$ I can waste at Radioshack (The Source,
here, actually) during a project....absolutely despise that place.
-ben