I assume that the transformer is at least 10VAC/0.5A so that the input to the 9V regulator is at least 11V.
If 300mA flows through the 30 ohm resistor then the output from the resistor will be only 0V.
Why do you use the 30 ohm resistor to waste the entire 9V? The 30 ohm resistor is simply a 2.7W heater.
What is the load? A DC motor?
I assume that the transformer is at least 10VAC/0.5A so that the input to the 9V regulator is at least 11V.
If 300mA flows through the 30 ohm resistor then the output from the resistor will be only 0V.
Why do you use the 30 ohm resistor to waste the entire 9V? The 30 ohm resistor is simply a 2.7W heater.
What is the load? A DC motor?
3 W will give you about 250 mA max and 6 W will give you about 500 mA max. So if you have a transformer capable of 12 VAC @ 500 mA the 9 volt 7809 regulator should work fine.
3 W will give you about 250 mA max and 6 W will give you about 500 mA max. So if you have a transformer capable of 12 VAC @ 500 mA the 9 volt 7809 regulator should work fine.
I go to my st. viny store and went through the used ac adapters for any voltage and current. be sure to read the info on the unit. you may have to o-scope the output to view the ripple. there is a rule of thumb for the amount of cap. needed. do not recall the ratio/ripple. this might be late. just signed on. hope this help is still viable?
Something that wasn't mentioned was to make sure you use a heatsink for the regulator .....
Taking a 12 volt AC, rectifying it to DC with a full wave bridge will give you 16.92 volts. Dropping this to 9 volts at 300ma will mean you need to dissipate around 2.4 watts of power as heat. It may not seem a lot but things can soon start to get hot !