I read this post about Dumbles swithing power supply scheme by funk:
"Dumble changed it many times over the years, but the power supply described in the "ODS Notes" described a 12V transformer, full wave bridge rectified, into a 1000uf decoupling cap, an NTE 966 3 pin 12v regulator, and then another 1000uf decoupling cap. that b+ supply is then split 3 ways: 1 goes to the + side of both relays, the other 2 each go through a 1K resistor to supply the LED's. "
I thought, from my lame Electrical Engineering (minor and a major in math) training, that a voltage regulator wants around 2V higher than the expected output of the regulator.  In the circuit above the regulator will be fed 10.6V due to the diode drops.  Does it work because the switches only need around 9V to do their thang?  I'm just trying to understand a little more instead of copying so much.
			
			
									
									
						Voltage regulators and switching
Moderators: pompeiisneaks, Colossal
Re: Voltage regulators and switching
Errr... read those school books carefully, partner.tonejunkie wrote:I read this post about Dumbles swithing power supply scheme by funk:
"Dumble changed it many times over the years, but the power supply described in the "ODS Notes" described a 12V transformer, full wave bridge rectified, into a 1000uf decoupling cap, an NTE 966 3 pin 12v regulator, and then another 1000uf decoupling cap. that b+ supply is then split 3 ways: 1 goes to the + side of both relays, the other 2 each go through a 1K resistor to supply the LED's. "
I thought, from my lame Electrical Engineering (minor and a major in math) training, that a voltage regulator wants around 2V higher than the expected output of the regulator. In the circuit above the regulator will be fed 10.6V due to the diode drops. Does it work because the switches only need around 9V to do their thang? I'm just trying to understand a little more instead of copying so much.
 As for the engineering part, when you rectify 12VAC you need to muplitly that number by 1.4 (SQRT 2) to get the peak voltage.  The math then says that 12 x 1.4 is about 17 VDC.  Subtract the diode bias voltage drop and you still have plenty of headroom for the regulator to work with.
   As for the engineering part, when you rectify 12VAC you need to muplitly that number by 1.4 (SQRT 2) to get the peak voltage.  The math then says that 12 x 1.4 is about 17 VDC.  Subtract the diode bias voltage drop and you still have plenty of headroom for the regulator to work with.Cheers,
Gil
