I'm guessing the clock uses about a hundred milliamp although it has six 7 segment displays and I'm sure the clock can handle up to 10v safety since it uses CMOS ICs. It seems to be a phenomenon that the clock would use less than a hundred milliamp considering the use of the 6 displays for Hour, minutes, and seconds but anyway this would be irrelevant with the main power off and the clock pulse source uses about 20 milliamps with a transistor inverter and diode OR gate using about another 25 milliamps along with a couple of 7555 timers in monostable mode used for increased pulse width because there was some clocking problems. I'm guessing on battery the clock will use about 50 milliamps max. Sorry I don't have a completed schematic to post. I like your circuit but it's a little too complicated for me.1. How's the draw current of the Clock?
2. Are you sure that the Clock can be supply up to 9V?
3. The circuit used Schottky diode to reduce the voltage drop to 0.3~0.4V.
4. The Output current setup to around 50mA, R2 and R5 use 2K then the output current will be about 20mA.
5. Q1 used for cutoff the base current of Q2 when the city power is apply to the adapter and output 5V.
6. If you have try this circuit then you can adjust the R4 more higher to try does the Q1 could turn on to cut off the base of Q2, because it will draw the current from battery when Q2 doesn't turn on, so you can make R3 and R4 as higher as you can.
View attachment 162510
I'm not sure what a Schottky diode does but I am sure the Clock will safely operate up to 10v. 4.5v might be too low.Will the Schottky diode drop of about 0.5V for the 5V supply be a problem for your circuit?
I fear that switch over to battery reaction time would be too slow with a relay although I could probably put a 470 microfarad capacitor from the clock power to ground to cater for the switch over time. Thank you for your effort.
I attached another circuit which is much simpler for me.Will this circuit switch to battery with a power failure. I plan to power my clock. Would D1 overload the base of Q1? The battery is 9v and the power supply is 5v.
View attachment 162498
The circuit should cut off the common cathode display with a power failure to conserve battery
I was just going to suggest that!! CrutsChow is brilliant indeed.Instead of using a 9V battery, why not use four AAA batteries or two 3V lithium cells (such as CR2032) to give you 6V.
That way you don't need so many diodes to drop the voltage for your diode isolation scheme.
I used 4 diodes with 0.8v drop each and a 9v battery giving 8v. Adapter out was 6.6v and the clock power out was 5.7v. With a used alkaline battery at 8v the clock power out is 5v. Good, but the count was sporadic as I interrupted battery power by applying the adapter power of a prototype mod 60 counter as a test. I think I need a 5v regulator to regulate adapter power and battery. I think the 7805 regulator has low power loss at no load (5mA?)
by Duane Benson
by Jake Hertz