I could use some help figuring what resistor to use to drop auto voltage (13 to 14) down to 8.8 volts to run an electric alarm clock by bypassing the 110 input and feeding power direct to the circuit board.
Fred

 

You will need to know about how much current (amps) the clock needs. Once you have that, use ohms law to figure out how much resistance will be needed. Let's say the clock needs 0.3 amps (300 mA), and you want to operate 5 volts less than supply, then 5/0.3 = 16 (easy to find value = 18 ohms). It may get kind of warm, that is 5 x 0.3 = 1.5 watts. Make sure you use a 3 to 5 watt resistor.

Or, you could use a 3 pin voltage regulator LM7809 to give you 9 volts,
Or you could use an LM 317 (variable output) and a few resistors to get exactly 8.8 if that is what you want.

Actually, if you look a little deeper into your circuit, the 8.8V supply is likely dropped down to 5 v with a LM7805. If that is the case, you can feed that chip 8.8v or 12 v or 15v with no problem except for heat that needs to be released as the wattage of dissipation (as calculated for the resistor wattage above). Make sure the LM7805 (05 means 5 volts on this chip and the 09 means 9 volts on the LM7809 suggested above).

I assume you are just working with a 12 v battery power supply and charger since vehicle modifications and corresponding discussions are strictly forbidden on this site and any sniff of automotive will result in the thread getting locked.

Good luck.

 

The only info it gives is 120 volt at 2 watts for the power supply

 

a loot of electric clocks use the ac line frequency for timing. 60 hz devided down to 1 hz for seconds on clock, 50 devided down to 1 second and such. clocks with motors use synchronous motors locked to the line frequency. a dc supply might not work.

 

The only info it gives is 120 volt at 2 watts for the power supply

I was thinking you would be committed to the project and clip the wire to then use the amperage setting on your meter to complete the circuit so you could measure amps.

On the other hand, 2 watts is two watts. 1.5 may be heat (8.8 volts brought down to 5v) and 0.5 may be to drive the clock circuitry and display. Start with a 18 or 22 ohm resistor if you want to try it.

Otherwise, another clock can be obtained from Goodwill for a buck or two.

 

Just tried pluging 9 v dc battery to the wires from transformer and got what looked like an error display with polarity one way just the two dots between hour and min lit up, with polaity reversed the display read 1 L 7.
So my thoughts of an easy reroute are for naught. Any suggestions

 

unless your clock has a battery to keep the tme running when the power is off, it cant be done that way.

 

It does have a battery backup, but does not light display. So I would think if we can figure out to power the display this could work.

 

Back to the original question, The clock draws 85 mA and a 5 volt drop would be 5/.o85 = 58.8 so about a 60 ohm resistor would do.
And 5 x .085 = .425 so 1 watt or better will do. Have I got that right?
There appears to be a standard double rectifier bridge with parts marked M1 and when I apply voltage to various points on that I get a partial display.

 

A fixed resistor won't work if the display is LED. What type of display is it?

 

It doesn't say if it is LED, I'll try to attach a picture. And was my math correct in the last post?

 

This clock depends on the 60Hz power line frequency to count cycles to keep time. If you isolate it from the AC power line, it will not keep time.

There are all kinds of clocks out there that are designed to run on 12Vdc from the get go...

 

Thanks guys for the help.
I wasn't happy with a clock I got from jc whitney many years back and was hoping this would work out, I liked the display size and buttons location. The best laid plans of mice and man, oh well.
I did order another clock today that is 12 volt and I'll just have to work out a new mount for it in the truck.
Thanks,
Fred