Quartz Clock Power Supply Hack (AA Battery to AC Power)

I had a quartz wall clock in my workshop that ran a AA battery power.  The battery died so I decided that rather than buy a battery, it would be more fun to convert the clock from battery power to AC power.

Step 1: Tools

For this project, the only tools that you'll need are a soldering iron and (optionally) a hot glue gun.

Step 2: Parts

You will need:

(1) 5V power supply .. I used an old phone-charger .. these can be readily found at any thrift store
(2) 22uF caps rated for 10V or better
(3) silicon diodes .. I used 1N4936 because it's what I had lying around .. but most any silicon diode will do .. the forward voltage drop will vary slightly for different diode types, but this circuit is fairly tolerant of small voltage differences.
(1) 1K resistor .. the power rating doesn't matter much, the resistor will only need to handle milliwatts of power

Step 3: Schematic

Here is the schematic ..

Step 4: Point-to-Point Wire the Circuit

Just connect up the circuit as per the schematic.  Keep the leads relatively short so the entire circuit will fit in the space that was previously occupied by the AA battery.  Once wired, you should check the output voltage with a multimeter.  It should be somewhere between 1.5V and 1.6V

Step 5: Hot-Glue the Circuit in Place

This is somewhat optional, but it's nice to encapsulate the circuit ..

Step 6: We're Done ..

Sit back and admire your work :)

Comments

[akjayak]

I used in 1990--- 1444 ic for quartz clock coil drive now that is not available
Any idea for that purpose,any ic or any circuit please reply

[JanosM8]

I think your schematic is not doing what you wanted it to do. First and most important: it does not power the clock with 1.5V but instead your clock is running at 5V. Maybe you draw it wrong, because the way you draw it, the clock is in parallel with the diodes. Now, with voltage, every parallel line uses the same voltage, so while the line where the diodes are will produce 1.5V, the next line where the clock is, will run on 5V. With voltage, in parallel every line uses the same IN voltage, when in series they ALL use the IN voltage. With amperage, it's the other way round.
So, I checked, some clock can work on 5V or any voltage, no need to reduce to 1.5V. Some can work without capacitor. I don't know why my tests show this way. Maybe it was the way I tested it, maybe it was the age of the clock, inner problems (none of them was new or up to date).
Anyway, if you want to run them on 1.5V, you need to connect the clock in series with the diodes. The capacitor always goes from positive to negative, so that is good.

[gamerchris555]

uhh you neglect that a diode on a current limited supply (resistive limiter of 1k) will have a drop of it's forward or reverse bias voltage depending on direction
zeners are very often used in reverse as a resistive limited zener capped input is basically used in tons of switch mode feedback lines to run a 2v optocoupler from way above that
the 3x 0.5-0.8v drop of diodes with a limited current input would tend to become a cheap low power linear regulator at about 1.5-2.0v
his circuit works by abusing the forward voltage drop and a limited supply
I still would add in a NiMh cell for power outages and use maybe a 2v zener or something in case the cell is removed under power to replace with new cells
but The OP is correct that it will be ROUGHLY 1.5v on the diode stack and the resitor will drop a few mW and the extra ~3.5v
a quartz clock is 1.5-3v not many will tolerate 5v without limiting resistors(to the quartz timing circuit)
at 1kohm you get about 3.5ma of max current before the diodes will shunt to gnd
this means you are pulling 3,5ma from the 5v at all times so the actual power even with the clock pulling power will end up being 3.5ma x 5v = 17.5mw
not bad for power bill
I would use a NiMh cell for backup and likely ~40ma charge to keep it at 100%
so a 470ohm resistor from a usb supply and a simple NiMh cell that is about 10 years old but still runs stuff well
if I cared more I would add a diode stack for the voltage capping without a cell

[Makedougall]

I know I am coming very late to this (very old) party, but I had to add a comment for this great, excellent project.

A previous commenter asked whether "...this is a good idea other than just for the fun of it."

I once had the misfortune of experiencing a house fire while no one was home. I arrived at the house to find firemen cutting through the roof to let smoke out. One of the questions asked by the fire chief was whether or not there was a plug-in wall clock in the house, so they could determine the time the power went out (i.e. when fire reached the circuit panel, which would give them a good estimate of when the fire started). Unfortunately, all my clocks in the house were battery operated! (I later realised that the clock on the oven/stove range was powered from the mains, but was not sure that it was set to the correct time, as we often ignored that clock.)

So this project would be PERFECT to power a wall clock that could tell when the power went out.

Question: I am wondering about the use of 3 diodes to provide the "1.5V" drop, to power the clock. Is not the "standard" forward voltage drop of a silicon rectifier or small-signal diode, 0.7V? (I do not know the specific characteristics of the 1N4936 types, though.) So three "standard" silicon diodes would provide 2.1V, would they not? I am thinking that two diodes should be enough, providing 1.4V to the clock. (These clocks are generally designed to power on less than 1.5V, to allow for battery depletion.)

@scd: Brilliant project. Thank you for sharing it.

Edit: Referring to a Fairchild Semiconductor datasheet for the 1N4933 - 1N4937 (including the 1N4936), I see a "Forward Characteristics" graph of "Forward Current" vs. "Forward Voltage." (On the second of three unnumbered pages, upper-right corner of the page.) Unfortunately, the graph cuts off at 10 mA, showing a voltage of 0.6V. I expect the current through the diodes would be smaller (about 5 mA?), so not sure what Vf should be. So maybe using three diodes is not so bad, after all. The cap would serve to mitigate the drop in forward current through the diodes during the clock motor pulses, so the forward current/voltage should be fairly stable. Gonna have to do some measurement. :)

[scd]

The forward voltage drop of a silicon diode is current dependent. For low current values such as in this project, the voltage drop is closer to .5V. This was kind of a parts-bin project and I used what I had lying around. A better way to generate a 1.5V supply would be to use a 1.5V Zener diode.

[Makedougall]

Thanks, SCD. After I first found your Instructable, early this morning, I also thought of using a Zener instead of forward-biased "regular" diodes. So I Googled "standard zener diode values", but the smallest "standard" Zener voltage I could find was something like 2.0V or 2.2V. I then tried Mouser Canada (since I live in Canada), and the smallest Zener voltage I could find there was 1.8V. So I don't know where I could get one at 1.5V. (Maybe there is some physics law or property that prevents Zener voltages below a certain value?)

So I think your forward biased "regular" diodes is a good solution.

BTW, I have also seen circuits where folks used a single LED (forward biased), but the smallest red LED Vf I found was also 1.8V (but that is probably at a current that is big enough to light it up, so perhaps it is at a more acceptable, lower voltage for smaller current where the LED is not emitting light?)

Well, I guess this will require a little more experimentation. :-)

[mgnscott]

Fantastic! This is exactly what I am trying to do! But I sure wish a kind soul would explain the schematic to me, I know roughly what the schematic is and could make it on a breadboard but I don't really know WHY I would be using each of these components. I really like to understand on a basic level what each part is doing before making. If anybody feels like dedicating some time to someone who needs to know things in too much detail please reply. (ps yes googling individual parts gives an idea but not really, example, why 2 caps, why 3 diodes etc...I have a very basic understanding of electronics so feel free to talk to me like a child)

[NigelL12]

Probably too late to help you now but I will explain anyway. The clock motor and its electronics require about 1.2 to 1.8 volts. The current requirement is variable as the electronic part takes literally only a few micro amps. Once a second for about 30ms a pulse of current is supplied to the motor and this is about 6mA in most clocks. The diodes need to be connected, forward biased, across the battery terminals. Most silicon diodes have a voltage drop of about 0.7 volts so I would use TWO. This will give 1.4 volts which is plenty. Three diodes at 0.7 volts will damage the circuitry. The 1k ohm resistor limits the current to about 3.6mA ( Ohms law: I = V/R. (5 - 1.4)/1000 = 0.0036) which is far more than the ic requires but is insufficient for the solenoid pulse. The capacitor stores charge and can supply the pulse for 30ms and recharge during the following 970ms. The capacitor has to be large enough to hold the adequate charge without too large a voltage drop but small enough for the voltage to recover. Anything in the range of 47 to 150uF should be ok for this. Obviously the capacitor is connected across the battery terminals in parallel with the two diodes which are just voltage limiters. If you keep to the smaller end of the capacitor range you will probably get away with a 2.2k ohm resistor which will reduce current consumption to about 1.6mA. It can be plugged into a usb socket if you have one and the running cost will be negligible. Another idea is to drop the voltage with a zener diode, in this case about 3.6 volts.

[brilliankitamboli]

Can i use voltage divider using resisitor to reduce voltage
Should i compulsory use capacitor for current

[scd]

The circuit was a hack way to generate a 1.5V power for the clock with parts that I happened to have on hand. The 3-forward-biased diodes creates an approximate 1.5V reference, and the caps store enough charge to supply current for when the clock ticks. A better way to generate 1.5V would be a buck converter. If you search on Google, or eBay or Amazon for "1.5V Buck converter" you can find a part that will do the job for less than $1.

[mgnscott]

Tried your way and with a converter and still no joy. Thanks anyway.

[NigelL12]

Probably cooked the clock circuit with three diodes. Many have a voltage drop of 0.7V so three could have given 2.1V. If you inadvertently reverse biased the diodes you would have forced 5V through the clock. Ouch!

[brilliankitamboli]

Can i use voltage divider using resistor to convert into 1.3 volt
Should i do something about current also

[SreenivasaR3]

Instead of in4936 can I use in 4001 series pl reply

[scd]

Yes

[SreenivasaR3]

Can I use in4001 diodes instead of the one specified

[nathanziarek1]

(Hopefully someone is still reading these comments!)

I have a project where I'm building something like clock wall they show in movies involving the President (you know, the ones that show DC time, London, Shanghai, etc).

I have four clocks that I bought from a thrift store, so there's no information on them. Each take a single AA battery and are just the basic second / minute / hour hand configuration like the one shown in this instructable.

I have a 6v adapter providing 500 mA. My questions:

1. I have no idea what the draw is of these clocks, but I have to assume its low. Will 500mA be enough to run all four of these?

2. If this will even work, I assume I'll be wiring these in series, not in parallel, correct?

Appreciate your time!

[scd]

The current draw for each clock is very low and it is not constant; most of the current draw happens once per second when the second hand ticks and (if I recall correctly) it was less than 1 mA for the clock that I used.

If you are going to build the power supply hack (as described in this instructable) for each clock, then you would run the 6V to each clock (in parallel).

It may be simpler to build a single 1.5V power supply using a MCP1702T voltage regulator and then run the 1.5V to each clock (in parallel). As an experiment you could wire all 4 clocks (in parallel) and try running them from a single AA battery. If this works, then great. If it acts funky, then you may need to add a cap to each clock to help supply the instantaneous current draw when the second hand ticks. Some experimenting will be needed on you part, but that's all part of the fun :)

Here is a digikey link for the MCP1702T 1.5V regulator:
http://www.digikey.com/product-detail/en/MCP1702T-1502E%2FCB/MCP1702T-1502E%2FCBCT-ND/2179250

And here is a link to the MCP1702T data sheet:
http://ww1.microchip.com/downloads/en/DeviceDoc/22008E.pdf

[nathanziarek1]

I really appreciate your taking the time to respond.

I'm really new to this, so I was making the mistake that, because 4 clocks needed 1.5v each, I could use my 6v supply and just "spread" it among them.

Seems like this project might be a bit beyond me, but your instructable and response really helped! Thanks much!

[NigelL12]

You have no doubt either done or abandoned this by now but if I was doing this I would put the four clocks in series across a 6 volt supply. Only trouble being that the current varies from a few micro amps to about 6mA for a short burst. To even this out requires four equal resistors, about 470 ohm should be fine, all in series. Across each resistor you need a capacitor, 47 to 150uF should be fine, to supply the surge and then recharge. Each one will charge to one quarter of the supply voltage. I would solder the resistor and capacitor in parallel across the battery terminals and loop the supply wire from clock to clock.