# Seeking help with (low) voltage limiter

#### gtc

Joined Jul 7, 2007
3
Project: a 1.5 volt battery eliminator for my garage wall clock.

Background: I love the old clock, but I hate replacing the AA cell in it, so I want to connect it to the mains via a power pack.

I have a DC power pack that provides these nominal DC voltages (at 150 mA) via a slide switch: 1.5, 3, 4.5, 6, 9, 12.

The open circuit DC voltages I measured are: 3.02, 4.87, 6.71, 8.54, 12.63, 16.82 respectively.

When I connect the power pack to the clock in lieu of the battery, using the 1.5 volt switch position, the clock's second hand runs at about twice the correct speed. I'm supposing that the impedance of the clock mechanism is too high to drag the power pack's output down to 1.5 volts, thus I'm putting 3.02 volts onto the clock.

So, I'm seeking suggestions as to how to cheaply regulate the power pack's output voltage to 1.5. I'm pretty rusty on electronics -- I left tech school in 1970 -- but I recall enough to know that my options include: a zener diode circuit or a resistive voltage divider.

I don't know if there's such a thing as a 1.5 volt zener (I think the lowest is 3.3 volts), and I need help with the current/power/resistance calculations for a simple voltage divider.

#### hgmjr

Joined Jan 28, 2005
9,027
You could use a couple of diodes in series with each other (either 1n914 or 1n4148) each with a forward voltage drop of around 0.7 volts. That will give you a voltage of 1.4 volts.

Put a series current limiting resistor of around 1kohms in series with the two diodes. Connect the cathode of the lower diode to ground and take your voltage output for the clock's voltage supply from the anode of the upper diode.

hgmjr

#### gtc

Joined Jul 7, 2007
3
Good thinking! Thanks

Actually, given the extremely low current draw of the clock mechanism, could I not get away with just the 2 diodes in series in the positive leg of the supply?

PS (3v+) ---> D1 ---> D2 -----> (1.6v) Clock
PS (Gnd) --------------------------- > Clock

#### hgmjr

Joined Jan 28, 2005
9,027
Good thinking! Thanks

Actually, given the extremely low current draw of the clock mechanism, could I not get away with just the 2 diodes in series in the positive leg of the supply?

PS (3v+) ---> D1 ---> D2 -----> (1.6v) Clock
PS (Gnd) --------------------------- > Clock
You can try that hookup to see how it behaves but I believe you will have better luck with regulation if you use the 2 series diodes and the 1K resistor scheme.

hgmjr

#### mozikluv

Joined Jan 22, 2004
1,435
why, do have a problem just by adding a series resistor just as what "hgmjr" suggested? that resistor is there to protect your clock.

moz

#### gtc

Joined Jul 7, 2007
3
why, do have a problem just by adding a series resistor just as what "hgmjr" suggested? that resistor is there to protect your clock. moz
No I don't have a problem as such with the idea, I just like to use the minimum necessary components as a matter of principle.

All fixed now.

#### bloguetronica

Joined Apr 27, 2007
1,541
why, do have a problem just by adding a series resistor just as what "hgmjr" suggested? that resistor is there to protect your clock.

moz
Resistors are good if you have a constant current circuit. This is not the case of the clock, since it will pull current in pulses. Also, a resistor won't protect the clock, since the voltage will be bigger when the clock is sinking less current.

Consider using a LM317 regulator. It will supply 1.25V minimum, but you can adjust it to give 1.5V by using a resistor network of only two resistors. Don't forget to use caps in the input voltage pin and output voltage pin, connected to ground. Should be simple.