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¿1 Hz oscillator?

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The TTL gates have a low impedance which loads the CMOS gates down.

To be honest, I'm not sure that this is the problem here, a CMOS IC should be able to drive a single LS TTL load.
 
I was trying to drive a lot more then a single LS TTL load, six of them to be exact. So I'm going to rebuild the circuit using SN74HC4060N IC's and 74HC74 flip flops. I was wondering if there are any special considerations when building the same cicrcuit as above, but with the IC's I mentioned?

Many thanks
 
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I don't now what you are doing wrong with the $5.00 clock. Use either of the outputs plus the negative of the battery and use a 2k2 resistor to the base of an NPN transistor. Connect a LED to the collector via a 330R resistor and connect the resistor to positive of a 6v battery and the negative of the battery goes to the negative of the 1.5v battery powering the clock. The LED will flash at 1Hz
 
I was trying to drive a lot more then a single LS TTL load, six of them to be exact.
Then that's your problem.

So I'm going to rebuild the circuit using SN74HC4060N IC's and 74HC74 flip flops. I was wondering if there are any special considerations when building the same cicrcuit as above, but with the IC's I mentioned?
No need, just replace the CD4013 with a 74HC4013 which has an output current of 25mA so can drive the loads.
 
I don't now what you are doing wrong with the $5.00 clock. Use either of the outputs plus the negative of the battery and use a 2k2 resistor to the base of an NPN transistor. Connect a LED to the collector via a 330R resistor and connect the resistor to positive of a 6v battery and the negative of the battery goes to the negative of the 1.5v battery powering the clock. The LED will flash at 1Hz

So I'll be using 2 batteries? One 6 volt and one 1.5V? What about the postive terminal of the timepiece? Is that connected to the 1.5 battery? Where does the emitter go?
 
There's no need to use two batteries.

The quartz movement can be powered from 1.2V generated by two silicon diodes connected in series.

It depends on the clock movement IC.

The output of the IC will connect to a coil which is used to turn the motor. If the coil connects from the 0V to the output of the IC, use a NPN transistor with a pull up resistor and a suitable base resistor.

If the coil is connected from +V to the input of the IC, then the IC should be connected between 5V and 3.8V and a PNP transistor used. That might sound confusing but it's no harder.

If you tell me how the movement is configured, I'll show you how to connect it up.
 
Quick correction, I drew the buzzer polarities in reverse. Positive should be negative, etc....

Actually, I can't be 100% sure which of the buzzer leads are + or -. All I know is that the BLUE buzzer wire is connected to the terminal which connects directly to the - battery terminal, and BLACK wire connects to the other one. I'm guessing the black was meant to be -.

I drew a little picture of the pcb. I hope you can make it out. To be honest, it doesn't look like the coil is connected to neither the - or +.

**broken link removed**

What kind of diodes would I need? Thanks Hero, you're a.....HERO! lol!
 
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This is what I did with the battery clock circuit. Outputs 1 and 2 are the coil connections. The resistor , two diodes, and capacitor on the left form a 1.5V regulator to replace the battery. There are a lot of different versions of the circuit layout/board in the battery clocks, but all have two battery connections and two coil connections. The ones with alarms also have a two connections for the buzzer and two for the alarm switch. You can ignore these for your clock.

Ken
 

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That will work but I think the base resistor needs reducing.

[latex]Ib = \frac{1.2-0.3-0.6}{10 \times 10^{-3}} = 30 \times 10^{-6}A \\
Ic = \frac {5}{1000} = 0.005A
[/latex]

With a collector current of 5mA and a base current of 30µA the forced beta will need to be 167.

The minimum gain of the 2N3904 is 100 at 10mA and will be less at 5mA.
https://www.electro-tech-online.com/custompdfs/2010/05/2N3904.pdf

I'd recommend reducing R2 to 4.7k or lower.
 
Woohoo! Works like a charm! :))) Thanks KM and Hero. I took your advice Hero and put a 4.7K on the collector. All's well :)

Quick question if that's ok, can I drive cmos counters with this circuit, or is the output too high? Sorry for all the mega "noob" questions!
 
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Providing the circuit is run off the same power supply as the CMOS IC, there won't be a problem.
 
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