2
\$\begingroup\$

I have built a digital alarm clock using an LM8560 chip. I tried powering the circuit with DC and the 7 segment display lights up, but none of the buttons worked and the numbers on the 7 segment display did not change.

So today I tried powering the circuit with a 11.8V AC power supply and one of the capacitors popped, even though it was rated a 16V.

Should I proceed in trying to make the circuit work on DC or should I try to make it work with AC. From looking at the circuit diagram, only the 50/60Hz input needs AC, so would connecting pin 3 of a 555 astable circuit do the job?

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ It looks like the LED display also relies on the 50/60Hz AC to multiplex the digits. \$\endgroup\$
    – Majenko
    Commented Nov 18, 2014 at 17:08
  • \$\begingroup\$ could you explain how that works please? \$\endgroup\$ Commented Nov 18, 2014 at 17:12
  • \$\begingroup\$ You do understand that 11.68VAC yields 16.8V peak, which is what most likely killed the e-cap? \$\endgroup\$ Commented Nov 18, 2014 at 17:14
  • \$\begingroup\$ It's quite a clever circuit, but the designer really ought to have specified what voltage should be applied! \$\endgroup\$
    – pjc50
    Commented Nov 18, 2014 at 17:14

1 Answer 1

1
\$\begingroup\$

With that circuit the AC is absolutely 100% required. There is no getting away from it (short of generating your own AC signal).

The AC:

  • Provides all the timing for the clock.
  • Provides the duplexing for the display.
  • Provides the main power through a very crude rectification system.
  • Provides the "buzz" of the alarm.

The "50/60Hz" input is used as a clock input, which is divided by either 50 or 60 to create a 1 second pulse against which the clock's timing is counted. A slight variation in that input frequency will mean clock drift over time.

The display output is multiplexed - it does hours in one half of the AC cycle, and then minutes in the next. As the AC wave changes the diodes connected to the transformer alternate as to which are sinking the current from the display.

If you can generate an exactly 50Hz or 60Hz signal with enough precision to not cause the clock to drift (or it would be a bit useless), and that signal can sink enough current to handle the display (100mA or so), then you may be able to get away without the AC transformer.

As for why your caps blew... well, an 11.8V transformer, after rectification and smoothing, would be nearer 16.7V. Best to use a smaller transformer, like 7-9V.

\$\endgroup\$
5
  • \$\begingroup\$ Thanks for the help, I have installed 25V capacitors now, all I have to do is rebuild part of the circuit to make it work. How would rectification increase the voltage? \$\endgroup\$ Commented Nov 18, 2014 at 18:32
  • \$\begingroup\$ 11.8V is the "RMS" value. When rectified and smoothed you get the "peak to peak" value. RMS is about 70% of the peak value. \$\endgroup\$
    – Majenko
    Commented Nov 18, 2014 at 18:52
  • \$\begingroup\$ Just nearly blew a 25V capacitor, what the HELL? \$\endgroup\$ Commented Nov 18, 2014 at 19:09
  • \$\begingroup\$ Measure the output voltage of the transformer. Are you sure you have an 11.8V transformer, and you're not using two windings together or something to give a much higher voltage? Or you have it wired backwards so it gives a couple of thousand volts? \$\endgroup\$
    – Majenko
    Commented Nov 18, 2014 at 19:18
  • \$\begingroup\$ The Transformer I am using is one that has been manufactured so the AC it is outputting is definitely 11.8V \$\endgroup\$ Commented Nov 19, 2014 at 7:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.