I'm trying to build a simple LED blinker using a 4060BP binary counter, as per the following: this one http://img.bhs4.com/ca/6/ca6021e3fd1d873281438d1bf721ef4e27ceaa0b_large.jpg and this

(Originals here and here). When I have built either circuit, the LED (I'm just using one for now) is permanently on. Looking at the circuits this seems logical, since because the negative terminal of the power source is connected to ground, there is a closed circuit between the positive terminal, the LED and the negative terminal. In fact, the LED stays on even if I remove the power from the 4060 IC! Should these circuits as per the two diagrams work as LED blinkers, or am I missing something? Thanks!

  • \$\begingroup\$ The bottom one might. The top one certainly won't. \$\endgroup\$ – Majenko Apr 12 '15 at 22:42

This circuit was probably not built and tested by the people who published it.
It can be made to work with a simple addition but, as is, it is almost useless.

The circuit as you are implementing it has two major problems

(1) The per-leg sum of LED forwards voltages MUST be less than Vcc and more than Vcc/2 (Ideally Vfsum < Vcc-3 & Vfsum > (Vcc/2 + 2))

If Vfsum is too low current will flow from VCC through upper and lower legs even if the 4060 is removed and all LEDs will always be on.
If Vfsum is too high there will never be enough voltage available to turn them on.

You do not explain clearly what "I am only using one (LED) for now" means.
One LED is not what the circuit "requires". If you use eg 2 x White LEDs in series - one each in the high and low legs and 12V supply - current will flow directly through the resistor and both LEDs and turn both on always.

"Vfsum range condition":

Vf = LED forward voltage.
Vfsum = sum of all Vfs in one half leg (upper or lower).

Calculate Vf_sum = sum of all Vfs for the LEDs in each half string.
Vcc must be greater than Vfsum and less than 2 x Vfsum (Ideally Vfsum < Vcc-3 & Vfsum > (Vcc/2 + 2))

(2) A 4060B data sheet here does not have enough drive to pull the skin off a rice pudding. AT 15V it has maybe 3 mA low drive and 2 mA high drive per pin. LEDs may glimmer but not much more.

Making it work:

A probably useful result can be obtained by adding a very simple and low cost driver per LED drive pin. Either of the two circuits below will work. The left hand one with two cheap bipolar transistors is probably easiest to implement. Almost any NPN (upper) and PNP lower) transistor can be used but higher current rated one are better.
In your application no resistors need to be added to this circuit as the 1k LED resistors shown limit current already.

I use BC337-40 (NPN) and BC327-40 (PNP) as general purpose transistors (or the SMD equivalents BC817-40, BC807-40) - these are high current, very high current gain (use the -40 version) and often low cost. BUT most types will work.

This circuit is more capable than may be obvious. It is 2 x emitter followers combined. When Vin is high the output is 1 Vbe = 0.7V below Vin and when Vin is low Vout is about 0.7V above Vin. This means that with a CD4060 with 12V supply you have over 11V of drive.

LED Vf is about 3V for blue, and 2V for red and green LEDs.
So for 1 each of Red + Green + blue in series Vfsum = 2 + 2 + 3 = 7V.
A 7V < Vcc and > Vcc/2 this meets my "Vfsum range condition" given above.

For 3 x red or 3 x green Vfsum = 6v - which is marginal as Vcc/2 = 12v/2 = 6v = Vfsum.
If using 12V supply and all red or all green LEDs use 4 LEDs per leg.

LED current will be about (Vcc-Vbe -Vf_sum)/ Rseries.
For say Vcc=12, Vbe=0.7, Vfsum (R+G+B) = 7V, Rseries = 1000 Ohms
I_LEDS = (12-0.7-7)/1000 = 4.3/1000 = 4.3 mA.
The LEDs will be visible but not as bright as they could be.
Up to 20 mA is OK with most small LEDs. Using 270 Ohms per Rseries gives I_LED ~= 4.3/270 ~= 16 mA hich is 'OK' and allows Vcc to be a bit higher and still "safe".

enter image description here

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When choosing LED strings note as above that Vfsum MUST be > Vcc/2 and < Vcc.
(Ideally < Vcc-3 & > (Vcc/2 + 2))


The CD4060 outputs should be driving some sort of switch to close the LED circuit.enter image description here

  • \$\begingroup\$ A buffer is needed BUT it MUST be "bipolar" \$\endgroup\$ – Russell McMahon Apr 13 '15 at 0:48

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