Let's Make Robots!

Project with 4060 Counter IC

Hi guys!

I can't understand why this project works as the guy states: http://www.kpsec.freeuk.com/projects/xmastree.htm

basically, by what i understood, the 3 columns of LEDs blink at different Hertz, but i don't understand why. Since they are connected to V+ and V0, should they stay firmly switched on? How can connecting a wire in between a column with an output which can be high or low affect this?

 This is what i think can happen: when the outputs from the 4060 go low, they sink the current so the three LEDs of that column (at the top of the column) light up.

FIRST QUESTION: why does it get sunk by the 4060 (if that is what happens) instead of going directly the the negative pole of the battery? i suppose because there is less resistance that way?

 When the outputs from the 4060 go high, erm...all the LEDs of that column light up? but then the LEDs on the lower part of the column would have more voltage right?

 

Thanks for helping :=) 

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A mate of mine was a technician for a pay TV company. He fitted one of these IC's into a TV remote used by the company and set it up in a test room up high on a shelf where the consoles and remotes were tested for faults. It was set up to change the channel every minute or so. When one of his fellow technicians went to test out a faulty console he couldn't work out why it had started changing channels all the time. He ended up with half the team trying to work out the fault before my mate quietly removed the modified remote from the room. This confused them more as to why the new fault had suddenly dissapeared.

Hey thanks for the explanation anachro! So basically in the conditions the unit is Volts while in the other parts it is the one listed on the right side, didn't notice that at first.

So if i well understand, at 25°C (typical conditions) and having a 9V (i'll take the 10V row for ease) it is gonna sink some 2.6 mA. But how is this gonna help?

Suppose the 3 LEDs have a 2V voltage drop across each, and the output is low, i am still gonna have 9V-2V-2V-2V=3V to deal with... 

OK, if the output is low, then the upper three LEDs will, as you say, drop about 6V.  That means that there will be 3V from the CMOS output to ground.  Now, a CMOS output would normally drop all the way to 0V when it is driven low by the chip.  But, we can see from the data sheet that it cannot sink more than (typically) 2.6mA.  That current limit is equivalent to a resistor, of just the right number of Ohms to make 2.6mA flow, whatever voltage we apply (within reason).  So, when the output is driven low, we'll get 2.6mA flowing through the upper LEDs.  The lower LEDs will see about 3V, but that's not enough to light them up.  The CMOS output pin behaves as if it has a small resistor in series.

A similar thing happens when the output is driven high, and again, about 2.6mA flows from the pin, through the lower three LEDs, to ground.  Of course, this circuit relies upon the chip supplying its maximum current in or out of the pin, all the time.  It will probably work, but the chip might warm up and may even fail prematurely.  Also, the exact maximum current is not specified in the data sheet, so in reality anything can happen. Most likely, it'll be OK, but this is one of those cases where the chip maker will not give a guarantee.

ooh. So that 2.6mA basically means, in this case, that a max of 2.6mA will be flowing in the circuit, and thus, if we have a 10V battery it will behave as if it had a 3.8KOhm (10V/2.6mA) in series with it. 

If we take the Output low (sink) current rows as an example, there are indeed three of them.  Look in the three columns headed "Conditions", where you'll see some voltages as headings, including the crucial Vdd.  Vdd ranges from 5V to 10V and then 15V, showing the device used with three different power supply voltages.  Then, carrying on across the data sheet, we have various operating temperatures, including some unreasonably high and low values (for mil-spec chips).  Near the right-hand end, we have the 25 degree Celcius cases, which are further split into guaranteed minimum and maximum values, and a typical value.  The very right-hand column shows the units, milliamps.

Remember that the data sheet is a sort of contract between the maker of the chip and the user.  Lots of detail is included that might seem excessive, but is there for a reason.  Sometimes, that reason is to avoid getting sued when a chip's behaviour changes at a certain temperature!

Ok so you are basically saying that when the output is low, the 4060 acts as a sort or resistor, right? But from which values can you understand how much it limits? The datasheet has tons of values that look arabian to me!

(the "output low (sink) current" for example relates to three different rows, what's the difference for them?)

Thank you for the answers :)  

"The datasheet has tons of values that look arabian to me!"

You mean all the numerals? (Har har, nerdy linguist joke) :)

Dan

Remember that the forward voltage drop for a red LED is nearly 2V.  With six red LEDs in series, you'd need nearly 12V to light them (as OddBot said).  A 9V battery won't be enough, until the CMOS chip effectively shorts out three LEDs.  But it doesn't quite short them out, because of the inherent current limit of the CMOS output stage.  So, you get a safe current through three LEDs at a time.  Clever, although I'm still not sure I like it...

When the outputs go high, the Leds between the output and the +V have little or no voltage across them while the the leds connected to ground have virtually the whole supply across them with the current limitted by the IC. When the output goes low, the opposite happens. Even if there is some voltage across the leds it has to exceed the forward voltage drop of the leds to turn them on. Each led depending on colour will have about 2.1V to 2.4V forward voltage drop. If the outputs were removed then none of the leds would glow as the forward voltage require to turn them on would be over 12V.

 

Looking at a 4060 datasheet, the I-output-high (sourcing) and I-output-low (sinking) of the pins is only a few mA, not enough to cook the LEDs. So instead of a current limiting resistor, there's an active current limiting semiconductor.

Clever circuit, should be cute.