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?
They blink at different frequencies becuase the 4060 is a binary output. The three bits are:
000 001 010 011 100 101 110 111
The left column keeps it state for three clock cycles, the middle for two and right for one clock cycle.
There is a hint on the page as to how it works: "You can see from the circuit diagram (below) that 6 LEDs are connected in series between the +9V supply and 0V. Each LED requires about 2V across it to light, so using a voltage of about 12V (= 6 × 2V) or more will make the LEDs conduct directly, regardless of the 4060B IC. With no series resistor to limit the current this will destroy the LEDs."
When the 4060 output is 0, the power will flow through the upper three diodes and to ground through the 4060. When the output is 1, it will go from the 4060 through the lower diodes. Power should go through all 6 diodes in a column, but it doesn’t, because the voltage is not high enough, as stated in the quote from the page.
What bugs me is, that the diodes doesn’t burn out due to lack of a current limiting resistor.
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.
Forward voltage drop 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…
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?)
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!
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 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.
