I am planning to build a new clock divider, using a 4040 IC. Looking at different schematics for inspiration, I came across Benji Jiao’s MiniDiv, where he uses a 40106 inverter IC after the 4040 outputs: https://benjiaomodular.com/post/2024-11-22-minidiv-pcb/
Could anyone enlighten me as to why the 40106 inverters are used after the 4040 outputs? Why would you want to invert the 4040 outputs?
Thanks!
Possibly the 106 was thought to be used as a buffer for the LEDs. Had he connecting the LEDs to the plus, that would have been ok (because then the LEDs would light up if the buffers would be presented with a 1, resulting in a 0 at their outputs making the LEDs light up). But using the inverted signal as output does not make much sense unless he uses negative logic. But why not ask him directly?
Because the output you get straight from the 4040 looks like this:
clk ▎▎▎▎▎▎▎▎▎▎▎▎▎▎▎▎
/2 ▁▆▁▆▁▆▁▆▁▆▁▆▁▆▁▆
/4 ▁▁▆▆▁▁▆▆▁▁▆▆▁▁▆▆
/8 ▁▁▁▁▆▆▆▆▁▁▁▁▆▆▆▆
/16 ▁▁▁▁▁▁▁▁▆▆▆▆▆▆▆▆
...
As you see, the rising edges of the outputs do not line up with eachother, so all of your divisions will be “off-beat”. By inverting each output you make the rising edges line up, which is what you expect for most musical division purposes.
It is actually useful to have both and I have both “on-beat” and “off-beat” outputs in my clock divider. The non-inverted outputs can be used to trigger snares and hi-hats for example.
I don’t understand why the LEDs are connected directly to the outputs. Doesn’t this cause a significant voltage drop at the output?
AFAIK it is better to use a decoupled LED output like this:
This stage takes a minimum current from the signal to display (“OUT”). R1 could even be enlarged if necessary. Important is to place the LED in the collector path of the transistor, not in the emitter path, as the base2emitter current creates an LED glim even if “OUT” is “0”.
R2 determines the brightness.
This is the stage I use for all of my LED indicators.
If you use a PNP transistor you can display “0” as on and “1” as off without the necessity of using an inverter. I used it in displaying S-GATE outs and ins of Moog circuits, see the S-TRIGGER display on
click
Ah, ofcourse! I was only looking at it from a technical perspective, and I could not find a reasoning for it.
Do you have on- and off-bwat clock dividers in the same module? Do you use separate output sockets, or some sort of switch to alternate between both options? If a switch, did you use multi pole switch to route the path past or across inverters?
Thanks, this is exactly what I concluded after breadboarding with a 4040! Having an LED directly driven by the 4040 outputs, makes the output too weak to activate some modules. For instance my quad gate/trigger sequencer, that uses the clock input to activate the 4017 counter and 4081 AND-gates simultanuously, did not pass through the steps when triggered by the 4040 outputs because the LEDs made them too weak.
I am definately going to power the 4040 clock dividers LEDs via transistors.
The sequencer is controlled by a microcontroller, so everything is software driven, but of course you can use any logic circuitry you like, but as you have noted as well outputs should be buffered and not stressed by voltage loss for LED displays.
On the other hand I have to admit that I just love LEDs and put them just anywhere 
, but with those transistor buffers of course.
And if I use NPN and PNP transistors I don’t have to invert at all.
Yes, highly recommended 
Do you have on- and off-bwat clock dividers in the same module? Do you use separate output sockets, or some sort of switch to alternate between both options?
I have individual outputs for /2 /4 /8 /16 /32 and their inverted versions so 10 output jacks total
