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enter image description here The very cool circuit is from here: http://engmousaalkaabi.blogspot.com/2016/01/81-led-chaser-using-double-ic-4017.html

Here are my few questions regarding it.

1, It has 18 transistors. are they really required for lighting up some small LEDs? I think 9 transistors should work.

2, It's about the 4017. When powering on and without a clock signal from 14, does Pin 3 (first output) goes HIGH or LOW? (according to this circuit, Pin 3 much go HIGH for it to work.)

3, what is 3.3k doing in this circuit?

4, last and my main question will be. Can I replace a 555 ic without a micro-controller such as a small attiny 95 and turn this array of leds into a clock? will the CD4017 fast enough for that? according to the datasheet, it has 2.5mhz clock input.

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1) indeed, if you run the LEDs on a small current such that the ratings of the left 4017 are not exceeded, you could remove those 9 emitter followers.

2) according to the datasheet (the one from Texas Instruments) output 0 (pin 3) is high even when in reset. Only after a clock will it go low. Look at the timing diagram in the datasheet and all will be clear (I hope).

3) it is in series with a input (RESET, pin 15) and as this is CMOS logic and I see no capacitor that could lift a voltage above the supply (near this pin) I would say: you don't need the 3.3 k resistor.

4) Yes you can use almost any oscillator or clock generator instead of the 555. The 4017 will indeed be capable to keep up up to a few MHz. But the LEDs would blink so fast that they would all light up. The idea is that the clock is only a couple of Hz which you can easily do with any micro.

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  • \$\begingroup\$ 1) Try again. The 100 ohm current resistors suggest a current in the range of 10 to 30 mA, depending on color. Data sheet says ~0.36 mA for Ioh for a few tenths of a volt output reduction. Those followers are necessary. \$\endgroup\$ – WhatRoughBeast Feb 5 '16 at 15:31
  • \$\begingroup\$ I'm not trying again because there is no need. The emitter followers are not needed as long as you do not exceed the ratings of the 4017. This can be achieved by changing the values of the 100 ohm resistors or lowering the supply voltage. Maybe I do not care about a few tenths of a volt drop, those LEDs don't care either, they will still light up. You seem to suggest it won't work without the emitter followers. I'm saying it will if you keep the currents low. There are LEDs that light up brightly at 1mA, if you used those, it would work fine. But watch the dissipation in the 4017. \$\endgroup\$ – Bimpelrekkie Feb 5 '16 at 15:50
  • \$\begingroup\$ Also, the emitter followers will also drop at least 0.6 V between supply and emitter. So with low current LEDs and no emitter follower it is possible to drop less voltage on the positive side ! \$\endgroup\$ – Bimpelrekkie Feb 5 '16 at 21:47
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I made this circuit about 1 month ago, in first I was thinking as you and I removed the 9 transistors that connected to positive line, but the circuit did not work. I was thinking my IC burned so change it, but did not work again. So I was forced to use all 18 transistors.

I don't know why but you have to use all 18 transistor, that not because of power consumption.

The resistor 3.3k reset the first IC. When I remove it, circuit stopped working. You can use any other IC for clock pulse generation, I've used 4060.

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You could use 74HC4017s instead of CD4017s, which can easily drive 5mA directly and get rid of the 9 of the transistors. You could replace the bottom 9 transistors with a couple ULN2003A drivers. Recalculate the 100 ohm resistor values to something a bit higher (probably a few hundred ohms).

There is no way to tell what output 0 does at power-up since this circuit has no reset circuit.

The 3.3K does nothing (short it and the behavior will be similar). Probably it was intended to connect a reset circuit to that pin.

Yes you can drive the clock input (and reset inputs) with a microcontroller. For the reset input drive the output high to reset the counters and high-Z (make it an input) after reset. You'll need to use two outputs and add another resistor to reset the second counter.

Or (and I would recommend) use a better circuit such as two OR gates (eg. 74HC02 with two gates connected as inverters) and then you don't need the resistors and can use a single output from the micro (active high, low for normal operation) to reset both counters. Like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Net effect is a reduction by ~26 parts (use networks for the resistors and knock another 7 or 8 off), improved functionality (proper reset) and lower overall cost. Engineering- booyah!

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