Sequencer LED Indicators

[mth5044]

I've been looking at the Vanishing Point V2.0 sequencer by the Tone God as it has a schematic similar to what I'm looking for (simple 4017 with Speed control). In the notes, TTG mentions that the indicator LED's off of each pin will vary in brightness based on the pot setting, which makes sense. I'm looking to have the LED's turn on to a set brightness independent of the pot value, so it just cycles through them.

My thought was to drive transistor switches to turn the indicator LED's on and off independantly of the LED/LDR combo. I've attached a schematic with two of the complete 4017 outs, but imagine it repeated 6 more times. Post-it note!



This would cause for an extra 16 resistors and 8 transistors. Is there a better way?

[merlinb]

Replace TTG's LEDs with ordinary diodes. Then also attach an entirely separate string of LEDs to the 4017 outputs, all sharing one limiting resistor? Not sure if the 4017 has enough output current though. Depends on how bright you want them.

[mth5044]

Thanks for the reply, Merlin.

I had seen a sequencer with 4148's, but didn't think about applying them here. Good call. I had a similar thought after I posted about just driving both LED's, but wasn't sure of the current. The data sheet says -0.9mA, but I am not sure what that could possibly mean as that's not enough to even drive an LED.

http://www.ti.com.cn/cn/lit/ds/symlink/cd4017bc.pdf

[mth5044]

What if two 4017 chips were literally piggbacked. Pin for pin matched, solder ontop of on another. Running off the same single clock and single reset parameters, but double the output current? No extra board space, just another quarter for the chip. Have the set up as Merlin outlined.

[duck_arse]

you could use a cmos hex buffer 4050 (or hex inverters 4049 wired as current sinks). you would want buffered versions and I suppose you'd need 2 chips for 10 outputs.

[mth5044]

4050... how does that work? The datasheet blew me away. I see there are six buffers that you put something and something comes out... 

[PRR]

> 4050... how does that work?

You go to the Guitar Boutique and buy one Audio Buffer for $99. (Or DIY  it here.)

Digital logic doesn't have to be fancy. You get *six* logic buffers in one chip for 19 cents. Your weak-armed 4017 tells the buffer to yank the load, up to maybe a dozen mA?

OTOH, any TO92 MOSFET will interface the 4017 to a load up to hundreds of mA.

[mth5044]

So.. back to this again. The 4050 duck posted blew my mind, again, so I'm still trying to figure out the transistor method. Behold, two new ideas and one old, none making sense.

1. (the original one)




2.




3.




In all cases, it doesn't make sense. In the 1st and 3rd pictures, won't there always be voltage passing through the LED, resistor an pot (1) or just through the pot (3) leaking into the CV dependent on where the pot is set, but independent on what transistor switch is open to ground?

The second one would be just the same, but with a resistance to ground constantly on the CV from all sections.

Is there a configuration I'm blanking on?  

[PRR]

> In the 1st ... pictures, won't there always be voltage passing through the LED, resistor an pot (1)

If the transistor Base is driven HARD, its Collector will fall to ZERO (typically 0.050V).

You may need less than 100K to Base to qualify as HARD. CMOS can usually deliver 1mA. No transistor will be destroyed by 1mA Base current. Assume 8V lost in resistor. So a 10K resistor is safe. Assume transistor hFE at very low Vce is as low as 10. So you can deliver 10mA to the LEDs (the CV demand will be negligible). Taking 7 V across LED limiting resistor, 1K or 720r is a safe LED resistor (and 470r may not be).

[alanp]

You'll need steering diodes for each step, going from pin 2 of the pot to the CV bus. This stops all the pots from interacting (otherwise, even though one step is the only one with voltage, the OTHER pots are still providing resistance to ground, or shorting CV to ground entirely if one is all the way down.)

[Quackzed]

-you could run a second 4017 in parallel with the first, that just does the leds... same shared clock...

[mth5044]

> In the 1st ... pictures, won't there always be voltage passing through the LED, resistor an pot (1)

If the transistor Base is driven HARD, its Collector will fall to ZERO (typically 0.050V).

You may need less than 100K to Base to qualify as HARD. CMOS can usually deliver 1mA. No transistor will be destroyed by 1mA Base current. Assume 8V lost in resistor. So a 10K resistor is safe. Assume transistor hFE at very low Vce is as low as 10. So you can deliver 10mA to the LEDs (the CV demand will be negligible). Taking 7 V across LED limiting resistor, 1K or 720r is a safe LED resistor (and 470r may not be).

Thank you sir! I will change out the values.

You'll need steering diodes for each step, going from pin 2 of the pot to the CV bus. This stops all the pots from interacting (otherwise, even though one step is the only one with voltage, the OTHER pots are still providing resistance to ground, or shorting CV to ground entirely if one is all the way down.)

-you could run a second 4017 in parallel with the first, that just does the leds... same shared clock...

Steering diodes makes sense! That was the piece that was missing, I suppose. However, using a second 4017 just seems too fool proof. I think that is likely the way I will go!