4017 IC Driving LEDs Using Only One Resistor

[HandyAndy]

Hello Everyone!  Long time subscriber, first time on the forums.

I created a basic LED sequence flasher using the 4017 IC chip and a 555 timer to provide a clock signal to it.
Please see attached photo.  The circuit on the left works.  By works I mean it starts by turning on the first led then turning it off and turning on the next led in sequence and repeats the loop.  The circuit on the right will only light the first led and keeps it lit and nothing else happens.  The only differences between them is that instead of using resistors on each LED I replaced them with a single resistor on the top right labeled R13.  I relocated the ground as well to make sure the circuit would have a common ground that wasn't effected by inserting that resistor.  I'm scratching my head now with this and appreciate any help I can get.
Thanks in advance mates and here is the photo:

[HandyAndy]

Just thought I would add another photo from Proteus Design Suite with the simulator running to show all the signal markers.  Anything I can do to help you guys help me let me know and thanks again!

[oPossum]

You are missing R12 in the second schematic.

[HandyAndy]

Thanks, right you are but it doesn't make a difference.  The first pic i posted does the same thing.

[HandyAndy]

Correction.  When I add back that resistor now instead of the first LED staying on solid it flashes real fast and never comes back on and none of the others do either.

[oPossum]

Are you referring to simulation or have you built this? What voltage is Vcc?

[ArthurDent]

If you look at the original design with separate resistors, when Q0 switches high, the anode side of R1, the first 1K resistor on the first LED, will go from around zero to around 2 volts depending on the LED, and none of the other LEDs are affected which is what you want .

In the single resistor version when Q0 goes high the anode of the 1st LED goes close to Vcc and the cathode, and all the other cathodes, goes to around Vcc - about 2 volts, depending on the LED, and this can upset the cathode voltage the other LEDs see when they try to turn on. Use the first version with separate resistors.

This post was edited because mikej spotted an obvious mistake. 

[Audioguru]

Your schematic is missing an extremely important supply voltage VCC. If VCC is low enough (less than 9V if the LEDs are 2V red ones) then the current-limiting resistors for the LEDs are not needed. The datasheet from Texas Instruments shows the output current vs supply voltage. It says the maximum allowed heating of each output transistor is 100mW (7V across it and 14mA through it).

Using a single current-limiting resistor is bad because it causes all the LEDs that are turned off to have a reverse voltage and many LEDs have a maximum allowed reverse voltage of only 5V.

[HandyAndy]

Thanks guys.  In the meantime I tossed this together and it works, but why does this work?  lol
Supply voltage is 12v by the way and I don't have all these parts to try it outside of the simulator so I am just trying to make it work in the simulator for now.

[Audioguru]

The CD4017 will be erratic with its Master reset pin 15 floating.

[Audioguru]

Your simulator program might not know what happens to your LEDs with 10V of reverse voltage.

[HandyAndy]

Yeah that's why in the others I usually ground the MR pin.  That doesn't effect the simulator by the way but good tip I remember seeing that in the data sheet.
Thanks I am also thinking something with the simulator just isn't right all of a sudden. 

[HandyAndy]

Ah-ha!  Eureka!  I figured out the issue.   So I had a working circuit and then decided to select the whole thing and right-click and then clicked on "Block Copy" to place another one next to it.  Then i ran the simulation and had issues.  Then I deleted what I had just copied and started over only this time I used "Copy to Clipboard" instead of "Block Copy" and then both worked just fine.  What exactly is the purpose of "Block Copy" vs. "Copy to Clipboard"?  All I know is I gotta remember never to do that again.
 

[HandyAndy]

Have a look it works!    Thanks anyway guys.

[digsys]

... What exactly is the purpose of "Block Copy" vs. "Copy to Clipboard"? ....

Easy. "Block copy" will automatically increment all part Ref#s to next available, so no duplicates. "Copy to clipboard" will duplicate all part Ref #s

[HandyAndy]

Thanks, but how would that cause issues with the simulator?  If I Block Copy it breaks the simulation the next time i try to play it nothing works, but if I Copy to Clipboard and paste it then both circuits still work fine when in start the simulation.

[Zero999]

Using a single current-limiting resistor is bad because it causes all the LEDs that are turned off to have a reverse voltage and many LEDs have a maximum allowed reverse voltage of only 5V.

It depends on the LED. Old red LEDs should be fine with more than 5V of reverse bias, even 20V or so won't do any harm, especially if the current is limited. In this case you'll get the supply voltage, minus one LED forward voltage drop, which will be about 7V, for a 9V supply and shouldn't be a problem with red LEDs. On the other hand more modern blue LEDs are much more sensitive and can be destroyed by reverse bias. In this case it would give 5V to 6V of reverse bias, which will be on the upper limit of what a blue LED can stand.

[mikerj]

In the single resistor version when Q0 goes high the anode of the 1st LED goes close to Vcc and the cathode goes to around Vcc - about 2 volts, depending on the LED, and this voltage is fed through all the other LEDs (which act as forward biased regular diodes) to the 4017 with most output pins at near zero volts, which isn't good. Use the first version.

The LEDs on the non-active outputs will be reverse biased, not forwards.

[ArthurDent]

mikerj - "The LEDs on the non-active outputs will be reverse biased, not forwards."

You are absolutely correct. I shouldn't be posting late at night. Thanks for catching that-I have corrected my original post.

[Audioguru]

How can your simulation program work when your schematic shows no supply voltage?

Years ago I made some chasers using 74HC4017 ICs, Hewlett Packard 3.2V blue and bright green LEDs and powered from 6V. A problem was caused since only one current-limiting resistor was used for 10 LEDs. Some of the LEDs leaked current when they were reverse biased (only 2.8V of reverse bias) causing some of the other LEDs to dimly glow when they should be turned off.

[Zero999]

How can your simulation program work when your schematic shows no supply voltage?

It looks like Proteus, which I've used for PCB design, but not the simulation, as like LTSpice too much. I believe it's possible to high the power supply pins, in order to make the schematic easier to follow, but I agree it can be confusing too.

Years ago I made some chasers using 74HC4017 ICs, Hewlett Packard 3.2V blue and bright green LEDs and powered from 6V. A problem was caused since only one current-limiting resistor was used for 10 LEDs. Some of the LEDs leaked current when they were reverse biased (only 2.8V of reverse bias) causing some of the other LEDs to dimly glow when they should be turned off.

Yes, tiny leakage currents can be a problem with blue and green LEDs. Not only do they start to glow at very low currents, but the human eyes is more sensitive to the shorter wavelengths of the spectrum, at lower intensities. In some cases a high value resistor in parallel can help, but the obvious solution in this case is one series resistor per LED.

[HandyAndy]

Sorry wasn't able to reply sooner.
That is in fact Proteus.
The VCC lines have 9V on them but for some reason Proteus hides that in its settings.  Might be set to hide it but I don't see why you would really want to hide that from view.
To share the simulation animation I used a program called ScreenToGif.
I hope I answered all your questions as you have answered all of mine as well as provided me with some great info.
Thank you all!