Problem with 555+4017 circuit (LEDs always on)

[colin55]

I have a circuit that uses just a 4017, 555 and a single transistor. It needs 9v to 12v operation of you want 15mA drive for the LEDs.

 

[Boncuk]

I dunno, but I have been to Pataya Beach and lots of nice looking ladies there...

Believe it or not, I never was in Pattaya, but climbed Mt. Inthanon (near Chiang Mai). No nice looking girls there, but lots of snow.

I know what those girls usually say if they speak English: "Daaaling, I love you!. Daaaling I take care you!. Daling, give me money!!!"

 

[Mikebits]

Believe it or not, I never was in Pattaya, but climbed Mt. Inthanon (near Chiang Mai). No nice looking girls there, but lots of snow.

I know what those girls usually say if they speak English: "Daaaling, I love you!. Daaaling I take care you!. Daling, give me money!!!"

No, it was, Hey baby I love you long time, buy me drink

 

[Boncuk]

No, it was, Hey baby I love you long time, buy me drink

That's a direct approach.

 

[Mikebits]

That's a direct approach.

I was in Thai a long time ago, around 1984, sure things have changed since then. Do they still have all the gem dealers in Thailand? Best jewelry deal ever.

 

[Boncuk]

I was in Thai a long time ago, around 1984, sure things have changed since then. Do they still have all the gem dealers in Thailand? Best jewelry deal ever.

Things have changed a lot. The entire jewelry business is in chinese hands now.

 

[Mikebits]

Things have changed a lot. The entire jewelry business is in chinese hands now.

That's too bad. I remember the jewelry dealers were real proud of their shops and had a very high integrity for their product...

 

[Mikebits]

I have a circuit that uses just a 4017, 555 and a single transistor. It needs 9v to 12v operation of you want 15mA drive for the LEDs.

Based on the data sheet, I would not feel confident in your claim. I thought you did not use 555 chips?

 

[madmazda86]

Wow, thankyou very much for all your help - it looks like I've bitten off a bit more than I can chew because that latest circuit diagram looks pretty complex. But I am determined! On that note, I have some more questions...

1) Somebody mentioned I need to have diodes attached before the LEDs to make sure they are switched off when the output is low. Do I still need those?

2) Is it possible to drive my LEDs without that extra chip? I've currently got 2.5V 25mA ones and I'm using a 9V battery, but I can get some lower current ones. It's just that all this circuitry is fitting inside a jacket, so the more compact the circuit the better - as it is I'll have a lot of trailing wires as the LEDs are fitted to buttons on the jacket. The total number of LEDs I was planning to run was about 16:
**broken link removed**
The ones at the top would be connected in series to make sure they all go off at the same time. So if I need the extra chip to run that many then I'll get one, but anything that makes it simpler for my stupid brain is a good thing! If I enter any construction competitions at the convention I'm attending I'll be asked how I made stuff like this, so I'm trying my best to understand it all XD

3) Are the 330R resistors there because of the forward current of the LEDs you've picked? Just I was using an online LED resistor calculator and it told me I needed 270R for a single LED, but that was for the values of LED I was currently using. Just to make sure I don't buy the wrong stuff!

I am so grateful to you all for your help!

 

[Boncuk]

If you are planning to use LEDs with a forward current of 25mA each you better prepare for a heavy battery pack instead of a lousy weak 100mA/h 9V block battery.

As I can see from the animated picture in your post there are just seven channels necessary to have the same effect as shown.

Just connect LEDs in series and parallel to achieve that effect. The ULN2004 can handle loads up to 250mA per channel (as far as I can remember).

Again I must point out that it is impossible to squeeze such a current out of the CD4017! 16 LEDs of 2mA forward current will deplete the 9V battery in less than two hours.

The resistor values given in my schematic are claerly described and suit only for the kind of LED posted. For any other kind of LED you must calculate their specific restistance value.

The schematic is pretty simple and logic. So you shouln't have any problems to wire the device properly.

Boncuk

 

[madmazda86]

Right, gotcha. I'll buy some LEDs to the specifications you've described, because using a heavier battery will mean I'll have problems carrying it around. I'd be wearing it for a whole day so using your circuit will mean I won't need to change the batteries so often! Thankyou very much for taking the time to create that schematic for me - you saved me a lot of frustration and hours scouring the internet

Time to hit Ebay then!

 

[madmazda86]

Actually, a final question - sorry. You've said ULN2804A on your schematic, but in one of your posts you called it a ULN2004A. To make sure I've got the right one, which one is it?

Edit: Scrap that, just realised the 2004A only has 16 pins whereas the 2804 has 18.

 

[audioguru]

With a 9V supply and 2.0V red LEDs the CD4017 will have a typical output current of about 12mA which lights modern LEDs brightly. It is shown on a graph on the datasheet from Texas Instruments.
The output transistor in the CD4017 dissipates only 7V x 12mA= 84mW (it is allowed to dissipate more than 100mW) so current-limiting resistors are not needed.
The output will go from 0V to completely turn off the LED to +2V to brightly light the LED.

The logic level output current is only 1mA with a 5V supply and only a 0.4V voltage drop as shown in the text of the datasheet. The output current into a dead short with a 5V supply is typically 4mA.

The chaser circuit with the steering diodes use the diodes as OR gates to allow the LEDs to sequence up then sequence back down. The diodes add to the voltage drop so the LED currents will be a little less than 12mA without current-limiting resistors and a 9V supply.

With a 12V supply, the LED current will try to be about 23mA which would make each output transistor in the CD4017 dissipate 10V x 23mA= 230mW which would destroy them so current-limiting resistors are needed.

 

[colin55]

Do you still want a circuit to do your original request:

"I'm trying to build an LED chaser to go on a costume I'm making. The LEDs need to do a single chase, then stay off for about 5-6 seconds, then do another."

or have you wandered off in another direction?

 

[Boncuk]

Hi madmazda86,

how about switching the entire circuit to SMD?

Instead of using two separate NE555s you could use one NE556 (dual timer IC). The ULN2003A has seven channels while the ULN2803A has eight, hence two more pins (one in and one out). Electrically they're just the same.

I suggest not to omit the LED driver (occupying just 6 square mm of board), mainly because the LEDs in the collar have to be connected in series and even then must be equally bright as the single triggered LEDs.

(You certainly don't want to make it look like junky chinese production.)

The inverters can be made using two twin inverters as they are made by a japanese semiconductor manufacturer, again saving space.

For a battery pack I suggest a 9V pack of NiMH-batteries with a capacity of 2000mA/h. They fit easily in a pocket of your jacket.

Boncuk

 

[colin55]

More overdesign. You wouldn't last one day in business.

With current draw of approx 20mA, a set of cheap AA cells will last 50 hours. How many parties do you want to go to?

 

[audioguru]

Hardly anybody knows that a CD4017 lights only one LED at a time. If the LED current is 12mA then a 150mAh Ni-MH rechargeable battery will power it for 12.5hrs.
The LEDs will probably dim in the last hour.

My LED Chaser projects brightly light LEDs for months on a set of AA alkaline battery cells. They chase around and around a few times then pause before continuing to chase. Each blink is for a very short duration so that the battery lasts a long time.

 

[colin55]

Every one of my readers know the 4017 lights only one LED at a time because they can see from the animation that the 4017 is a "chaser."

 

[ccurtis]

1) Somebody mentioned I need to have diodes attached before the LEDs to make sure they are switched off when the output is low. Do I still need those?

The diodes, in addition to their use in an ORing function, may help with the specific observation you give concerning LEDs that are lit dimly when they should be off, as well as the variance you observe in the brightness of the LEDs that are expected to be off from one count to another. Using them though, even if they do serve to help, would be covering up for the supposition that output currents beyond which the 4017 is intended for are cross-coupling into other areas of the die, adversely affecting operation of those other areas. You state you are using 270 ohm series resistors, while the schematic you linked to shows 680 ohm resistors. 270 ohm resistors will increase the current load and aggravate any supposed cross-coupling. I suggest changing the value of the series resistors before adding band-aides. The 4017 will supply several milliamps, but differing internal voltage drops with their associated internal heating will develop, as the device is not intended for operation with such loading. The incidental ability to supply several milliamps is, obviously, exploited in cheap toys and novelty products where reliability, adverse effects, and undefined parameters that vary from one device manufacturer to the next, and even from one device to another, are not primary concerns. The fact of the matter is, the 4017 and most of the 4000 series devices are not suitable for driving multi-milliampere loads in a professional design.

 

[audioguru]

The maximum output current of a CD4xxx IC is determined by the load resistance and supply voltage as is shown in the datasheets from Texas Instruments. It is enough to brightly light modern LEDs with a 9V supply without current-limiting resistors.
There is no "cross-coupling on the chip". Each output will go high to turn on an LED then the output will go to 0.000000000V (or less) and the LED will be off.