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Unique LED light chaser questions, w/ schematic

R2-D2

Member
Hello All! :)

This is my first post here and I'm glad to have discovered this forum for its wealth of electronics information and help! I hope I can get some feedback for an electronics novelty project I'm working on that I intend to market soon.

I am designing an elaborate LED chaser circuit but have a few questions about it.
Please refer to attached schematic "LED Group Chaser.jpg" I drew up.

The circuit is a sequential light chaser for three groups of LEDs using the 555 timer and 4017 decade counter. Only three outputs are used from the 4017 counter for each group of LEDs. Individual LEDs require 3.4 volts and 24mA each. Therefore, each group of LEDs (consisting of 18 LEDs) requires 10.2 volts at 144mA. (Each branch of three LEDs requires 10.2 volts, 24mA multiplied by six branches for each LED group = 10.2 volts at 144mA.)

I am using a wall adapter as the source of 12 volts DC, 500mA.

What transistors (T1, T2 and T3) will work to handle the current required to drive the LED groups from the 4017 counter outputs?

What values do I need for the resistors (R4, R5 and R6) connected to the transistor bases?

Can I use just one resistor for each group of LEDs (R7, R8 and R9) instead of one for each of the six branches in the LED groups?

I have calculated the resistance for these resistors (R7, R8 and R9) to be 12.5Ω R = (1.8 volts dropped / 0.144 Amps) = 12.5Ω

If I need a resistor for each branch (of three LEDs), I have calculated them to be at 75Ω each. R = (1.8 volts dropped / 0.024 Amps) = 75Ω

I would also like to be able to turn ALL LEDs on (no blinking, counting, etc.) To do this, I have included a DPST switch (shown in blue) to create a connection to all bases of the transistors so that when ANY of the three outputs from the 4017 counter is active, it triggers on all three transistors. I'm not sure if this method works. Diodes (D1, D2 and D3) are connected to the 4017 outputs to prevent current flowing back into it when the switch is closed. When this switch is closed, the current from any of the three outputs from the 4017 gets split three ways to each transistor. Therefore, the current is 1/3 less to each transistor base. The transistors will still need
enough gain to power the LED groups with this decreased current after the switch is closed. Will this method work?

Another method I came up with to turn on ALL LEDs is to use a 3PST switch (shown in purple) to allow current from each of the LED groups to bypass the transistors when this switch is closed. I'm not sure if this method will work either.

Thanks for any help or suggestions!:)
 

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audioguru

Well-Known Member
Most Helpful Member
Your LEDs are in parallel. This is bad unless you test them and sort them so that all their voltages are exactly the same. To avoid testing them all, connect a current-limiting resistor in series with each string of 3 LEDs, instead of the single resistor you have.

The 4017 has an output current of about 21mA when it has a 12V supply and its output is pulled down to 6V by the transistor. Then its output transistor dissipates too much heat. So the current must be reduced to about 15mA then the output voltage will be about +8V and the output transistor will dissipate 60mW which is fine.

The transistors will saturate well with a base current of 15mA and a collector current of 144mA.

The value of the base resistor is (8V - 0.7V - 0.7V)/15ma= 440 ohms. Use 470 ohms/1/2W.
 
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R2-D2

Member
Use different transistor?

(Oops, sorry about that - I'm a newbie here.)

Hi audioguru,

Thanks for that info! I'll take your advice connecting resistors to each parallel branch in the circuit.

About the transistor: could I use a higher-powered transistor, (maybe MOSFET) to achieve a higher voltage output. (I'm not really familiar with how transistors work. I need to read up about them more:eek: )
I see that I need to lower the current out from the 4017 counter to gain higher voltage output from the transistor. I'd like to achieve at least 10V.

One of these from RadioShack might work?
RadioShack.com

Thanks!
Jeff
 

R2-D2

Member
Use different transistor?

Hi audioguru,

Thanks for that info! I'll take your advice connecting resistors to each parallel branch in the circuit.

About the transistor: could I use a higher-powered transistor, (maybe MOSFET) to achieve a higher voltage output. (I'm not really familiar with how transistors work. I need to read up about them more:eek: )
I see that I need to lower the current out from the 4017 counter to gain higher voltage output from the transistor. I'd like to achieve at least 10V.

Thanks!
Jeff
 

audioguru

Well-Known Member
Most Helpful Member
I was talking about the output transistors inside the CD4017 IC.
Their max allowed heat dissipation is only 100mW so I selected the base resistor of the driver transistor to allow a dissipation in the output transistors inside the CD4017 IC to be only 60mW.

The current in the driver transistor is only 144mA so you don't need a power transistor. Use a little 2N4401 or little BC337 transistor.
 

R2-D2

Member
The 4017 has an output current of about 21mA when it has a 12V supply and its output is pulled down to 6V by the transistor. Then its output transistor dissipates too much heat. So the current must be reduced to about 15mA then the output voltage will be about +8V and the output transistor will dissipate 60mW which is fine.

The transistors will saturate well with a base current of 15mA and a collector current of 144mA.
Hi audioguru, me again...

Do you mean I need to use an 8V power source (instead of 12V) to obtain 15mA from the 4017 counter outputs? If I use an 8V source, I will need at least 216mA collector current from each transistor to power 18 LEDs for each LED group.
LED series parallel array wizard
I will need transistors that can supply that amount of current when they are saturated at 15mA to their bases and need to make sure I have the right type. I guess I could use a heatsink for the 4017.

Thanks!
Happy New Years! :)
 

R2-D2

Member
Transistor and output

The 4017 has an output current of about 21mA when it has a 12V supply and its output is pulled down to 6V by the transistor. Then its output transistor dissipates too much heat. So the current must be reduced to about 15mA then the output voltage will be about +8V and the output transistor will dissipate 60mW which is fine.

The transistors will saturate well with a base current of 15mA and a collector current of 144mA.
Hi audioguru, me again...

Do you mean that I need a power source of 8V (instead of 12V) to obtain a 15mA output from the 4017? If so, with an 8V source, I will need to supply 216mA of collector current out of each drive transistor to power the 18 LEDs. I will then need a transistor to supply that amount of current when saturated at its base of 15mA.

Thanks - and Happy New Years!
Jeff
 

audioguru

Well-Known Member
Most Helpful Member
Do you mean that I need a power source of 8V (instead of 12V) to obtain a 15mA output from the 4017?
No.
With a 12V supply, the output of a CD4017 is 15mA at 8V. It has a 4V voltage drop so its output transistor dissipates 4V x 15mA= 60mW.

With a lower supply voltage the output current is less which is not enough to turn on a driver transistor completely.

... with an 8V source, I will need to supply 216mA of collector current out of each drive transistor to power the 18 LEDs.
Maybe you were thinking of two LEDs in series and using 9 strings of them where each string has 24mA. Why bother, use a 12V supply and three LEDs in series for 6 strings and a total of 144mA.
 

R2-D2

Member
Maybe you were thinking of two LEDs in series and using 9 strings of them where each string has 24mA. Why bother, use a 12V supply and three LEDs in series for 6 strings and a total of 144mA.
Yes, I was thinking of that previous LED arrangement. But I'll go with 12V instead for the three LEDs in each of six strings.
Thanks for the help! :)

Jeff
 

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