i need help in my counter 4017 (led bright)

Thread Starter

superayman

Joined Apr 1, 2013
22
well the title sais it all i made a 555 timer with 4017 counter to make a math project for sin(x) and i placed all the led's on the graph and made it look like walkin but the only problem is the led's there not bright as they should be
am thinking of using a relay but i red its slow 50ms
but some said transistors are good
i got three outputs of my counter and i want to make each one give more power how?????:(
 

Tealc

Joined Jun 30, 2011
140
Feed your 4017 output to the base of a transistor, add a current limiting resistor and your LED to the emitter and connect the collector to the power source.

I used some 2N3904 when I did this as they were cheap and sufficient for the job.
 

Thread Starter

superayman

Joined Apr 1, 2013
22
the only problem is that the led's are welded and all the ground are common and the vcc is the only thing that separates them and the transistor gives me ground
 

Thread Starter

superayman

Joined Apr 1, 2013
22
Feed your 4017 output to the base of a transistor, add a current limiting resistor and your LED to the emitter and connect the collector to the power source.

I used some 2N3904 when I did this as they were cheap and sufficient for the job.
where should i put the res ???????
 

timescope

Joined Dec 14, 2011
298
Connect one end of a 560 ohm resistor to each of the leds and connect the other end to the emitter of the transistor as suggested by Tealc in post #2. This will give a current of about 10mA.

Timescope.
 

elec_mech

Joined Nov 12, 2008
1,500
Here's a quick and dirty schematic.

R1 is chosen somewhat randomly to allow the transistor to act as a switch. You can also use a 2N3904 or 2N4401.

R2 will depend on the forward voltage of the LED. Assuming you're using a basic diffused red, yellow, or green LED, then 560Ω as Timescope suggested will work fine and provide ~10mA.
 

Attachments

Ron H

Joined Apr 14, 2005
7,063
Here's a quick and dirty schematic.

R1 is chosen somewhat randomly to allow the transistor to act as a switch. You can also use a 2N3904 or 2N4401.

R2 will depend on the forward voltage of the LED. Assuming you're using a basic diffused red, yellow, or green LED, then 560Ω as Timescope suggested will work fine and provide ~10mA.
R1 is unnecessary. Connect the base directly to the 4017 output.
 

elec_mech

Joined Nov 12, 2008
1,500
RonH,

If you will excuse my ignorance, why? Is the output current from the 4017 so low that it cannot damage the transistor thus negating the need for a base resistor or is there a different reason?
 

Ron H

Joined Apr 14, 2005
7,063
RonH,

If you will excuse my ignorance, why? Is the output current from the 4017 so low that it cannot damage the transistor thus negating the need for a base resistor or is there a different reason?
The transistor is an emitter follower. It will not saturate, because the base voltage cannot be higher than the collector voltage. The transistor will have Vce≈0.7V. The LED current will be Ie=(Vce+Vfwd)/Re, Where Ie is the emitter current, Vfwd is the forward voltage of the LED, and Re is the current limiting resistor. Ib=Ie/(β+1), where β is the current gain of the transistor.
The base current is therefore limited by the emitter current and β. A series base resistor just wastes voltage that could be applied to the LED circuit. Of course, 100Ω wouldn't waste much voltage (or power), but it is an extraneous part.
 

elec_mech

Joined Nov 12, 2008
1,500
I follow the math and that makes sense, but I'm still at a loss as to the need or lack thereof for a base resistor.

I don't want to hijack the thread (much further :rolleyes:), so allow me to ask another ignorant question: The base resistor is not needed because a) the base voltage will never exceed the collector voltage, b) the current draw from the load (the LED in this case, controlled by R2) is well below the max current limits of the transistor, or c) both?
 

@android

Joined Dec 15, 2011
178
Do not connect led's directly to 9V battery. You really don't need 9V to glow 3mm LED(if that is what you are using!). How many leds are there and how they are wired?
 

Ron H

Joined Apr 14, 2005
7,063
I follow the math and that makes sense, but I'm still at a loss as to the need or lack thereof for a base resistor.

I don't want to hijack the thread (much further :rolleyes:), so allow me to ask another ignorant question: The base resistor is not needed because a) the base voltage will never exceed the collector voltage, b) the current draw from the load (the LED in this case, controlled by R2) is well below the max current limits of the transistor, or c) both?
I guess you could say it's a), because a base resistor would not help for b).
I don't think this answer helps you understand what's really going on, though. Until you understand the emitter follower (common collector), all you have is another unexplained fact to remember.
Some experiments might help. Make an emitter follower with a fixed value of emitter resistance, and an LED, if you want. Try different values of base resistor, from zero to whatever. Measure emitter current. Measure base current, or the voltage across the base resistor.
 

sheldons

Joined Oct 26, 2011
613
Here are 3 examples,which work well-driving leds with transistors from a 4017 output is not at all complicated-if you have your counter set as a 3 stage counter,with the reset on output 4 use 0ne transistor stage per output and away you go....also theres a fet op stage which is very useful when controlling quite a few leds per op stage....
 

Attachments

Top