Modify this diagram...

[thejuggla1]

Ok so I am modifying this

https://www.qsl.net/yo5ofh/hobby circuits/circuit pictures/traffic.gif

As it shows only 1 'set' of Red Yellow and Green LEDs for 2 directions, I want to make it 4. I am not sure if adding more LEDs changes any of the resistor values or not and would like some help with that. I have modded it into 2 possible Schematics that I hope are right.

First one

https://i233.photobucket.com/albums/ee268/Allan-2greasy/TrafficLightsMod.jpg?t=1263104616

I just basically duped N/S and E/W but I believe there is a way to do it that will use less parts....

2nd Diagram, if my logic is correct(I am pretty sure, but then again I have no idea...) then this should be a cheaper and more efficient way of doing this.

https://i233.photobucket.com/albums/ee268/Allan-2greasy/TrafficLightsMod2.jpg?t=1263104753

First thing I just noticed is I got a ground sign overlaping the set of diodes, just ignore that I got it uploaded already don't want to reupload.

Anyways with this here I believe I can bypass the extra set of transistors saving me some time and money, am not 100% if this works and I also have no idea what the new resistor values should be. I am thinking the 2 red LEDs will be a problem, should I series them?

One thing why I am having troubles calculating the LED voltage is, I really don't know what voltages comes out of the 4017 or the 555, is it 9Vs all around the diagram? Do I calculate the LED's resistors like I normally would off a 9V circuit then?

 

[Dragon Tamer]

have you made this circuit? it looks pretty faulty to me. if i'm right, then the green LED is on for most of the time (ie: during the red or yellow light). when i get to another computer later, then i may be of more help.

 

[thejuggla1]

I dunno maybe I am seeing it wrong, but in either diagram either of the Greens should only be on for 4 'ticks' right?

And some of the parts don't get stocked till tuesday so I don't have the necessary parts to test it out.

 

[KMoffett]

Velleman makes a kit for that does that: **broken link removed** Check the manual. I actually used the kit to drive LEDs in SSRs to sequence a full sized traffic light.

Ken

 

[thejuggla1]

The kit looks nice, but I rather make it myself.

 

[vne147]

I would try this:



The 2N3094 has a max continuos collector current of 200 mA which will be enough to drive the extra LEDs. You might need to put resistors in between the output pins of the 4017 and the bases of the NPNs. I'm not sure if they are necessary to limit the current output from the 4017 or not.

 

[mneary]

@vne147: Excellent solution!

Your question: You don't need base resistors. Resistance in the emitter circuit takes care of current limiting.

 

[vne147]

@vne147: Excellent solution!

Your question: You don't need base resistors. Resistance in the emitter circuit takes care of current limiting.

Ahhh, I understand. Should have figured that out. I tend to err on the side of caution when I'm unsure so I thought I'd mention it. Thanks.

Another question, are the diodes on the output pins of the 4017 really necessary?

 

[thejuggla1]

I am not sure about the diode question, I just kept them there because they were in the original, the question should be simple to answer of course if you know the answer to a different question, does the 4017 have polarity?(Which I think it doesn't) If not then they would be needed. Because then the power could go back into the 4017 possibly damaging it?

I like the diagram you posted makes a long more sense paralleling each 'set' of leds but at the cost of only 2 NPNs per set and 2 resistors.

It says I can increase the time from changing the 47k Resistor, I am not sure but I would increase its value if I wanted the light to be longer correct? But that increases the yellow lights duration also though?(Which could cause some balances problems) but no big deal its just a lego model if I have to go completely changing the diagram.

One thing is since I am doubling the LEDs, should I also double the uF rating of the cap? I am hoping to run this off a steady power source(From a house outlet using a step down) eventually, what is the min voltage you would recommend for the Cap?

Another question, though it doesn't matter now just for personal growth in electronics, the Voltage stays 9 throughout the whole circuit correct? Even after it passes through the 555 and 4017? Would there be any exceptions to it?

 

[Dragon Tamer]

i would include the diodes just as a protection device, some of my 4017 are fried because one output fed into another with no protection at all.

 

[vne147]

It says I can increase the time from changing the 47k Resistor, I am not sure but I would increase its value if I wanted the light to be longer correct? But that increases the yellow lights duration also though?(Which could cause some balances problems) but no big deal its just a lego model if I have to go completely changing the diagram.

Increasing the 47 KΩ resistor would decrease the frequency output of the 555 and would act to slow down the changing of the lights. Decreasing the resistance would have the opposite effect. If you want you can replace the 47 kΩ resistor with a pot so you can change the timing whenever. Any change in frequency output of the 555 would change all the lights (red, yellow, and green) but not equally. For example, if you were to change the 555 output such that it added 1 second to its period the red and green lights would stay on for 4 seconds longer when compared to before but the yellow light would only stay on for 1 second longer. The reason is that the output of the 555 becomes the clock input to the 4017. With the way the outputs of the 4017 are connected, the red and green lights stay on for 4 clock cycles while the yellow only stays on for 1.

One thing is since I am doubling the LEDs, should I also double the uF rating of the cap? I am hoping to run this off a steady power source(From a house outlet using a step down) eventually, what is the min voltage you would recommend for the Cap?

The extra LEDs don't affect the cap. The cap is strictly for the 555 portion of the circuit. As for the voltage rating of the cap, you always want to make sure that it is higher than the supply voltage. If you keep the supply voltage at 9V then a cap rating of 16V should be more than adequate. If you're uncertain what the voltage will be after you modify the circuit to run off mains and you want to allow for this now, you can look up the data sheets for your 555 and 4017 and find out what the maximum allowable supply voltage is. Since you know you can't have a higher supply voltage than that, rating the caps for the maximum supply voltage will cover you. One word of caution though, if you significantly change the supply voltage, you may end up having to change the current limiting 100Ω resistors that are down stream from the red LEDs.

Another question, though it doesn't matter now just for personal growth in electronics, the Voltage stays 9 throughout the whole circuit correct? Even after it passes through the 555 and 4017? Would there be any exceptions to it?

The supply voltage is 9V throughout the circuit however the voltage is different at different points in the circuit. The outputs of the 555 and the 4017 will be close to 9V when they are high and close to 0V when they are low. The voltages at the trigger, threshold, and discharge inputs of the 555 are constantly changing. The value of voltage all depends of where in the circuit you are looking at.

 

[vne147]

i would include the diodes just as a protection device, some of my 4017 are fried because one output fed into another with no protection at all.

I was unsure because I have made 4017 circuits before where I directly connected one of the outputs to the reset pin and that caused no problems. If I were building this circuit I'd probably just buy several 4017s and test it with and without the diodes. If I fried one of them oh well, they are cheap.

 

[Boncuk]

Another question, are the diodes on the output pins of the 4017 really necessary?

I guess they are.

Would you like to poop against pressure?

(if one output of the 4017 goes high the other one is low. That's pooping against pressure.)

No IC likes its output forced high if it is low.

Boncuk

 

[vne147]

I guess they are.

Would you like to poop against pressure?

(if one output of the 4017 goes high the other one is low. That's pooping against pressure.)

No IC likes its output forced high if it is low.

Boncuk

I'm not sure I understand your analogy but it doesn't sound pleasant. I'll make sure to use a diode next time I poop.

 

[mneary]

The diodes are necessary to prevent a high on one pin feeding into other pins which are trying to stay low.

 

[thejuggla1]

Alright boys here is the bread board

YouTube - Led traffic light

I increased the resistors for the LEDs to 220 because I thought 180 was a tiny bit to low, it also changes very fast I might even want to double or triple the 47K but atm that is the highest resistor I have. I ran it at about 7Vs in the vid for safety and even then the LEDs looked pretty bright. I think I should up it to maybe 300 Ohms of resistance before putting it up to 9Vs?

One problem is I tested the LEDs I got, by tested I just mean I hooked them up by themselves to the power source and increased it as high as I thought it could go without changing its color, yellow and Red are about 2Vs and Green is 3 which is a problem when the light turns Yellow the red becomes really bright, well thats not really the problem, its just when the light is green it drains the red and the green is bright.

Solution to this? I was thinking I could add maybe 100 Ohm resistor or something between yellow to red, that might balance it out? Or should I try and find a 2V Green LED? Then when I turn it up to 9Vs red should get more power when it is green the other direction, but the extra resistance on the yellow won't make those LEDs break.

 

[vne147]

Solution to this? I was thinking I could add maybe 100 Ohm resistor or something between yellow to red, that might balance it out? Or should I try and find a 2V Green LED? Then when I turn it up to 9Vs red should get more power when it is green the other direction, but the extra resistance on the yellow won't make those LEDs break.

This is a good solution. The reason red is getting brighter is as you suspected, the lower forward voltage of the yellow LED compared to the green. When it changes from green to yellow, the volatge drop across the resistor downstream of the red LED increases. Since the resistance stays the same, the current must increase causing the red to get brighter. You could measure the volatge drop across each component and then calculate an appropriate value for the extra resistor instead of just guessing but 100Ω won't be a bad starting point and it won't damage anything.

Edit: In hindsight an easier way to determine the proper value of the extra resistor would be to measure the voltage drop across the current resistor. Record the voltage drop with the green LED and then again with the yellow. Then measure the current through it with the green LED lit.
The difference in voltages will be what you must drop across the additional resistor to maintain a constant red LED brightness. You can then calculate the necessary resistor value by using:

R = V/I

 

[thejuggla1]

How do I measure the voltage drop between LEDs? An ohms meter?

Oh and to solve my lack of resistors higher than 47k I read something each resistor automatically uses up 2Vs or something?(read it on the LED page on the putting LEDs in series part http://www.kpsec.freeuk.com/components/led.htm I dunno if I understood it wrong?) Could I possibly keep adding more 47k Resistors in the point till I get the desired light change speed?

 

[vne147]

How do I measure the voltage drop between LEDs? An ohms meter?

You can measure voltage by using the voltage setting on your multimeter. I'm assuming you have a multimeter. If you don't have one, get one as you will surely need one eventually.

For your purpose, it is not the voltage drop BETWEEN LEDs that you should be concerned with. In between the LEDs is just wire which will have a negligible voltage drop for your circuit. What you need to determine is the voltage drop ACROSS the current limiting resistor (shown as 180 in the attached schematic). I know there are 4 of them. Just pic one. The results will be nearly the same for all.

Refer to the attached schematic. You need to measure the voltage between points A and B. You do this by placing the multimeter in parallel with the resistor. Once you select the voltage setting on your multimeter place the positive (red) lead on point A and the negative (black) lead on point B. Now power up the circuit. The voltage you measure will change depending on whether the green or yellow LED is lit. Record both voltages you see and calculate the difference. The difference is what you must account for with your extra resistor.

You're not done yet. You still need to measure the current through the resistor. You'll have to measure the current differently than you did the voltage. First select the current measurement on your multimeter. It will most likely be marked "mA" or "A". You'll also have to change where the positive (red) lead plugs into your multimeter as most multimeters have one jack for voltage measurement and another for current.

Once your multimeter is reconfigured, hook it up like it shows in the schematic. You'll need to measure the current by placing the multimeter in series with the resistor. You should still do this with the resistor you measured the voltage at, I just showed it with a different resistor in the schematic for clarity.

Like with the voltage measurement, the current measurement will change depending on whether the green LED or yellow LED is lit. Record the current that you measure when the green LED is lit.

Once you measure the current, calculate the necessary resistor value using Ohms law.

V = IR or R = V/I since you want to find R. Pay attention to the units and remember that 1 mA = .001 A. The units you should use when you calculate the reststance is Volts and Amps.

I read something each resistor automatically uses up 2Vs or something?(read it on the LED page on the putting LEDs in series part Light Emitting Diodes (LEDs) I dunno if I understood it wrong?)

There is no standard voltage drop across a resistor. I think you misunderstood what was written on that site. The voltage drop across the resistor totally depends on the supply voltage, the number of LEDs, and the color of the LEDs. Each LED has a set voltage drop depending on its color and how much current it's conducting. That is most likely what they are talking about.

Could I possibly keep adding more 47k Resistors in the point till I get the desired light change speed?

Yes, 2 47KΩ resistors in series is equivalent to one 94KΩ resistor. 3 would be equivalent to a 137 KΩ and so on.

 

[thejuggla1]

Atm don't have a DMM, I also don't have a lot of cash to spare for a bit, how much does one go for?